NCERT CBSE Standard 11 Chemistry Chapter 7 Chemical Equilibrium SKMClasses South Bangalore Subhashish Sir

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Solutions to Chapter 7 :

Chemical Equilibrium

Must see https://zookeepersblog.wordpress.com/some-points-which-i-wish-all-my-new-prospective-students-know/

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The previous chapter Solution is at https://zookeepersblog.wordpress.com/ncert-cbse-standard-11-chemistry-chapter-6-thermodynamics-thermochemistry/
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Q : What is hydrolysis ?

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Kp changes with

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Relationship between Kp and Kc

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Le Chatelier’s Principle

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Exothermic reaction releases heat. Heating will decrease value of Kp

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Endothermic reaction absorbs heat. Heating will increase value of Kp

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The equilibrium constant is independent of individual concentrations

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The equilibrium constant is independent of the volume of the container

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Chemical equilibria are important in numerous biological and environmental processes. For example, equilibria involving O₂ molecules and the protein hemoglobin play a crucial role in the transport and delivery of O₂ from our lungs to our muscles. Similar equilibria involving CO molecules and hemoglobin account for the toxicity of CO.

When a liquid evaporates in a closed container, molecules with relatively higher kinetic energy escape the liquid surface into the vapour phase and number of liquid molecules from the vapour phase strike the liquid surface and are retained in the liquid phase. It gives rise to a constant vapour pressure because of an equilibrium in which the number of molecules leaving the liquid equals the number returning to liquid from the vapour. We say that the system has reached equilibrium state at this stage. However, this is not static equilibrium and there is a lot of activity at the boundary between the liquid and the vapour. Thus, at equilibrium, the rate of evaporation is equal to the rate of condensation. It may be represented by

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The double half arrows indicate that the processes in both the directions are going on simultaneously. The mixture of reactants and products in the equilibrium state is called an equilibrium mixture.

Equilibrium can be established for both physical processes and chemical reactions. The reaction may be fast or slow depending on the experimental conditions and the nature of the reactants. When the reactants in a closed vessel at a particular temperature react to give products, the concentrations of the reactants keep on decreasing, while those of products keep on increasing for some time after which there is no change in the concentrations of either of the reactants or products. This stage of the system is the dynamic equilibrium and the rates of the forward and reverse reactions become equal. It is due to this dynamic equilibrium stage that there is no change in the concentrations of various species in the reaction mixture. Based on the extent to which the reactions proceed to reach the state of chemical equilibrium, these may be classified in three groups.

(i) The reactions that proceed nearly to completion and only negligible concentrations of the reactants are left. In some cases, it may not be even possible to detect these experimentally.

(ii) The reactions in which only small amounts of products are formed and most of the reactants remain unchanged at equilibrium stage.

(iii) The reactions in which the concentrations of the reactants and products are comparable, when the system is in equilibrium.

The extent of a reaction in equilibrium varies with the experimental conditions such as concentrations of reactants, temperature, etc. Optimisation of the operational conditions is very important in industry and laboratory so that equilibrium is favorable in the direction of the desired product. Some important aspects of equilibrium involving physical and chemical processes are dealt in this unit along with the equilibrium involving ions in aqueous solutions which is called as ionic equilibrium.

7.1 EQUILIBRIUM IN PHYSICAL PROCESSES

The characteristics of system at equilibrium are better understood if we examine some physical processes. The most familiar examples are phase transformation processes, e.g.,

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7.1.1 Solid-Liquid Equilibrium

Ice and water kept in a perfectly insulated thermos flask (no exchange of heat between its contents and the surroundings) at 273K and the atmospheric pressure are in equilibrium state and the system shows interesting characteristic features. We observe that the mass of ice and water do not change with time and the temperature remains constant. However, the equilibrium is not static. The intense activity can be noticed at the boundary between ice and water. Molecules from the liquid water collide against ice and adhere to it and some molecules of ice escape into liquid phase. There is no change of mass of ice and water, as the rates of transfer of molecules from ice into water and of reverse transfer from water into ice are equal at atmospheric pressure and 273 K.

It is obvious that ice and water are in equilibrium only at particular temperature and pressure. For any pure substance at atmospheric pressure, the temperature at which the solid and liquid phases are at equilibrium is called the normal melting point or normal freezing point of the substance. The system here is in dynamic equilibrium and we can infer the following:

(i) Both the opposing processes occur simultaneously.

(ii) Both the processes occur at the same rate so that the amount of ice and water remains constant.

7.1.2 Liquid-Vapour Equilibrium

This equilibrium can be better understood if we consider the example of a transparent box carrying a U-tube with mercury (manometer). Drying agent like anhydrous calcium chloride (or phosphorus penta-oxide) is placed for a few hours in the box. After removing the drying agent by tilting the box on one side, a watch glass (or petri dish) containing water is quickly placed inside the box. It will be observed that the mercury level in the right limb of the manometer slowly increases and finally attains a constant value, that is, the pressure inside the box increases and reaches a constant value. Also the volume of water in the watch glass decreases (Fig. 7.1). Initially there was no water vapour (or very less) inside the box. As water evaporated the pressure in the box increased due to addition of water molecules into the gaseous phase inside the box. The rate of evaporation is constant. However, the rate of increase in pressure decreases with time due to condensation of vapour into water. Finally it leads to an equilibrium condition when there is no net evaporation. This implies that the number of water molecules from the gaseous state into the liquid state also increases till the equilibrium is attained i.e.,

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At equilibrium the pressure exerted by the water molecules at a given temperature remains constant and is called the equilibrium vapour pressure of water (or just vapour pressure of water); vapour pressure of water increases with temperature. If the above experiment is repeated with methyl alcohol, acetone and ether, it is observed that different liquids have different equilibrium vapour pressures at the same temperature, and the liquid which has a higher vapour pressure is more volatile and has a lower boiling point.

If we expose three watch glasses containing separately 1mL each of acetone, ethyl alcohol, and water to atmosphere and repeat the experiment with different volumes of the liquids in a warmer room, it is observed that in all such cases the liquid eventually disappears and the time taken for complete evaporation depends on (i) the nature of the liquid, (ii) the amount of the liquid and (iii) the temperature. When the watch glass is open to the atmosphere, the rate of evaporation remains constant but the molecules are dispersed into large volume of the room. As a consequence the rate of condensation from vapour to liquid state is much less than the rate of evaporation. These are open systems and it is not possible to reach equilibrium in an open system.

Water and water vapour are in equilibrium position at atmospheric pressure (1.013 bar) and at 100°C in a closed vessel. The boiling point of water is 100°C at 1.013 bar pressure. For any pure liquid at one atmospheric pressure (1.013 bar) the temperature at which the liquid and vapours are at equilibrium is called normal boiling point of the liquid. Boiling point of the liquid depends on the atmospheric pressure. It depends on the altitude of the place; at high altitude the boiling point decreases.

7.1.3 Solid — Vapour Equilibrium

Let us now consider the systems where solids sublime to vapour phase. If we place solid iodine in a closed vessel, after sometime the vessel gets filled up with violet vapour and the intensity of colour increases with time. After certain time the intensity of colour becomes constant and at this stage equilibrium is attained. Hence solid iodine sublimes to give iodine vapour and the iodine vapour condenses to give solid iodine. The equilibrium can be represented as,

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and the rate of dissolution of sugar = rate of crystallisation of sugar.

Equality of the two rates and dynamic nature of equilibrium has been confirmed with the help of radioactive sugar. If we drop some radioactive sugar into saturated solution of non-radioactive sugar, then after some time radioactivity is observed both in the solution and in the solid sugar. Initially there were no radioactive sugar molecules in the solution but due to dynamic nature of equilibrium, there is exchange between the radioactive and non-radioactive sugar molecules between the two phases. The ratio of the radioactive to nonradioactive molecules in the solution increases till it attains a constant value.

Gases in liquids

When a soda water bottle is opened, some of the carbon dioxide gas dissolved in it fizzes out rapidly. The phenomenon arises due to difference in solubility of carbon dioxide at different pressures. There is equilibrium between the molecules in the gaseous state and the molecules dissolved in the liquid under pressure i.e.,

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(iii) For dissolution of solids in liquids, the solubility is constant at a given temperature.

(iv) For dissolution of gases in liquids, the concentration of a gas in liquid is proportional to the pressure (concentration) of the gas over the liquid. These observations are summarised in Table 7.1

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7.1.5 General Characteristics of Equilibria Involving Physical Processes

For the physical processes discussed above, following characteristics are common to the system at equilibrium:

(i) Equilibrium is possible only in a closed system at a given temperature.

(ii) Both the opposing processes occur at the same rate and there is a dynamic but stable condition.

(iii) All measurable properties of the system remain constant.

(iv) When equilibrium is attained for a physical process, it is characterised by constant value of one of its parameters at a given temperature. Table 7.1 lists such quantities.

(v) The magnitude of such quantities at any stage indicates the extent to which the physical process has proceeded before reaching equilibrium.

7.2 EQUILIBRIUM IN CHEMICAL PROCESSES ? DYNAMIC EQUILIBRIUM

Analogous to the physical systems chemical reactions also attain a state of equilibrium. These reactions can occur both in forward and backward directions. When the rates of the forward and reverse reactions become equal, the concentrations of the reactants and the products remain constant. This is the stage of chemical equilibrium. This equilibrium is dynamic in nature as it consists of a forward reaction in which the reactants give product(s) and reverse reaction in which product(s) gives the original reactants.

For a better comprehension, let us consider a general case of a reversible reaction,

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Eventually, the two reactions occur at the same rate and the system reaches a state of equilibrium.

Similarly, the reaction can reach the state of equilibrium even if we start with only C and D; that is, no A and B being present initially, as the equilibrium can be reached from either direction.

The dynamic nature of chemical equilibrium can be demonstrated in the synthesis of ammonia by Haber’s process. In a series of experiments, Haber started with known amounts of dinitrogen and dihydrogen maintained at high temperature and pressure and at regular intervals determined the amount of ammonia present. He was successful in determining also the concentration of unreacted dihydrogen and dinitrogen. Fig. 7.4 (page 191) shows that after a certain time the composition of the mixture remains the same even though some of the reactants are still present. This constancy in composition indicates that the reaction has reached equilibrium. In order to understand the dynamic nature of the reaction, synthesis of ammonia is carried out with exactly the same starting conditions (of partial pressure and temperature) but using D₂ (deuterium) in place of H₂. The reaction mixtures starting either with H₂ or D₂ reach equilibrium with the same composition, except that D2 and ND₃ are present instead of H₂ and NH₃. After equilibrium is attained, these two mixtures (H₂, N₂, NH₃ and D₂, N₂, ND₃) are mixed together and left for a while. Later, when this mixture is analysed, it is found that the concentration of ammonia is just the same as before. However, when this mixture is analysed by a mass spectrometer, it is found that ammonia and all deuterium containing forms of ammonia (NH₃, NH₂D, NHD₂ and ND₃) and dihydrogen and its deutrated forms (H₂, HD and D₂) are present. Thus one can conclude that scrambling of H and D atoms in the molecules must result from a continuation of the forward and reverse reactions in the mixture. If the reaction had simply stopped when they reached equilibrium, then there would have been no mixing of isotopes in this way.

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Use of isotope (deuterium) in the formation of ammonia clearly indicates that chemical reactions reach a state of dynamic equilibrium in which the rates of forward and reverse reactions are equal and there is no net change in composition.

Equilibrium can be attained from both sides, whether we start reaction by taking, H₂(g) and N₂(g) and get NH₃(g) or by taking NH₃(g) and decomposing it into N₂(g) and H₂(g).

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If we start with equal initial concentration of H₂ and I₂, the reaction proceeds in the forward direction and the concentration of H₂ and I₂ decreases while that of HI increases, until all of these become constant at equilibrium (Fig. 7.5). We can also start with HI alone and make the reaction to proceed in the reverse direction; the concentration of HI will decrease and concentration of H₂ and I₂ will increase until they all become constant when equilibrium is

reached (Fig.7.5). If total number of H and I atoms are same in a given volume, the same equilibrium mixture is obtained whether we start it from pure reactants or pure product.

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Dynamic Equilibrium – A students Activity

Equilibrium whether in a physical or in a chemical system, is always of dynamic nature. This can be demonstrated by the use of radioactive isotopes. This is not feasible in a school laboratory. However this concept can be easily comprehended by performing the following activity. The activity can be performed in a group of 5 or 6 students.Take two 100mL measuring cylinders (marked as 1 and 2) and two glass tubes each of 30 cm length. Diameter of the tubes may be same or different in the range of 3-5mm. Fill nearly half of the measuring cylinder-1 with coloured water (for this purpose add a crystal of potassium permanganate to water) and keep second cylinder (number 2) empty.Put one tube in cylinder 1 and second in cylinder 2. Immerse one tube in cylinder 1, close its upper tip with a finger and transfer the coloured water contained in its lower portion to cylinder 2. Using second tube, kept in 2nd cylinder, transfer the coloured water in a similar manner from cylinder 2 to cylinder 1. In this way keep on transferring coloured water using the two glass tubes from cylinder 1 to 2 and from 2 to 1 till you notice that the level of coloured water in both the cylinders becomes constant.If you continue intertransferring coloured solution between the cylinders, there will not be any further change in the levels of coloured water in two cylinders. If we take analogy of ?level? of coloured water with ?concentration? of reactants and products in the two cylinders, we can say the process of transfer, which continues even after the constancy of level, is indicative of dynamic nature of the process. If we repeat the experiment taking two tubes of different diameters we find that at equilibrium the level of coloured water in two cylinders is different. How far diameters are responsible for change in levels in two cylinders? Empty cylinder (2) is an indicator of no product in it at the beginning.

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7.3 LAW OF CHEMICAL EQUILIBRIUM AND EQUILIBRIUM CONSTANT

A mixture of reactants and products in the equilibrium state is called an equilibrium mixture. In this section we shall address a number of important questions about the composition of equilibrium mixtures: What is the relationship between the concentrations of reactants and products in an equilibrium mixture? How can we determine equilibrium concentrations from initial concentrations?

What factors can be exploited to alter the composition of an equilibrium mixture? The last question in particular is important when choosing conditions for synthesis of industrial chemicals such as H₂, NH₃, CaO etc.

To answer these questions, let us consider a general reversible reaction:

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Six sets of experiments with varying initial conditions were performed, starting with only gaseous H₂ and I₂ in a sealed reaction vessel in first four experiments (1, 2, 3 and 4) and only HI in other two experiments (5 and 6). Experiment 1, 2, 3 and 4 were performed taking different concentrations of H₂ and / or I₂, and with time it was observed that intensity of the purple colour remained constant and equilibrium was attained. Similarly, for experiments 5 and 6, the equilibrium was attained from the opposite direction.

Data obtained from all six sets of experiments are given in Table 7.2.

Table 7.2 Initial and Equilibrium Concentrations of H₂, I₂ and HI

Experiment number	Initial concentration/mol L^-1			Equilibrium concentration/mol L^-1
	[ H₂ (g) ]	[ I₂ (g) ]	[ HI (g) ]	[ H₂ (g) ]	[ I₂ (g) ]	[ HI (g) ]
1	2.4 x 10^-2	1.38 x 10^-2	0	1.14 x 10^-2	0.12 x 10^-2	2.52 x 10^-2
2	2.4 x 10^-2	1.68 x 10^-2	0	0.92 x 10^-2	0.20 x 10^-2	2.96 x 10^-2
3	2.44 x 10^-2	1.98 x 10^-2	0	0.77 x 10^-2	0.31 x 10^-2	3.34 x 10^-2
4	2.46 x 10^-2	1.76 x 10	0	0.92 x 10^-2	0.22 x 10^-2	3.08 x 10^-2
5	0	0	3.04 x 10^-2	0.345 x 10^-2	0.345 x 10^-2	2.35 x 10^-2
6	0	0	7.58 x 10^-2	0.86 x 10^-2	0.86 x 10^-2	5.86 x 10^-2

It is evident from the experiments 1, 2, 3 and 4 that number of moles of dihydrogen reacted = number of moles of iodine reacted = 1/2(number of moles of HI formed). Also, experiments 5 and 6 indicate that,

[H₂(g)]_eq = [I₂(g)]_eq

Knowing the above facts, in order to establish a relationship between

concentrations of the reactants and products, several combinations can be tried. Let us consider the simple expression,

[HI(g)]_eq / [H₂(g)] _eq [I₂(g)] _eq

It can be seen from Table 7.3 that if we put the equilibrium concentrations of the reactants and products, the above expression is far from constant. However, if we consider the expression,

[HI(g)]_eq² / [H₂(g)]_eq [I₂(g)]_eq

Table 7.3 Expression Involving the Equilibrium Concentration of Reactants

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Generally the subscript ‘eq’ (used for equilibrium) is omitted from the concentration terms. It is taken for granted that the concentrations in the expression for K_c are equilibrium values. We, therefore, write,

K_c = [HI(g)]² / [H₂(g)] [I₂(g)] —————————————————————-(7.3)

The subscript ‘c’ indicates that K_c is expressed in concentrations of mol L^-1.

At a given temperature, the product of concentrations of the reaction products raised to the respective stoichiometric coefficient in the balanced chemical equation divided by the product of concentrations of the reactants raised to their individual stoichiometric coefficients has a constant value. This is known as the Equilibrium Law or Law of Chemical Equilibrium.

The equilibrium constant for a general reaction,

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Molar concentration of different species is indicated by enclosing these in square bracket and, as mentioned above, it is implied that these are equilibrium concentrations. While writing expression for equilibrium constant, symbol for phases (s, l, g) are generally ignored.

Let us write equilibrium constant for the reaction,

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the equilibrium constant for the above reaction is given by

K″_c = [HI] / [H₂]^1/2[I₂]^1/2 = {[HI]² / [H₂][I₂]}^1/2 = x^1/2 = K_c^1/2 ————————————————————————————(7.10)

On multiplying the equation (7.5) by n, we get

nH2 (g) + nI2 ( g ) = 2nHI ( g )

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Therefore, equilibrium constant for the reaction is equal to K_cⁿ. These findings are summarised in Table 7.4. It should be noted that because the equilibrium constants K_c and K′_c have different numerical values, it is important to specify the form of the balanced chemical equation when quoting the value of an equilibrium constant.

Table 7.4 Relations between Equilibrium Constants for a General Reaction and its Multiples.

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Problem 7.1

The following concentrations were obtained for the formation of NH₃ from N₂ and H₂ at equilibrium at 500K. [N₂] = 1.5 x10^-2M. [H₂] = 3.0 x10^-2 M and [NH₃] = 1.2 x 10^-2M. Calculate equilibrium constant.

Solution

The equilibrium constant for the reaction,

N2 ( g ) + 3 H2 ( g ) = 2NH3 ( g )

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can be written as,

K_c = [NH₃(g)]²/[N₂(g)][H₂(g)]³

= (1.2 x 10^-2)²/(1.5×10^-2)(3.0×10^-2)³

= 0.106 x 10⁴ = 1.06 x 10³

Problem 7.2

At equilibrium, the concentrations of

N₂=3.0 x10^-3M, O₂ = 4.2 x 10^-3M and NO= 2.8 x10^-3M

in a sealed vessel at 800K. What will be K_c for the reaction

N2 ( g ) + O2 ( g ) = 2 NO ( g )

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Solution

For the reaction equilibrium constant, K_c can be written as,

K_c = [NO]²/[N₂][O₂]

= (2.8 x 10^-3)²/(3.0×10^-3M)(4.2×10^-3M)

= 0.622

7.4 HOMOGENEOUS EQUILIBRIA

In a homogeneous system, all the reactants and products are in the same phase. For example, in the gaseous reaction,

N2 ( g ) + 3 H2 ( g ) = 2 N H3 ( g )

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, reactants and products are in the homogeneous phase. Similarly, for the reactions,

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all the reactants and products are in homogeneous solution phase. We shall now consider equilibrium constant for some homogeneous reactions.

7.4.1 Equilibrium Constant in Gaseous Systems

So far we have expressed equilibrium constant of the reactions in terms of molar concentration of the reactants and products, and used symbol, Kc for it. For reactions involving gases, however, it is usually more convenient to express the equilibrium constant in terms of partial pressure.

The ideal gas equation is written as,

pV = nRT

⇒ p = (n/v) RT

Here, p is the pressure in Pa, n is the number of moles of the gas, V is the volume in m³ and T is the temperature in Kelvin

Therefore,

n/V is concentration expressed in mol/m³

If concentration c, is in mol/L or mol/dm³, and p is in bar then

p = cRT,

We can also write p = [gas]RT.

Here, R= 0.0831 bar litre/mol K

At constant temperature, the pressure of the gas is proportional to its concentration i.e., p ∝ [gas]

For reaction in equilibrium

H2 ( g ) + I2 ( g ) = 2 HI ( g )

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We can write either

K_c = [HI(g)]²/[H₂(g)][I₂(g)]

or K_c = (P_HI)²/(P_H₂)(P_I₂) ———————————————————————————————(7.12)

Further, since P_HI = [HI(g)]RT

P_I₂ = [I₂(g)] RT

P_H₂ = [H₂(g)] RT

Therefore,

K_p = (P_HI)²/(P_H₂)(P_I₂) = [HI(g)]² [RT]²/[H₂(g)] RT. [I₂(g)] RT

= [HI(g)]²/[H₂(g)][I₂(g)] = K_c —————————————————————————————-(7.13)

In this example, K_p = K_c i.e., both equilibrium constants are equal. However, this is not always the case. For example in reaction

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N2 ( g ) + 3 H2 ( g ) = 2 NH3 ( g )

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K_p = (P_NH₃)²/(P_N₂)(P_H₂)³

= [NH₃(g)]²[RT]²/[N₂(g)]RT . [H₂(g)]³ (RT)³

=[NH₃(g)]²[RT]^-2/[N₂(g)]RT . [H₂(g)]³ = K_c(RT)^-2

or K_p = K_c (RT)^-2 ————————————————————————-(7.14)

Similarly, for a general reaction

aA + bB = cC + dD

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K_p = (p_C^c)(p_D^d)/(p_A^a)(p_B^b)

= [C]^c [D]^d (RT)^c+d/[A]^a [B]^b (RT)^a+b

= [C]^c [D]^d /[A]^a [B]^b (RT)^{(c+d) – (a+b)}

= [C]^c [D]^d/[A]^a [B]^b (RT)^Δn = K_c (RT)^Δn —————————————————————————————(7.15)

where Δn = (number of moles of gaseous products) – (number of moles of gaseous reactants) in the balanced chemical equation. (It is necessary that while calculating the value of K_p, pressure should be expressed in bar as standard state is 1bar). We have known from Unit 1,

1pascal, Pa=1Nm^-2, and 1bar = 105 Pa

K_p values for a few selected reactions at different temperatures are given in Table 7.5

Table 7.5 Equilibrium Constants, K_p for a Few Selected Reactions

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Problem 7.3

PCl₅, PCl₃ and Cl₂ are at equilibrium at 500 K and having concentration 1.59M PCl₃, 1.59M Cl₂ and 1.41 M PCl₅. Calculate K_c for the reaction,

PCl5 = PCl3 + Cl2

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Solution

The equilibrium constant K_c for the above reaction can be written as,

K_c = [PCl₅][Cl₂]/[PCl₅] = (1.59)²/(1.41) = 1.79

Problem 7.4

The value of K_c = 4.24 at 800K for the reaction,

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CO ( g ) + H2O ( g ) = CO2 ( g ) + H2 ( g )

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Calculate equilibrium concentrations of CO₂, H₂, CO and H₂O at 800 K, if only CO and H₂O are present initially at concentrations of 0.10M each.

Solution

For the reaction

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where x is the amount of CO₂ and H₂ at equilibrium.

Hence, equilibrium constant can be written as,

K_c = x²/(0.1-x)² = 4.24

x² = 4.24(0.01 + x²-0.2x)

x² = 0.0424 + 4.24x²-0.848x

3.24x² – 0.848x + 0.0424 = 0

a = 3.24, b = – 0.848, c = 0.0424

(for quadratic equation ax² + bx + c = 0,

x = ( -b ± √(b2 – 4ac) )/2a

x = 0.848±√(0.848)²– 4(3.24)(0.0424)/(3.24 x 2)

x = (0.848 ± 0.4118)/ 6.48

x₁ = (0.848 – 0.4118)/6.48 = 0.067

x₂ = (0.848 + 0.4118)/6.48 = 0.194

the value 0.194 should be neglected because it will give concentration of the

reactant which is more than initial concentration.

Hence the equilibrium concentrations are,

[CO₂] = [H_2-] = x = 0.067 M

[CO] = [H₂O] = 0.1 – 0.067 = 0.033 M

Problem 7.5

For the equilibrium,

2NOCl ( g ) = 2NO ( g ) + Cl2 ( g )

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the value of the equilibrium constant, K_c is 3.75 x 10^-6 at 1069 K. Calculate the K_p for the reaction at this temperature?

Solution

We know that,

K_p = K_c(RT)^Δn

For the above reaction,

Δn = (2+1) – 2 = 1

K_p = 3.75 x 10^-6 (0.0831 x 1069)

K_p = 0.033

7.5 HETEROGENEOUS EQUILIBRIA

Equilibrium in a system having more than one phase is called heterogeneous equilibrium. The equilibrium between water vapour and liquid water in a closed container is an example of heterogeneous equilibrium.

H2O ( l ) = H2O ( g )

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In this example, there is a gas phase and a liquid phase. In the same way, equilibrium between a solid and its saturated solution,

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On the basis of the stoichiometric equation, we can write,

K_c = [CaO(s)][CO₂(g)]/[CaCO₃(s)]

Since [CaCO₃(s)] and [CaO(s)] are both constant, therefore modified equilibrium constant for the thermal decomposition of calcium carbonate will be

K´_c = [CO₂(g)] ———————————————————————————–(7.17)

or K_c = p_CO₂ ———————————————————————————–(7.18)

Units of Equilibrium Constant

The value of equilibrium constant K_c can be calculated by substituting the concentration terms in mol/L and for K_p partial pressure is substituted in Pa, kPa, bar or atm. This results in units of equilibrium constant based on molarity or pressure, unless the exponents of both the numerator and denominator are same.For the reactions,H2(g ) + I2 ( g ) = 2HI, K_c and K_p have no unit.N2O4 ( g ) = 2NO2 ( g ), K_c has unit mol/L and K_p has unit barEquilibrium constants can also be expressed as dimensionless quantities if the standard state of reactants and products are specified. For a pure gas, the standard state is 1bar. Therefore a pressure of 4 bar in standard state can be expressed as 4 bar/1 bar = 4, which is a dimensionless number. Standard state (c₀) for a solute is 1 molar solution and all concentrations can be measured with respect to it. The numerical value of equilibrium constant depends on the standard state chosen. Thus, in this system both K_p and K_c are dimensionless quantities but have different numerical values due to different standard states.

This shows that at a particular temperature, there is a constant concentration or pressure of CO₂ in equilibrium with CaO(s) and CaCO₃(s). Experimentally it has been found that at 1100 K, the pressure of CO₂ in equilibrium with CaCO₃(s) and CaO(s), is 2.0 105 Pa. Therefore, equilibrium constant at 1100K for the above reaction is:

K_p = P_CO₂ = 2 ×10⁵ Pa/10⁵Pa = 2.00

Similarly, in the equilibrium between nickel, carbon monoxide and nickel carbonyl (used in the purification of nickel),

Ni ( s ) + 4 CO ( g ) = Ni ( CO)4 ( g )

the equilibrium constant is written as

K_c = [Ni(CO)₄]/[CO]⁴

It must be remembered that in heterogeneous equilibrium pure solids or liquids must be present (however small the amount may be) for the equilibrium to exist, but their concentrations or partial pressure do not appear in the expression of the equilibrium constant. In the reaction,

Ag2O ( s ) + 2 HNO3 ( aq ) = 2AgNO3 (aq ) + H2O ( l )

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K_c = [AgNO₃]²/[HNO₃]²

Problem 7.6

The value of K_p for the reaction,

CO2 ( g ) + C ( s ) = 2CO ( g )

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is 3.0 at 1000 K. If initially P_CO₂= 0.48 bar and P_CO = 0 bar and pure graphite is present, calculate the equilibrium partial pressures of CO and CO₂.

Solution

For the reaction,

let ‘x’ be the decrease in pressure of CO₂, then

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K_p = P²_CO/P_CO²

K_p = (2x)²/(0.48 – x) = 3

4x² = 3(0.48 – x)

4x² = 1.44 – x

4x² + 3x – 1.44 = 0

a = 4, b = 3, c = -1.44

x = ( -b ± √(b2 – 4ac) )/2a

= [-3 ± √(3)² – 4(4)(-1.44)]/2 x 4

= (-3 ± 5.66)/8

= (-3 + 5.66)/ 8 (as value of x cannot be negative hence we neglect that value)

x = 2.66/8 = 0.33

The equilibrium partial pressures are,

P_CO = 2x = 2 0.33 = 0.66 bar

P_CO₂ = 0.48 – x = 0.48 – 0.33 = 0.15 bar

7.6 APPLICATIONS OF EQUILIBRIUM CONSTANTS

Before considering the applications of equilibrium constants, let us summarise the important features of equilibrium constants as follows:

1. Equilibrium constant is applicable only when concentrations of the reactants and products have attained their equilibrium state.

2. The value of equilibrium constant is independent of initial concentrations of the reactants and products.

3. Equilibrium constant is temperature dependent having one unique value for a particular reaction represented by a balanced equation at a given temperature.

4. The equilibrium constant for the reverse reaction is equal to the inverse of the equilibrium constant for the forward reaction.

5. The equilibrium constant K for a reaction is related to the equilibrium constant of the corresponding reaction, whose equation is obtained by multiplying or dividing the equation for the original reaction by a small integer.

Let us consider applications of equilibrium constant to:

• predict the extent of a reaction on the basis of its magnitude,

• predict the direction of the reaction, and

• calculate equilibrium concentrations.

7.6.1 Predicting the Extent of a Reaction

The numerical value of the equilibrium constant for a reaction indicates the extent of the reaction. But it is important to note that an equilibrium constant does not give any information about the rate at which the equilibrium is reached. The magnitude of K_c or K_p is directly proportional to the concentrations of products (as these appear in the numerator of equilibrium constant expression) and inversely proportional to the concentrations of the reactants (these appear in the denominator). This implies that a high value of K is suggestive of a high concentration of products and vice-versa.

We can make the following generalisations concerning the composition of equilibrium mixtures:

• If K_c > 10³, products predominate over reactants, i.e., if K_c is very large, the reaction proceeds nearly to completion. Consider the following examples:

(a) The reaction of H₂ with O₂ at 500 K has a very large equilibrium constant ,

K_c = 2.4 x 10⁴⁷.

(b) H2 ( g ) + Cl2 ( g ) = 2HCl ( g )

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at 300K has K_c = 4.0 x 10³¹.

(c)

H2 ( g ) + Br2 ( g ) = 2HBr ( g )

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at 300 K, K_c = 5.4 c 10¹⁸

• If K_c < 10^-3, reactants predominate over products, i.e., if K_c is very small, the reaction proceeds rarely. Consider the following examples:

(a) The decomposition of H₂O into H₂ and O₂ at 500 K has a very small equilibrium constant, K_c = 4.1 x 10^-48

(b)

N2 ( g ) + O2 ( g ) = 2NO ( g )

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, at 298 K has K_c = 4.8 x 10³¹.

• If K_c is in the range of 10^-3 to 10³, appreciable concentrations of both reactants and products are present. Consider the following examples:

(a) For reaction of H₂ with I₂ to give HI, K_c = 57.0 at 700K.

(b) Also, gas phase decomposition of N₂O₄ to NO₂ is another reaction with a value of K_c = 4.64 x 10^-3 at 25°C which is neither too small nor too large. Hence, equilibrium mixtures contain appreciable concentrations of both N₂O₄ and NO₂.

These generarlisations are illustrated in Fig. 7.6

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7.6.2 Predicting the Direction of the Reaction

The equilibrium constant helps in predicting the direction in which a given reaction will proceed at any stage. For this purpose, we calculate the reaction quotient Q. The reaction quotient, Q (Q_c with molar concentrations and Q_p with partial pressures) is defined in the same way as the equilibrium constant K_c except that the concentrations in Q_c are not necessarily equilibrium values. For a general reaction:

aA + bB = cC + dD — ( 7.19 )

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Q_c = [C]^c[D]^d / [A]^a[B]^b ——————————————————————————————–(7.20)

Then,

If Q_c > K_c, the reaction will proceed in the direction of reactants (reverse reaction).

If Q_c < K_c, the reaction will proceed in the direction of the products (forward reaction).

If Q_c = K_c, the reaction mixture is already at equilibrium.

Consider the gaseous reaction of H₂ with I₂,

H2 ( g ) + I2 ( g ) = 2 HI ( g );

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K_c = 57.0 at 700 K.

Suppose we have molar concentrations [H₂]_t=0.10M, [I₂]_t = 0.20 M and [HI]_t = 0.40 M. (the subscript t on the concentration symbols means that the concentrations were measured at some arbitrary time t, not necessarily at equilibrium).

Thus, the reaction quotient, Q_c at this stage of the reaction is given by,

Q_c = [HI]²_t/[H₂]_t [I₂]_t = (0.40)²/ (0.10)x(0.20) = 8.0

Now, in this case, Q_c (8.0) does not equal K_c (57.0), so the mixture of H₂(g), I₂(g) and HI(g) is not at equilibrium; that is, more H₂(g) and I₂(g) will react to form more HI(g) and their concentrations will decrease till Q_c = K_c.

The reaction quotient, Q_c is useful in predicting the direction of reaction by comparing the values of Q_c and K_c.

Thus, we can make the following generalisations concerning the direction of the reaction (Fig. 7.7) :

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• If Q_c < K_c, net reaction goes from left to right

• If Q_c > K_c, net reaction goes from right to left.

• If Q_c = K_c, no net reaction occurs.

Problem 7.7

The value of K_c for the reaction

2A = B + C

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is 2 x 10^-3. At a given time, the composition of reaction mixture is [A] = [B] = [C] = 3 x 10^-4 M. In which direction the reaction will proceed?

Solution

For the reaction the reaction quotient Q_c is given by,

Q_c = [B][C]/ [A]

as [A] = [B] = [C] = 3 x10^-4M

Q_c = (3 x 10^-4)(3 x 10^-4) / (3 x 10^-4)2 = 1

as Q_c > K_c so the reaction will proceed in the reverse direction.

7.6.3 Calculating Equilibrium Concentrations

In case of a problem in which we know the initial concentrations but do not know any of the equilibrium concentrations, the following three steps shall be followed:

Step 1. Write the balanced equation for the reaction.

Step 2. Under the balanced equation, make a table that lists for each substance involved in the reaction:

(a) the initial concentration,

(b) the change in concentration on going to equilibrium, and

In constructing the table, define x as the concentration (mol/L) of one of the substances that reacts on going to equilibrium, then use the stoichiometry of the reaction to determine the concentrations of the other substances in terms of x.

Step 3. Substitute the equilibrium concentrations into the equilibrium equation for the reaction and solve for x. If you are to solve a quadratic equation choose the mathematical solution that makes chemical sense.

Step 4. Calculate the equilibrium concentrations from the calculated value of x.

Step 5. Check your results by substituting them into the equilibrium equation.

Problem 7.8

13.8g of N₂O₄ was placed in a 1L reaction vessel at 400K and allowed to attain equilibrium

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N2O4 ( g ) = 2NO2 ( g )

The total pressure at equilbrium was found to be 9.15 bar. Calculate K_c, K_p and partial pressure at equilibrium.

Solution

We know pV = nRT

Total volume (V ) = 1 L

Molecular mass of N₂O₄ = 92 g

Number of moles = 13.8g/92 g = 0.15 of the gas (n)

Gas constant (R) = 0.083 bar L mol^-1K^-1

Temperature (T ) = 400 K

pV = nRT

p 1L = 0.15 mol x 0.083 bar L mol^-1K^-1 x 400 K

p = 4.98 bar

Initial pressure: 4.98 bar 0

At equilibrium: (4.98 – x) bar 2x bar

Hence,

p_total at equilibrium = p_N₂O₄ + NO₂

9.15 = (4.98 – x) + 2x

9.15 = 4.98 + x

x = 9.15 – 4.98 = 4.17 bar

Partial pressures at equilibrium are,

p_N₂O₄ = 4.98 – 4.17 = 0.81bar

p_NO₂ = 2 x 4.17 = 8.34 bar

K_p=(p_NO₂)²/p_N₂O₄

= (8.34)²/0.81 = 85.87

K_p = K_c(RT)^Δn

85.87 = K_c(0.083 x 400)¹

K_c = 2.586 = 2.6

Problem 7.9

3.00 mol of PCl₅ kept in 1L closed reaction vessel was allowed to attain equilibrium at 380K. Calculate composition of the mixture at equilibrium. K_c= 1.80

Solution

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K_c = [PCl₃][Cl₂]/[PCl₅]

1.8 = x²/ (3 – x)

x² + 1.8x – 5.4 = 0

x = [-1.8 ± √(1.8)2 – 4(-5.4)]/2

x = [-1.8 ± √3.24 + 21.6]/2

x = [-1.8 ± 4.98]/2

x = [-1.8 + 4.98]/2 = 1.59

[PCl₅] = 3.0 – x = 3 -1.59 = 1.41 M

[PCl₃] = [Cl₂] = x = 1.59 M

7.7 RELATIONSHIP BETWEEN QUILIBRIUM CONSTANT K, REACTION QUOTIENT Q AND GIBBS ENERGY G

The value of K_c for a reaction does not depend on the rate of the reaction. However, as you have studied in Unit 6, it is directly related to the thermodynamics of the reaction and in particular, to the change in Gibbs energy, ΔG. If,

• ΔG is negative, then the reaction is spontaneous and proceeds in the forward

direction.

• ΔG is positive, then reaction is considered non-spontaneous. Instead, as reverse reaction would have a negative ΔG, the products of the forward reaction shall be converted to the reactants.

• ΔG is 0, reaction has achieved equilibrium; at this point, there is no longer any free energy left to drive the reaction.

A mathematical expression of this thermodynamic view of equilibrium can be

described by the following equation:

ΔG = ΔG^Θ + RT lnQ —————————————————————————————-(7.21)

where, G^Θ is standard Gibbs energy.

At equilibrium, when ΔG = 0 and Q = K_c, the equation (7.21) becomes,

ΔG = ΔG^Θ + RT ln K = 0

ΔG^Θ = – RT lnK —————————————————————————————(7.22)

lnK = – ΔG^Θ / RT

Taking antilog of both sides, we get,

K = e^{– ΔG}^Θ / RT ————————————————————————————– (7.23)

Hence, using the equation (7.23), the reaction spontaneity can be interpreted in terms of the value of ΔG^Θ.

• If ΔG^Θ < 0, then -ΔG^Θ/RT is positive, and e^{– ΔG}^Θ /RT >1, making K >1, which implies a spontaneous reaction or the reaction which proceeds in the forward direction to such an extent that the products are present predominantly.

• If ΔG^Θ > 0, then -ΔG^Θ/RT is negative, and e^{– ΔG}^Θ /RT < 1, that is , K < 1, which implies a non-spontaneous reaction or a reaction which proceeds in the forward direction to such a small degree that only a very minute quantity of product is formed.

Problem 7.10

The value of ΔG^Θfor the phosphorylation of glucose in glycolysis is 13.8 kJ/mol. Find the value of Kc at 298 K.

Solution

ΔG^Θ = 13.8 kJ/mol = 13.8 x 10³J/mol

Also, ΔG^Θ = – RT lnK_c

Hence, ln K_c = -13.8 x 10³J/mol (8.314 J mol^-1K^-1 298 K)

ln K_c = – 5.569

K_c = e^-5.569

K_c = 3.81 x 10^-3

Problem 7.11

Hydrolysis of sucrose gives,

Sucrose + H2O = Glucose + Fructrose

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Equilibrium constant K_c for the reaction is 2 x 10¹³ at 300K. Calculate ΔG^Θ at 300K.

Solution

ΔG^Θ = – RT lnKc

ΔG^Θ = – 8.314J mol^-1K^-1 x 300K ln(21013)

ΔG^Θ = – 7.64 104 J mol^-1

7.8 FACTORS AFFECTING EQUILIBRIA

One of the principal goals of chemical synthesis is to maximise the conversion of the reactants to products while minimizing the expenditure of energy. This implies maximum yield of products at mild temperature and pressure conditions. If it does not happen, then the experimental conditions need to be adjusted. For example, in the Haber process for the synthesis of ammonia from N₂ and H₂, the choice of experimental conditions is of real economic importance. Annual world production of ammonia is about hundred million tones, primarily for use as fertilizers.

Equilibrium constant, K_c is independent of initial concentrations. But if a system at equilibrium is subjected to a change in the concentration of one or more of the reacting substances, then the system is no longer at equilibrium; and net reaction takes place in some direction until the system returns to equilibrium once again. Similarly, a change in temperature or pressure of the system may also alter the equilibrium. In order to decide what course the reaction adopts and make a qualitative prediction about the effect of a change in conditions on equilibrium we use Le Chatelier’s principle. It states that a change in any of the factors that determine the equilibrium conditions of a system will cause the system to change in such a manner so as to reduce or to counteract the effect of the change. This is applicable to all physical and chemical equilibria.

We shall now be discussing factors which can influence the equilibrium.

7.8.1 Effect of Concentration Change

In general, when equilibrium is disturbed by the addition/removal of any reactant/products, Le Chatelier’s principle predicts that:

• The concentration stress of an added reactant/product is relieved by net reaction in the direction that consumes the added substance.

• The concentration stress of a removed reactant/product is relieved by net reaction in the direction that replenishes the removed substance.

or in other words,

“When the concentration of any of the reactants or products in a reaction at

equilibrium is changed, the composition of the equilibrium mixture changes so as to minimize the effect of concentration changes”.

Let us take the reaction,

H2 ( g ) + I2 ( g ) = 2HI ( g )

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If H₂ is added to the reaction mixture at equilibrium, then the equilibrium of the reaction is disturbed. In order to restore it, the reaction proceeds in a direction wherein H₂ is consumed, i.e., more of H₂ and I₂ react to form HI and finally the equilibrium shifts in right (forward) direction (Fig.7.8). This is in accordance with the Le Chatelier’s principle which implies that in case of addition of a reactant/product, a new equilibrium will be set up in which the concentration of the reactant/product should be less than what it was after the addition but more than what it was in the original mixture.

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The same point can be explained in terms of the reaction quotient, Q_c,

Q_c = [HI]²/ [H₂][I₂]

Addition of hydrogen at equilibrium results in value of Q_c being less than K_c . Thus, in order to attain equilibrium again reaction moves in

the forward direction. Similarly, we can say that removal of a product also boosts the forward reaction and increases the concentration of the products and this has great commercial application in cases of reactions, where the product is a gas or a volatile substance. In case of manufacture of ammonia, ammonia is liquified and removed from the reaction mixture so that reaction keeps moving in forward direction. Similarly, in the large scale production of CaO (used as important building material) from CaCO₃, constant removal of CO₂ from the kiln drives the reaction to completion. It should be remembered that continuous removal of a product maintains Q_c at a value less than K_c and reaction continues to move in the forward direction.

Effect of Concentration – An experiment

This can be demonstrated by the following reaction:

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K_c = [Fe(SCN)²⁺(aq)]/[Fe³⁺(aq)][SCN^–(aq)] ————————————————————————————— (7.25)

A reddish colour appears on adding two drops of 0.002 M potassium thiocynate solution to 1 mL of 0.2 M iron(III) nitrate solution due to the formation of [Fe(SCN)]²⁺. The intensity of the red colour becomes constant on attaining equilibrium. This equilibrium can be shifted in either forward or reverse directions depending on our choice of adding a reactant or a product. The equilibrium can be shifted in the opposite direction by adding reagents that remove Fe³⁺ or SCN^– ions. For example, oxalic acid (H₂C₂O₄), reacts with Fe³⁺ ions to form the stable complex ion [Fe(C₂O₄)₃]^3-, thus decreasing the concentration of free Fe³⁺(aq). In accordance with the Le Chatelier’s principle, the concentration stress of removed Fe³⁺ is relieved by dissociation of [Fe(SCN)]²⁺ to replenish the Fe³⁺ ions. Because the concentration of [Fe(SCN)]²⁺ decreases, the intensity of red colour decreases.

Addition of aq. HgCl₂ also decreases red colour because Hg²⁺ reacts with SCN^– ions to form stable complex ion [Hg(SCN)₄]^2-. Removal of free SCN^– (aq) shifts the equilibrium in equation (7.24) from right to left to replenish SCN^– ions. Addition of potassium thiocyanate on the other hand increases the colour intensity of the solution as it shift the equilibrium to right.

Experiments related to Chemical Equilibrium

AimStudy of shift in equilibrium in the reaction of ferric ions and thiocyanate ions by increasing the concentration of any one of these ions.TheoryThe equilibrium reaction between ferric chloride and potassium thiocyanate is conveniently studied through the change in the intensity of colour of the solution.Fe₃₊(aq) + SCN_–(aq)= [Fe(SCN)]²⁺ (aq)The equilibrium constant for the above reaction may be written as:K = [[Fe(SCN)]²⁺(aq)]/[Fe³⁺(aq)][SCN^–(aq)]Here K is constant at a constant temperature. Increasing the concentration of either Fe³⁺ ion or thiocyanate ion would result in a corresponding increase in the concentration of [Fe(SCN)]²⁺ ions. In order to keep the value of K constant, there is a shift in equilibrium, in the forward direction and consequently an increase in the intensity of the blood red colour which is due to [Fe(SCN)]²⁺ . At equilibrium colour intensity remains constant.Material required•Beakers (100ml) : Two•Beaker (250ml) : one•Boiling tubes : Six•Burettes : Four•Glass droppers : two•Test tube stand : One•Glass rod : One•Ferric chloride : 0.100g•Potassium thiocyanate : 0.100g

NOTE:-HAZARD WARNING
Ferric chloride
• Avoid contact with skin and eyes

Procedure

(i) Dissolve 0.100 g ferric chloride in 100 mL of water in a beaker and 0.100 g potassium thiocyanate in 100 mL of water in another beaker.

(ii) Mix 20 mL of ferric chloride solution with 20 mL of potassium thiocyanate solution. Blood red colour will be obtained. Fill this solution in a burette.

(iii) Take five boiling tubes of same size and mark them as a,b,c, d and e.

(iv) Add 2.5 mL of blood red solution to each of the boiling tubes from the burette.

(v) Add 17.5 mL of water to the boiling tube ‘a’ so that total volume of solution in the boiling tube ‘a’ is 20 mL. Keep it for reference.

(vi) Now take three burettes and label them as A, B, and C.

(vii) Fill burette A with ferric chloride solution, burette B with potassium thiocynate solution and burette C with water.

(viii) Add 1.0 mL, 2.0 mL, 3.0 mL and 4.0 mL of ferric chloride solution to boiling tubes b, c, d and e respectively from burette A.

(ix) Now add 16.5 mL, 15.5 mL, 14.5 mL, and 13.5 mL of water to boiling tubes b, c, d and e respectively from burette C so that total volume of solution in each boiling tube is 20 mL.

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(x) Compare the colour intensity of the solution in each boiling tube with the colour intensity of reference solution in boiling tube ‘a’.

(xi) Take another set of four clean boiling tubes. Add 2.5 mL of blood red solution to each of the boiling tubes from the burette. Repeat the experiment by adding 1.0 mL, 2.0 mL, 3.0 mL and 4.0 mL of potassium thiocynate solution from burette B to the boiling tubes b′, c′, d′, and e′ respectively followed by addition of 16.5 mL, 15.5 mL, 14.5 mL and 13.5 mL of water respectively to these test tubes. Again compare the colour intensity of the solution of these test tubes with reference equilibrium solution in boiling tube ‘a’.

(xii) Record your results in tabular form as in Tables 4.1 and 4.2.

(xiii) You may repeat the observations with different amounts of potassium thiocyanate and ferric chloride solution and compare with the reference solution.

Table 4.1:Equilibrium shift on increasing the concentration of ferric ions
Boiling Tube	Volume of ferric chloride solution taken in the system in ml	Direction of shift in equilibrium
a	Reference solution for matching color containing 2.5 ml blood red solution + 17.5 ml water (20 ml equilibrium mixture)	Equlibrium position
b	1.0
c	2.0
d	3.0
e	4.0

Table 4.2:Equilibrium shift on increasing the concentration of thiocyanate ions
Boiling Tube	Volume of thiocyanate solution taken in the system in ml	Direction of shift in equilibrium
a	Reference solution for matching color containing 2.5 ml blood red solution + 17.5 ml water (20 ml equilibrium mixture)	Equlibrium position
b’	1.0
c’	2.0
d’	3.0
e’	4.0

Note : • Colour intensity of the solution will decrease very much on dilution.

It will not be deep blood red colour.

• Total volume in each test tube is 20 mL.

• Each test tube has 2.5 mL equilibrium mixture.

• Amount of FeCl₃ is increasing from test tubes ‘b’ to ‘e’.

Precautions

(a)Use very dilute solutions of ferric chloride and potassium thiocyanate.

(b)Compare the colour of the solutions by keeping the boiling tube and the reference test tube side by side.

(c)To judge the change in colour of the solution in an effective manner, note the colour change in diffused sunlight.

(d)Use boiling tubes of the same size.

Discussion Questionns

(i) Explain why representing the ionic reaction between ferric and thiocyanate ions as given in the text viz.

Fe³⁺(aq) + SCN^– (aq) = [Fe(SCN)]²⁺(aq)

is more appropriate in the following form ?

[Fe (H²O)⁶]³⁺ + SCN^–(aq) = [Fe(H₂O)₅(SCN)]²⁺ + H₂O.

(ii) Does the constancy in colour intensity indicate the dynamic nature of equilibrium? Explain your answer with appropriate reasons.

(iii) What is equilibrium constant and how does it differ from the rate constant?

(iv) It is always advisable to carry out the present experiment with dilute solutions. Why?

(v) What will be the effect of adding solid potassium chloride to the system at equilibrium? Verify your answer experimentally.

(vi) Why boiling tubes of same size are used in the experiment?

EXPERIMENT4.2

Aim

Study of the shift in equilibrium in the reaction between [Co(H₂O)₆]M²⁺ and Cl^– ions, by changing the concentration of any one of these ions.

Theory

In the reaction between [Co (H₂O)₆]²⁺ and Cl^– ions, the following displacement reaction takes place.

[Co(H₂O)₆]²⁺+4Cl^{– =}[CoCl₄]^2-+6H₂O

Pink Blue

This reaction is known as ligand displacement reaction and the equilibrium constant, K, for this is written as follows:

K = [[CoCl₄]^2– / [[Co(H₂O)₆]²⁺] [Cl^–]⁴

Since the reaction occurs in the aqueous medium, it is believed that concentration of H ₂O is almost constant and is included in the value of K itself and is not shown separately in the expression for equilibrium constant.

Now if at equilibrium the concentration of either [Co (H₂O)₆²⁺] ion or Cl^– ions is increased, then this would result in an increase in [CoCl₄] ion concentration thus, maintaining the value of K as constant. In other words we can say that equilibrium will shift in the forward direction and will result in a corresponding change in colour.

Material Required

•Conical Flask(100ml) : One

•Beakers(100ml) : Three

•Burettes : Three

•Test tubes : Six

•Test tube stand : One

•Glass rod : One

•Acetone/alcohol : 60ml

•concentrated hydrochloric acid : 30ml

•Cobalt chloride : 0.6000g

NOTE:-HAZARD WARNING
Hydrochloride acid
Acetone
Alcohol
• Acetone and alcohol are inflamable, do not let the bottle open when not in use.•keep the bottle away from flames.•Wash your hands after use.•wear safety spectacles.

Procedure

(i) Take 60 mL of acetone in a 100 mL conical flask and dissolve 0.6000 g CoCl2 in it to get a blue solution.

(ii) Take 5 test tubes of same size and mark them as A, B, C, D and E. Add 3.0 mL of cobalt chloride solution in each of the test tubes from ‘A’ to ‘E’ respectively. Now add 1.0 mL, 0.8 mL, 0.6 mL, 0.4 mL and 0.2 mL of acetone respectively in these test tubes. Add 0.2 mL, 0.4 mL, 0.6 mL and 0.8 mL of water to test tubes B, C, D and E respectively, so that the total volume of solution in each of the test tubes is 4.0 mL.

(iii) Note the gradual change in colour of the mixture from blue to pink with an increase in the amount of water.

(iv) Take 10 mL cobalt chloride solution in acetone prepared above and add 5 mL distilled water to it. A solution of pink colour will be obtained.

(v) Take 1.5 mL of pink solution from step (iv) in five different test tubes labeled as A′ B′, C′, D′ and E′. Add 2.0 mL, 1.5 mL, 1.0 mL and 0.5 mL of water to the test tubes labelled from A′ to D′ and 0.5 mL, 1.0 mL, 1.5 mL, 2.0 mL and 2.5 mL concentrated HCl respectively in the test tubes A′ to E′ so that total volume of solution in the test tubes is 4 mL.

(vi)Note the gradual change in colour of pink solution to light blue with increasing amounts of hydrochloric acid. Record your observations in tabular form (Tables 4.3 and 4.4).

Table 4.3 : Shift in equilibrium on adding water
Sl. No.	Test tube	Volume of acetone added in ml	Volume of CoCl₂ solution added in ml	Volume of water added in ml
1	A	1.0	3.0	0.0
2	B	0.8	3.0	0.2
3	C	0.6	3.0	0.4
4	D	0.4	3.0	0.6
5	E	0.2	3.0	0.8

Table 4.4 : Shift in equilibrium on adding Cl^–ions
Sl. No.	Test tube	Volume of HCl added in ml	Volume of aquo complex solution added in ml	Volume of water added in ml
1	A’	0.5	1.5	2.0
2	B’	1.0	1.5	1.5
3	C’	1.5	1.5	1.0
4	D’	2.0	1.5	0.5
5	E’	2.5	1.5	0.0

Precautions

(a) Take all the precautions of experiment 4.1.

(b) Use distilled water for the experiment.

Discussion Questions

(i) What will be the effect of increasing the temperature of the reaction mixture at equilibrium?

(ii) Can an aqueous solution of sodium chloride replace concentrated HCl? Verify your answer experimentally.

(iii) Why should the total volume of the solution in each test tube be kept same?

7.8.2 Effect of Pressure Change

A pressure change obtained by changing the volume can affect the yield of products in case of a gaseous reaction where the total number of moles of gaseous reactants and total number of moles of gaseous products are different. In applying Le Chatelier’s principle to a heterogeneous equilibrium the effect of pressure changes on solids and liquids can be ignored because the volume (and concentration) of a solution/liquid is nearly independent of pressure.

Consider the reaction,

CO ( g ) + 3H2 (g ) = CH4 ( g ) + H2O ( g )

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Here, 4 mol of gaseous reactants (CO + 3H₂) become 2 mol of gaseous products (CH₄ + H₂O). Suppose equilibrium mixture (for above reaction) kept in a cylinder fitted with a piston at constant temperature is compressed to one half of its original volume. Then, total pressure will be doubled (according to pV = constant). The partial pressure and therefore, concentration of reactants and products have changed and the mixture is no longer at equilibrium. The direction in which the reaction goes to re-establish equilibrium can be predicted by applying the Le Chatelier’s principle. Since pressure has doubled, the equilibrium now shifts in the forward direction, a direction in which the number of moles of the gas or pressure decreases (we know pressure is proportional to moles of the gas). This can also be understood by using reaction quotient, Q_c. Let [CO], [H₂], [CH₄] and [H₂O] be the molar concentrations at equilibrium for methanation reaction. When volume of the reaction mixture is halved, the partial pressure and the concentration are doubled. We obtain the reaction quotient by replacing each equilibrium concentration by double its value.

Q_c = [CH₄(g)][H₂O(g)]/[CO(g)][H₂(g)]³

As Q_c < K_c , the reaction proceeds in the forward direction.

In reaction

C ( s ) + CO2 ( g ) = 2CO ( g )

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when pressure is increased, the reaction goes in the reverse direction because the number of moles of gas increases in the forward direction.

7.8.3 Effect of Inert Gas Addition

If the volume is kept constant and an inert gas such as argon is added which does not take part in the reaction, the equilibrium remains undisturbed. It is because the addition of an inert gas at constant volume does not change the partial pressures or the molar concentrations of the substance involved in the

reaction. The reaction quotient changes only if the added gas is a reactant or product involved in the reaction.

7.8.4 Effect of Temperature Change

Whenever an equilibrium is disturbed by a change in the concentration, pressure or volume, the composition of the equilibrium mixture changes because the reaction quotient, Q_c no longer equals the equilibrium constant, K_c. However, when a change in temperature occurs, the value of equilibrium constant, K_c is changed.

In general, the temperature dependence of the equilibrium constant depends on the sign of ΔH for the reaction.

• The equilibrium constant for an exothermic reaction (negative ΔH) decreases as the temperature increases.

• The equilibrium constant for an endothermic reaction (positive ΔH) increases as the temperature increases.

Temperature changes affect the equilibrium constant and rates of reactions.

Production of ammonia according to the reaction,

N2 ( g ) + 3H2 ( g ) = 2NH3 ( g )

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ΔH= – 92.38 kJ mol^-1

is an exothermic process. According to Le Chatelier’s principle, raising the

temperature shifts the equilibrium to left and decreases the equilibrium concentration of ammonia. In other words, low temperature is favourable for high yield of ammonia, but practically very low temperatures slow down the reaction and thus a catalyst is used.

Effect of Temperature — An experiment

Effect of temperature on equilibrium can be demonstrated by taking NO₂ gas (brown in colour) which dimerises into N₂O₄ gas (colourless).

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2NO2 ( g ) = N2O4 ( g )

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ΔH = -57.2 kJ mol^-1

NO₂ gas prepared by addition of Cu turnings to conc. HNO₃ is collected in two 5 mL test tubes (ensuring same intensity of colour of gas in each tube) and stopper sealed with araldite. Three 250 mL beakers 1, 2 and 3 containing freezing mixture, water at room temperature and hot water (36 3K), respectively, are taken (Fig. 7.9). Both the test tubes are placed in beaker 2 for 8-10 minutes. After this one is placed in beaker 1 and the other in beaker 3. The effect of temperature on direction of reaction is depicted very well in this experiment. At low temperatures in beaker 1, the forward reaction of formation of N₂O₄ is preferred, as reaction is exothermic, and thus, intensity of brown colour due to NO₂ decreases. While in beaker 3, high temperature favours the reverse reaction of formation of NO₂ and thus, the brown colour intensifies.

Effect of temperature can also be seen in an endothermic reaction,

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At room temperature, the equilibrium mixture is blue due to [CoCl₄]^2-. When cooled in a freezing mixture, the colour of the mixture turns pink due to [Co(H₂O)₆]³⁺ .

7.8.5 Effect of a Catalyst

A catalyst increases the rate of the chemical reaction by making available a new low energy pathway for the conversion of reactants to products. It increases the rate of forward and reverse reactions that pass through the same transition state and does not affect equilibrium. Catalyst lowers the activation energy for the forward and reverse reactions by exactly the same amount. Catalyst does not affect the equilibrium composition of a reaction mixture. It does not appear in the balanced chemical equation or in the equilibrium constant expression.

Let us consider the formation of NH from dinitrogen and dihydrogen which is highly exothermic reaction and proceeds with decrease in total number of moles formed as compared to the reactants. Equilibrium constant decreases with increase in temperature. At low temperature rate decreases and it takes long time to reach at equilibrium, whereas high temperatures give satisfactory rates but poor yields.

German chemist, Fritz Haber discovered that a catalyst consisting of iron catalyse the reaction to occur at a satisfactory rate at temperatures, where the equilibrium concentration of NH₃ is reasonably favourable. Since the number of moles formed in the reaction is less than those of reactants, the yield of NH₃ can be improved by increasing the pressure.

Optimum conditions of temperature and pressure for the synthesis of NH₃ using catalyst are around 500°C and 200 atm.

Similarly, in manufacture of sulphuric acid by contact process,

2SO2 ( g ) + O2 ( g ) = 2 SO3 ( g )

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K_c = 1.7 x 10²⁶

though the value of K is suggestive of reaction going to completion, but practically the oxidation of SO₂ to SO₃ is very slow. Thus, platinum or divanadium penta-oxide (V₂O₅) is used as catalyst to increase the rate of the reaction.

Note: If a reaction has an exceedingly small K, a catalyst would be of little help.

7.9 IONIC EQUILIBRIUM IN SOLUTION

Under the effect of change of concentration on the direction of equilibrium, you have incidently come across with the following equilibrium which involves ions:

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There are numerous equilibria that involve ions only. In the following sections we will study the equilibria involving ions. It is well known that the aqueous solution of sugar does not conduct electricity. However, when common salt (sodium chloride) is added to water it conducts electricity. Also, the conductance of electricity increases with an increase in concentration of common salt. Michael Faraday classified the substances into two categories based on their ability to conduct electricity. One category of substances conduct electricity in their aqueous solutions and are called electrolytes while the other do not and are thus, referred to as nonelectrolytes. Faraday further classified electrolytes into strong and weak electrolytes. Strong electrolytes on dissolution in water are ionized almost completely, while the weak electrolytes are only partially dissociated. For example, an aqueous solution of sodium chloride is comprised entirely of sodium ions and chloride ions, while that of acetic acid mainly contains unionized acetic acid molecules and only some acetate ions and hydronium ions. This is because there is almost 100% ionization in case of sodium chloride as compared to less than 5% ionization of acetic acid which is a weak electrolyte. It should be noted that in weak electrolytes, equilibrium is established between ions and the unionized molecules. This type of equilibrium involving ions in aqueous solution is called ionic equilibrium. Acids, bases and salts come under the category of electrolytes and may act as either strong or weak electrolytes.

7.10 ACIDS, BASES AND SALTS

Acids, bases and salts find widespread occurrence in nature. Hydrochloric acid present in the gastric juice is secreted by the lining of our stomach in a significant amount of 1.2-1.5 L/day and is essential for digestive processes. Acetic acid is known to be the main constituent of vinegar. Lemon and orange juices contain citric and ascorbic acids, and tartaric acid is found in tamarind paste. As most of the acids taste sour, the word ‘acid’ has been derived from a latin word ‘acidus’ meaning sour. Acids are known to turn blue litmus paper into red and liberate dihydrogen on reacting with some metals. Similarly, bases are known to turn red litmus paper blue, taste bitter and feel soapy. A common example of a base is washing soda used for washing purposes. When acids and bases are mixed in the right proportion they react with each other to give salts. Some commonly known examples of salts are sodium chloride, barium sulphate, sodium nitrate. Sodium chloride (common salt ) is an important component of our diet and is formed by reaction between hydrochloric acid and sodium hydroxide. It exists in solid state as a cluster of positively charged sodium ions and negatively charged chloride ions which are held together due to electrostatic interactions between oppositely charged species (Fig.7.10). The electrostatic forces between two charges are inversely proportional to dielectric constant of the medium. Water, a universal solvent, possesses a very high dielectric constant of 80. Thus, when sodium chloride is dissolved in water, the electrostatic interactions are reduced by a factor of 80 and this facilitates the ions to move freely in the solution. Also, they are wellseparated due to hydration with water molecules.

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Faraday was born near London into a family of very limited means. At the age of 14 he was an apprentice to a kind bookbinder who allowed Faraday to read the books he was binding. Through a fortunate chance he became laboratory assistant to Davy, and during 1813-4, Faraday accompanied him to the ontinent. During this trip he gained much from the experience of coming into contact with many of the leading scientists of the time. In 1825, he succeeded Davy as Director of the Royal Institution laboratories, and in 1833 he also became the first Fullerian Professor of Chemistry. Faraday’s first important work was on analytical chemistry. After 1821 much of his work was on electricity and agnetism and different electromagnetic phenomena. His ideas have led to the establishment of modern field theory. He discovered his two laws of electrolysis in 1834. Faraday was a very modest and kind hearted person. He declined all honours and avoided scientific controversies. He preferred to work alone and never had any assistant. He disseminated science in a variety of ways including his Friday evening discourses, which he founded at the Royal Institution. He has been very famous for his Christmas lecture on the ‘Chemical History of a Candle’. He published nearly 450 scientific papers.

Comparing, the ionization of hydrochloric acid with that of acetic acid in water we find that though both of them are polar covalent molecules, former is completely ionized into its constituent ions, while the latter is only partially ionized (< 5%). The extent to which ionization occurs depends upon the strength of the bond and the extent of solvation of ions produced. The terms dissociation and ionization have earlier been used with different meaning. Dissociation refers to the process of separation of ions in water already existing as such in the solid state of the solute, as in sodium chloride. On the other hand, ionization corresponds to a process in which a neutral molecule splits into charged ions in the solution. Here, we shall not distinguish between the two and use the two terms interchangeably.

7.10.1 Arrhenius Concept of Acids and Bases

According to Arrhenius theory, acids are substances that dissociates in water to give hydrogen ions H⁺(aq) and bases are substances that produce hydroxyl ions OH^–(aq). The ionization of an acid HX (aq) can be represented by the following equations:

HX (aq) → H⁺(aq) + X^– (aq)

HX(aq) + H₂O(l) → H₃O+(aq) + X^–(aq)

A bare proton, H⁺ is very reactive and cannot exist freely in aqueous solutions. Thus, it bonds to the oxygen atom of a solvent water molecule to give trigonal pyramidal hydronium ion, H₃O⁺ {[H (H₂O)]⁺} (see box). In this chapter we shall use H⁺(aq) and H₃O+(aq) interchangeably to mean the same i.e., a hydrated proton.

Similarly, a base molecule like MOH ionizes in aqueous solution according to the equation:

MOH(aq) → M⁺(aq) + OH^–(aq)

The hydroxyl ion also exists in the hydrated form in the aqueous solution. Arrhenius concept of acid and base, however, suffers from the limitation of being applicable only to aqueous solutions and also, does not account for the basicity of substances like, ammonia which do not possess a hydroxyl group.

Hydronium and Hydroxyl Ions

Hydrogen ion by itself is a bare proton with very small size (~10^-15 m radius) and intense electric field, binds itself with the water molecule at one of the two available lone pairs on it giving H₃O⁺. This species has been detected in many compounds (e.g., H₃O⁺Cl^–) in the solid state. In aqueous solution the hydronium ion is further hydrated to give species like H₅O₂⁺, H₇O₃⁺ and H₉O₄⁺. Similarly the hydroxyl ion is hydrated to give several ionic species like H₃O₂ ^–, H₅O₃^– and H₇O₄^– etc.

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7.10.2 The Brönsted-Lowry Acids and Bases

The Danish chemist, Johannes Brönsted and the English chemist, Thomas M. Lowry gave a more general definition of acids and bases. According to Brönsted-Lowry theory, acid is a substance that is capable of donating a

hydrogen ion H⁺ and bases are substances capable of accepting a hydrogen ion, H⁺. In short, acids are proton donors and bases are proton acceptors. Consider the example of dissolution of NH₃ in H₂O represented by the following equation:

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The basic solution is formed due to the presence of hydroxyl ions. In this reaction, water molecule acts as proton donor and ammonia molecule acts as proton acceptor and are thus, called Lowry-Brönsted acid and base, respectively. In the reverse reaction, H⁺ is transferred from NH₄⁺ to OH^–. In this case, NH₄⁺ acts as a Bronsted acid while OH^– acted as a Brönsted base. The acid-base pair that differs only by one proton is called a conjugate acid-base pair. Therefore, OH^– is called the conjugate base of an acid H₂O and NH₄⁺ is called conjugate acid of the base NH₃. If Brönsted acid is a strong acid then its conjugate base is a weak base and viceversa. It may be noted that conjugate acid has one extra proton and each conjugate base has one less proton.

Consider the example of ionization of hydrochloric acid in water. HCl(aq) acts as an acid by donating a proton to H₂O molecule which acts as a base.

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It can be seen in the above equation, that water acts as a base because it accepts the proton. The species H₃O⁺ is produced when water accepts a proton from HCl. Therefore, Cl^– is a conjugate base of HCl and HCl is the conjugate acid of base Cl^–. Similarly, H₂O is a conjugate base of an acid H₃O⁺ and H₃O⁺ is a conjugate acid of base H₂O.

It is interesting to observe the dual role of water as an acid and a base. In case of reaction with HCl water acts as a base while in case of ammonia it acts as an acid by donating a proton.

Arrhenius was born near Uppsala, Sweden. He presented his thesis, on the conductivities of electrolyte solutions, to the University of Uppsala in 1884. For the next five years he travelled extensively and visited a number of research centers in Europe. In 1895 he was appointed professor of physics at the newly formed University of Stockholm, serving its rector from 1897 to 1902. From 1905 until his death he was Director of physical chemistry at the Nobel Institute in Stockholm. He continued to work for many years on electrolytic solutions. In 1899 he discussed the temperature dependence of reaction rates on the basis of an equation, now usually known as Arrhenius equation.He worked in a variety of fields, and made important contributions to immunochemistry, cosmology, the origin of life, and the causes of ice age. He was the first to discuss the ‘green house effect’ calling by that name. He received Nobel Prize in Chemistry in 1903 for his theory of electrolytic dissociation and its use in the development of chemistry.

Problem 7.12

What will be the conjugate bases for the following Brönsted acids: HF, H₂SO₄ and HCO₃^– ?

Solution

The conjugate bases should have one proton less in each case and therefore the corresponding conjugate bases are: F^–, HSO₄^– and CO₃^2- respectively.

Problem 7.13

Write the conjugate acids for the following Brönsted bases: NH₂^–, NH₃ and HCOO^–.

Solution

The conjugate acid should have one extra proton in each case and therefore the corresponding conjugate acids are: NH₃, NH₄⁺ and HCOOH respectively.

Problem 7.14

The species: H₂O, HCO₃^–, HSO₄^– and NH₃ can act both as Bronsted acids and bases. For each case give the corresponding conjugate acid and conjugate base.

Solution

The answer is given in the following Table:

Species	Conjugate acid	Conjugate base
H₂O	H₃O⁺	OH^–
HCO₃^–	H₂CO₃	CO₃^2-
HSO₄^–	H₂SO₄	SO₄^2-
NH₃	NH₄⁺	NH₂^–

7.10.3 Lewis Acids and Bases

G.N. Lewis in 1923 defined an acid as a species which accepts electron pair and base which donates an electron pair. As far as bases are concerned, there is not much difference between Brönsted-Lowry and Lewis concepts, as the base provides a lone pair in both the cases. However, in Lewis concept many acids do not have proton. A typical example is reaction of electron deficient species BF₃ with NH₃.

BF₃ does not have a proton but still acts as an acid and reacts with NH₃ by accepting its lone pair of electrons. The reaction can be represented by,

BF₃ + :NH₃ → BF₃:NH₃

Electron deficient species like AlCl₃, Co³⁺, Mg²⁺, etc. can act as Lewis acids while species like H₂O, NH₃, OH^– etc. which can donate a pair of electrons, can act as Lewis bases.

Problem 7.15

Classify the following species into Lewis acids and Lewis bases and show how

these act as such:

(a) HO^–(b)F^– (c) H⁺ (d) BCl₃

Solution

(a) Hydroxyl ion is a Lewis base as it can donate an electron lone pair (:OH^– ).

(b) Flouride ion acts as a Lewis base as it can donate any one of its four electron lone pairs.

(d) BCl₃ acts as a Lewis acid as it can accept a lone pair of electrons from species like ammonia or amine molecules.

7.11 IONIZATION OF ACIDS AND BASES

Arrhenius concept of acids and bases becomes useful in case of ionization of acids and bases as mostly ionizations in chemical and biological systems occur in aqueous medium. Strong acids like perchloric acid (HClO₄), hydrochloric acid (HCl), hydrobromic acid (HBr), hyrdoiodic acid (HI), nitric acid (HNO₃) and sulphuric acid (H₂SO₄) are termed strong because they are almost completely dissociated into their constituent ions in an aqueous medium, thereby acting as proton (H⁺) donors. Similarly, strong bases like lithium hydroxide (LiOH), sodium hydroxide (NaOH), potassium hydroxide (KOH), caesium hydroxide (CsOH) and barium hydroxide Ba(OH)₂ are almost completely dissociated into ions in an aqueous medium giving hydroxyl ions, OH^–. According to Arrhenius concept they are strong acids and bases as they are able to completely dissociate and produce H₃O⁺ and OH^– ions respectively in the medium. Alternatively, the strength of an acid or base may also be gauged in terms of Brönsted- Lowry concept of acids and bases, wherein a strong acid means a good proton donor and a strong base implies a good proton acceptor. Consider, the acid-base dissociation equilibrium of a weak acid HA,

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In section 7.10.2 we saw that acid (or base) dissociation equilibrium is dynamic involving a transfer of proton in forward and reverse directions. Now, the question arises that if the equilibrium is dynamic then with passage of time which direction is favoured? What is the driving force behind it? In order to answer these questions we shall deal into the issue of comparing the strengths of the two acids (or bases) involved in the dissociation equilibrium. Consider the two acids HA and H₃O⁺ present in the above mentioned acid-dissociation equilibrium. We have to see which amongst them is a stronger proton donor. Whichever exceeds in its tendency of donating a proton over the other shall be termed as the stronger acid and the equilibrium will shift in the direction of weaker acid. Say, if HA is a stronger acid than H₃O⁺, then HA will donate protons and not H₃O⁺, and the solution will mainly contain A^– and H₃O⁺ ions. The equilibrium moves in the direction of formation of weaker acid and weaker base because the stronger acid donates a proton to the stronger base.

It follows that as a strong acid dissociates completely in water, the resulting base formed would be very weak i.e., strong acids have very weak conjugate bases. Strong acids like perchloric acid (HClO₄), hydrochloric acid (HCl), hydrobromic acid (HBr), hydroiodic acid (HI), nitric acid (HNO₃) and sulphuric acid (H₂SO₄) will give conjugate base ions ClO4 ^–, Cl, Br^–, I^–, NO3^– and HSO4^– , which are much weaker bases than H₂O. Similarly a very strong base would give a very weak conjugate acid. On the other hand, a weak acid say HA is only partially dissociated in aqueous medium and thus, the solution mainly contains undissociated HA molecules. Typical weak acids are nitrous acid (HNO₂), hydrofluoric acid (HF) and acetic acid (CH₃COOH). It should be noted that the weak acids have very strong conjugate bases. For example, NH₂^–, O^2- and H^– are very good proton acceptors and thus, much stronger bases than H₂O.

Certain water soluble organic compounds like phenolphthalein and bromothymol blue behave as weak acids and exhibit different colours in their acid (HIn) and conjugate base (In^– ) forms.

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Such compounds are useful as indicators in acid-base titrations, and finding out H⁺ ion concentration.

7.11.1 The Ionization Constant of Water and its Ionic Product

Some substances like water are unique in their ability of acting both as an acid and a base. We have seen this in case of water in section 7.10.2. In presence of an acid, HA it accepts a proton and acts as the base while in the presence of a base, B^– it acts as an acid by donating a proton. In pure water, one H₂O molecule donates proton and acts as an acid and another water molecules accepts a proton and acts as a base at the same time. The following equilibrium exists:

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The dissociation constant is represented by,

K = [H₃O⁺] [OH^–] / [H₂O] ——————————————————————————————-(7.26)

The concentration of water is omitted from the denominator as water is a pure liquid and its concentration remains constant. [H₂O] is incorporated within the equilibrium constant to give a new constant, K_w, which is called the ionic product of water.

K_w = [H⁺][OH^–] ————————————————————————————————(7.27)

The concentration of H⁺ has been found out experimentally as 1.0 x 10^-7 M at 298 K. And, as dissociation of water produces equal number of H⁺ and OH^– ions, the concentration of hydroxyl ions, [OH^–] = [H⁺] = 1.0 x 10^-7 M.

Thus, the value of Kw at 298K,

K_w = [H₃O⁺][OH^–] = (1 x 10^-7)² = 1 x 10^-14 M² ———————————————————————————(7.28)

The value of K_w is temperature dependent as it is an equilibrium constant. The density of pure water is 1000 g / L and its molar mass is 18.0 g /mol. From this the molarity of pure water can be given as, [H₂O] = (1000 g /L)(1 mol/18.0 g) = 55.55 M.

Therefore, the ratio of dissociated water to that of undissociated water can be given as:

10^-7/ (55.55) = 1.8 x 10^-9 or ~ 2 in 10^-9

(thus, equilibrium lies mainly towards undissociated water)

We can distinguish acidic, neutral and basic aqueous solutions by the relative values of the H₃O⁺ and OH^– concentrations:

Acidic: [H₃O⁺] > [OH^– ]

Neutral: [H₃O⁺] = [OH^– ]

Basic : [H₃O⁺] < [OH^–]

7.11.2 The pH Scale

Hydronium ion concentration in molarity is more conveniently expressed on a logarithmic scale known as the pH scale. The pH of a solution is defined as the negative logarithm to base 10 of the activity (a_H⁺) of hydrogen ion. In dilute solutions (< 0.01 M), activity of hydrogen ion (H⁺) is equal in magnitude to molarity represented by [H⁺]. It should be noted that activity has no units and is defined as:

a_H⁺ = [H⁺] / mol L^–1

From the definition of pH, the following can be written,

pH = – log a_H⁺ = – log {[H⁺] / mol L^–1}

Thus, an acidic solution of HCl (10^-2 M) will have a pH = 2. Similarly, a basic solution of NaOH having [OH^–] =10^-4 M and [H₃O⁺] = 10^-10 M will have a pH = 10. At 25 °C, pure water has a concentration of hydrogen ions, [H⁺] = 10^-7 M. Hence, the pH of pure water is given as:

pH = -log(10^-7) = 7

Acidic solutions possess a concentration of hydrogen ions, [H⁺] > 10^-7 M, while basic solutions possess a concentration of hydrogen ions, [H⁺] < 10^-7 M. thus, we can summarise that

Acidic solution has pH < 7 Basic solution has pH > 7 Neutral solution has pH = 7

Now again, consider the equation (7.28) at 298 K

K_w = [H₃O⁺] [OH^–] = 10^-14

Taking negative logarithm on both sides of equation, we obtain

-log Kw = -log {[H₃O⁺] [OH⁺]} = – log [H₃O⁺] – log [OH^–] = – log 10^-14

pK_w = pH + pOH = 14 —————————————————————————————–(7.29)

Note that although K_w may change with temperature the variations in pH with temperature are so small that we often ignore it.

pK_w is a very important quantity for aqueous solutions and controls the relative concentrations of hydrogen and hydroxyl ions as their product is a constant. It should be noted that as the pH scale is logarithmic, a change in pH by just one unit also means change in [H⁺] by a factor of 10. Similarly, when the hydrogen ion concentration, [H⁺] changes by a factor of 100, the value of pH changes by 2 units. Now you can realise why the change in pH with temperature is often ignored.

Measurement of pH of a solution is very essential as its value should be known when dealing with biological and cosmetic applications. The pH of a solution can be found roughly with the help of pH paper that has different colour in solutions of different pH. Now-a-days pH paper is available with four strips on it. The different strips have different colours (Fig. 7.11) at the same pH. The pH in the range of 1-14 can be determined with an accuracy of ~0.5 using pH paper.

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For greater accuracy pH meters are used. pH meter is a device that measures the pH-dependent electrical potential of the test solution within 0.001 precision. pH meters of the size of a writing pen are now available in the market. The pH of some very common substances are given in Table 7.5 (page 212).

Problem 7.16

The concentration of hydrogen ion in a sample of soft drink is 3.8 x 10^-3M. what is its pH ?

Solution

pH = – log[3.8 x 10^-3 ]

= – {log[3.8] + log[10^-3]}

= – {(0.58) + (- 3.0)} = – { – 2.42} = 2.42

Therefore, the pH of the soft drink is 2.42 and it can be inferred that it is acidic.

Problem 7.17

Calculate pH of a 1.0 x 10^-8 M solution of HCl.

Solution

2H2O ( l ) = H3O+ ( aq ) + OH- ( aq )

K_w = [OH^–][H₃O⁺] = 10^-14

Let, x = [OH^–] = [H₃O⁺] from H₂O. The H₃O⁺ concentration is generated (i) from the ionization of HCl dissolved i.e.,

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HCl ( aq ) + H2O ( l ) = H3O+ ( aq ) + Cl- ( aq )

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and (ii) from ionization of H₂O. In these very dilute solutions, both sources of H₃O⁺ must be considered:

[H₃O⁺] = 10^-8 + x

K_w = (10^-8 + x)(x) = 10^-14

or x² + 10^-8 x – 10^-14 = 0

[OH^– ] = x = 9.5 x 10^-8

So, pOH = 7.02 and pH = 6.98

7.11.3 Ionization Constants of Weak Acids

Consider a weak acid HX that is partially ionized in the aqueous solution. The equilibrium can be expressed by:

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Here, c = initial concentration of the undissociated acid, HX at time, t = 0. α = extent up to which HX is ionized into ions. Using these notations, we can derive the equilibrium constant for the above discussed aciddissociation equilibrium:

K_a = c²α² / c(1-α) = cα² / 1-α

K_a is called the dissociation or ionization constant of acid HX. It can be represented alternatively in terms of molar concentration as follows,

K_a = [H⁺][X^–] / [HX] —————————————————————————————(7.30)

At a given temperature T, K_a is a measure of the strength of the acid HX i.e., larger the value of Ka, the stronger is the acid. K_a is a dimensionless quantity with the understanding that the standard state concentration of all species is 1M. The values of the ionization constants of some selected weak acids are given in Table 7.6.

Table 7.6 The Ionization Constants of Some

Selected Weak Acids (at 298K)

Acid	Ionization Constant,K_a
Hydrofluoric Acid (HF)	3.5 x 10^-4
Nitrous Acid (HNO₂)	4.5 x 10^-4
Formic Acid (HCOOH)	1.8 x 10^-4
Niacin (C₅H₄NCOOH)	1.5 x 10^-5
Acetic Acid (CH₃COOH)	1.74 x 10^-5
Benzoic Acid (C₆H₅COOH)	6.5 x 10^-5
Hypochlorous Acid (HCIO)	3.0 x 10^-8
Hydrocyanic Acid (HCN)	4.9 x 10^-10
Phenol (C₆ H₅OH)	1.3 x 10^-10

The pH scale for the hydrogen ion concentration has been so useful that besides pK_w, it has been extended to other species and quantities. Thus, we have:

pK_a = -log (K_a) —————————————————————————————–(7.31)

Knowing the ionization constant, K_a of an acid and its initial concentration, c, it is possible to calculate the equilibrium concentration of all species and also the degree of ionization of the acid and the pH of the solution.

A general step-wise approach can be adopted to evaluate the pH of the weak

electrolyte as follows:

Step 1. The species present before dissociation are identified as Brönsted-Lowry

acids / bases.

Step 2. Balanced equations for all possible reactions i.e., with a species acting both as acid as well as base are written.

Step 3. The reaction with the higher K_a is identified as the primary reaction whilst the other is a subsidiary reaction.

Step 4. Enlist in a tabular form the following values for each of the species in the primary reaction

(a) Initial concentration, c.

(b) Change in concentration on proceeding to equilibrium in terms of α, degree of ionization.

Step 5. Substitute equilibrium concentrations into equilibrium constant equation for principal reaction and solve for α.

Step 6. Calculate the concentration of species in principal reaction.

Step 7. Calculate pH = – log[H₃O⁺]

The above mentioned methodology has been elucidated in the following examples.

Problem 7.18

The ionization constant of HF is 3.2 x 10^-4. Calculate the degree of

dissociation of HF in its 0.02 M solution. Calculate the concentration of all species present (H₃O⁺, F^– and HF) in the solution and its pH.

Solution

The following proton transfer reactions are possible:

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Substituting equilibrium concentrations in the equilibrium reaction for principal reaction gives:

K_a = (0.02α)² / (0.02 – 0.02α) = 0.02 α² / (1 -α) = 3.2 x 10^-14

We obtain the following quadratic equation:

α² + 1.6 x 10^-2 α – 1.6 x 10^-2 = 0

The quadratic equation in α can be solved and the two values of the roots are:

α = + 0.12 and – 0.12

The negative root is not acceptable and hence,

α = 0.12

This means that the degree of ionization, α = 0.12, then equilibrium concentrations of other species viz., HF, F ^– and H₃O⁺ are given by:

[H₃O⁺ ] = [F ^– ] = cα = 0.02 0.12

= 2.4 x 10^-3 M

[HF] = c(1 – α) = 0.02 (1 – 0.12) = 17.6 10-3 M

pH = – log[H⁺] = -log(2.4 x 10^-3 ) = 2.62

Problem 7.19

The pH of 0.1M monobasic acid is 4.50. Calculate the concentration of species H⁺, A^– and HA at equilibrium. Also, determine the value of K_a and pK_a of the monobasic acid.

Solution

pH = – log [H⁺ ]

Therefore, [H⁺ ] = 10 ^-pH = 10^-4.50 = 3.16 x 10 ^-5

[H⁺] = [A^–] = 3.16 x 10^-5

Thus, K_a = [H⁺][A⁺] / [HA]

[HA]_eqlbm = 0.1 -(3.16 x 10^-5) = 0.1

K_a = (3.16 x 10^-5)₂ / 0.1 = 1.0 x 10_-8

pK_a = – log(10^-8) = 8

Alternatively, ‘Percent dissociation’ is another useful method for measure of strength of a weak acid and is given as:

Percent dissociation = [HA]_dissociated/[HA]_initial x 100% ———————————————————————————–(7.32)

Problem 7.20

Calculate the pH of 0.08M solution of hypochlorous acid, HOCl. The ionization constant of the acid is 2.5 x 10^-5.Determine the percent issociation of HOCl.

Solution

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K_a = {[H₃O⁺][ClO^–] / [HOCl]}

= x² / (0.08 – x)

As x<<0.08, therefore 0.08 – x = 0.08

x² / 0.08 = 2.5 x 10^-5

x² = 2.0 x 10^-6, thus, x = 1.41 x 10^-34

[H⁺] = 1.41 x 10^-3 M.

Therefore,

Percent dissociation = {[HOCl]_dissociated / [HOCl]_initial } x 100 = 1.41 x 10^-3 / 0.08 = 1.76 %.

pH = -log(1.41 x 10^-3) = 2.85.

7.11.4 Ionization of Weak Bases

The ionization of base MOH can be represented by equation:

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MOH (aq ) = M+ (aq ) + OH- ( aq )

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In a weak base there is partial ionization of MOH into M⁺ and OH^–, the case is similar to that of acid-dissociation equilibrium. The equilibrium constant for base ionization is called base ionization constant and is represented by K_b. It can be expressed in terms of concentration in molarity of various species in equilibrium by the following equation:

K_b = [M⁺][OH^–] / [MOH] —————————————————————————————-(7.33)

Alternatively, if c = initial concentration of base and α = degree of ionization of base i.e. the extent to which the base ionizes. When equilibrium is reached, the equilibrium constant can be written as:

K_b = (cα)² / c (1-α) = cα² / (1-α)

The values of the ionization constants of some selected weak bases, K_b are given in Table 7.7.

Table 7.7 The Values of the Ionization Constant of Some Weak Bases a 298 K

Base	K_b
Dimethylamine, (CH₃)₂NH	5.4 x 10^-4
Triethylamine, (C₂H₅)₃N	6.45 x 10^-5
Ammonia, NH₃ or NH₄OH	1.77 x 10^-5
Quinine, (A plant product)	1.10 x 10^-6
Pyridine, C₅H₅N	1.77 x 10^-9
Aniline, C₆H₅NH₂	4.27 x 10^-10
Urea, CO (NH₂)₂	1.3 x 10^-14

Many organic compounds like amines are weak bases. Amines are derivatives of ammonia in which one or more hydrogen atoms are replaced by another group. For example, methylamine, codeine, quinine and nicotine all behave as very weak bases due to their very small K_b. Ammonia produces OH^– in aqueous solution:

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The pH scale for the hydrogen ion concentration has been extended to get:

pK_b = -log (K_b) ———————————————————————————–(7.34)

Problem 7.21

The pH of 0.004M hydrazine solution is 9.7. Calculate its ionization constant K_b and pK_b.

Solution

NH2NH2 + H2O = NH2NH3+ + OH-

From the pH we can calculate the hydrogen ion concentration. Knowing hydrogen ion concentration and the ionic product of water we can calculate the concentration of hydroxyl ions. Thus we have:

[H⁺] = antilog (-pH) = antilog (-9.7) = 1.67 x 10^-10

[OH^–] = K_w / [H⁺] = 1 x 10^-14 / 1.67 x 10^-10 = 5.98 x 10^-5

The concentration of the corresponding hydrazinium ion is also the same as that of hydroxyl ion. The concentration of both these ions is very small so the concentration of the undissociated base can be taken equal to 0.004M. Thus,

K_b = [NH₂NH₃ +][OH^–] / [NH₂NH₂] = (5.98 x 10^-5)2 / 0.004 = 8.96 x 10^-7

pK_b = -logK_b = -log(8.96 x 10^-7) = 6.04.

Problem 7.22

Calculate the pH of the solution in which 0.2M NH₄Cl and 0.1M NH₃ are present. The pK_b of ammonia solution is 4.75.

Solution

NH3 + H2O = NH4+ + OH-

The ionization constant of NH₃, K_b = antilog (-pK_b) i.e.

K_b = 10^-4.75 = 1.77 x 10^-5 M

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K_b = [NH₄⁺][OH^–] / [NH₃] = (0.20 + x)(x) / (0.1 – x) = 1.77 x 10^-5

As K_b is small, we can neglect x in comparison to 0.1M and 0.2M. Thus,

[OH^–] = x = 0.88 x 10^-5

Therefore, [H⁺] = 1.12 x 10^-9

pH = – log[H⁺] = 8.95.

7.11.5 Relation between K_a and K_b

As seen earlier in this chapter, K_a and K_b represent the strength of an acid and a base, respectively. In case of a conjugate acid-base pair, they are related in a simple manner so that if one is known, the other can be deduced. Considering the example of NH₄⁺ and NH₃ we see,

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K_w = [H₃O⁺][ OH^– ] = 1.0 x 10^-14 M

Where, K_a represents the strength of NH₄⁺ as an acid and K_b represents the strength of NH₃ as a base.

It can be seen from the net reaction that the equilibrium constant is equal to the product of equilibrium constants K_a and K_b for the reactions added. Thus,

K_a K_b = {[H₃O⁺][ NH₃] / [NH₄⁺ ]} {[NH₄⁺ ][ OH^–] / [NH₃]}

= [H₃O⁺][ OH^–] = K_w

= (5.6×10^-10) (1.8 x 10^-5) = 1.0 x 10^-14 M

This can be extended to make a generalisation. The equilibrium constant for

a net reaction obtained after adding two (or more) reactions equals the product of the equilibrium constants for individual reactions:

K_NET= K₁ x K₂ x ……. ————————————————————————————————(3.35)

Similarly, in case of a conjugate acid-base pair,

K_a x K_b = K_w ——————————————————————————————-(7.36)

Knowing one, the other can be obtained. It should be noted that a strong acid will have a weak conjugate base and vice-versa. Alternatively, the above expression K_w = K_a x K_b, can also be obtained by considering the base-dissociation equilibrium reaction:

B ( aq ) + H2O ( l ) = BH+ ( aq ) + OH- ( aq )

K_b = [BH⁺][OH^–] / [B]

As the concentration of water remains constant it has been omitted from the

denominator and incorporated within the dissociation constant. Then multiplying and dividing the above expression by [H⁺], we get:

K_b = [BH⁺][OH^–][H⁺] / [B][H⁺] ={[ OH^–][H⁺]}{[BH⁺] / [B][H⁺]} = K_w / K_a

or K_a K_b = K_w

It may be noted that if we take negative logarithm of both sides of the equation, then pK values of the conjugate acid and base are related to each other by the equation:

pK_a + pK_b = pK_w = 14 (at 298K)

Problem 7.23

Determine the degree of ionization and pH of a 0.05M of ammonia solution. The ionization constant of ammonia can be taken from Table 7.7. Also, calculate the ionization constant of the conjugate acid of ammonia.

Solution

The ionization of NH₃ in water is represented by equation:

NH3 + H2O = NH4+ + OH-

We use equation (7.33) to calculate hydroxyl ion concentration,

[OH^–] = c α = 0.05 α

K_b = 0.05 α² / (1 – α)

The value of α is small, therefore the quadratic equation can be simplified by neglecting α in comparison to 1 in the denominator on right hand side of the equation, Thus,

K_b = c α² or α = √ (1.77 x 10^-5 / 0.05)

= 0.018.

[OH^–] = c α = 0.05 x 0.018 = 9.4 x 10^-4M.

[H⁺] = K_w / [OH^–] = 10^-14 / (9.4 x 10^-4) = 1.06 x 10^-11

pH = -log(1.06 x 10^-11) = 10.97.

Now, using the relation for conjugate cid-base pair,

K_a x K_b = K_w

using the value of Kb of NH3 from Table 7.7.

We can determine the concentration of conjugate acid NH₄⁺

K_a = K_w / K_b = 10^-14 / 1.77 x 10^-5 = 5.64 x 10^-10.

7.11.6 Di- and Polybasic Acids and Di- and Polyacidic Bases

Some of the acids like oxalic acid, sulphuric acid and phosphoric acids have more than one ionizable proton per molecule of the acid. Such acids are known as polybasic or polyprotic acids. The ionization reactions for example for a dibasic acid H₂X are represented by the equations:

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And the corresponding equilibrium constants are given below:

K_a₁ = {[H⁺][HX^–]} / [H₂X]

and K_a₂ = {[H⁺][X^2-]} / [HX^–]

Here, K_a₁ and K_a₂ are called the first and second ionization constants respectively of the acid H₂ X. Similarly, for tribasic acids like H₃PO₄ we have three ionization constants. The values of the ionization constants for some common polyprotic acids are given in Table 7.8.

Table 7.8 The Ionization Constants of Some

Common Polyprotic Acids (298K)

Acid	K_a₁	K_a₂	K_a₃
Oxalic Acid	5.9 x 10^-2	6.4 x 10^-5
Ascorbic Acid	7.4 x 10^-4	1.6 x 10^-12
Sulphurous Acid	1.7 x 10^-2	6.4 x 10^-8
Sulphuric Acid	Very large	1.2 x 10^-2
Carbonic Acid	4.3 x 10^-7	5.6 x 10^-11
Citric Acid	7.4 x 10^-4	1.7 x 10^-5	4.0 x 10^-7
Phosphoric Acid	7.5 x 10^-3	6.2 x 10^-8	4.2 x 10^-13

It can be seen that higher order ionization constants (K_a₂, K_a₃ ) are smaller than the lower order ionization constant ( K_a₁ ) of a polyprotic acid. The reason for this is that it is more difficult to remove a positively charged proton from a negative ion due to electrostatic forces. This can be seen in the case of removing a proton from the uncharged H₂CO₃ as compared from a negatively charged HCO₃^–.

Similarly, it is more difficult to remove a proton from a doubly charged HPO₄^2- anion as compared to H₂PO₄^–.

Polyprotic acid solutions contain a mixture of acids like H₂A, HA^– and A^2- in case of a diprotic acid. H₂A being a strong acid, the primary reaction involves the dissociation of H₂ A, and H₃O⁺ in the solution comes mainly from the first dissociation step.

7.11.7 Factors Affecting Acid Strength

Having discussed quantitatively the strengths of acids and bases, we come to a stage where we can calculate the pH of a given acid solution. But, the curiosity rises about why should some acids be stronger than others? What factors are responsible for making them stronger? The answer lies in its being a complex phenomenon. But, broadly speaking we can say that the extent of dissociation of an acid depends on the strength and polarity of the H-A bond.

In general, when strength of H-A bond decreases, that is, the energy required to break the bond decreases, HA becomes a stronger acid. Also, when the H-A bond becomes more polar i.e., the electronegativity difference between the atoms H and A increases and there is marked charge separation, cleavage of the bond becomes easier thereby increasing the acidity.

But it should be noted that while comparing elements in the same group of the periodic table, H-A bond strength is a more important factor in determining acidity than its polar nature. As the size of A increases down the group, H-A bond strength decreases and so the acid strength increases. For example,

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K_a = [H⁺][Ac^– ] / [HAc]

Addition of acetate ions to an acetic acid solution results in decreasing the concentration of hydrogen ions, [H⁺]. Also, if H⁺ ions are added from an external source then the equilibrium moves in the direction of undissociated acetic acid i.e., in a direction of reducing the concentration of hydrogen ions, [H⁺]. This phenomenon is an example of common ion effect. It can be defined as a shift in equilibrium on adding a substance that provides more of an ionic species already present in the dissociation equilibrium. Thus, we can say that common ion effect is a phenomenon based on the Le Chatelier’s principle discussed in section 7.8.

In order to evaluate the pH of the solution resulting on addition of 0.05M acetate ion to 0.05M acetic acid solution, we shall consider the acetic acid dissociation equilibrium once again,

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Therefore,

K_a= [H⁺][Ac^– ]/[H Ac] = {(0.05+x)(x)}/(0.05-x)

As K_a is small for a very weak acid, x<<0.05.

Hence, (0.05 + x) ≈ (0.05 – x) ≈ 0.05

Thus,

1.8 x 10^-5 = (x) (0.05 + x) / (0.05 – x) = x(0.05) / (0.05) = x = [H⁺] = 1.8 x 10^-5M

pH = – log(1.8 x 10^-5) = 4.74

Problem 7.24

Calculate the pH of a 0.10M ammonia solution. Calculate the pH after 50.0 mL of this solution is treated with 25.0 mL of 0.10M HCl. The dissociation constant of ammonia, K_b = 1.77 x 10^-5

Solution

NH₃ + H₂O → NH₄⁺ + OH^–

K_b = [NH₄⁺][OH^–] / [NH₃] = 1.77 x 10^-5

Before neutralization,

[NH₄⁺] = [OH^–] = x

[NH₃] = 0.10 – x = 0.10

x² / 0.10 = 1.77 x 10^-5

Thus, x = 1.33 x 10^-3 = [OH^–]

Therefore,[H⁺] = Kw / [OH^–] = 10^-14 /

(1.33 x 10^-3) = 7.51 x 10^-12

pH = -log(7.5 x 10^-12) = 11.12

On addition of 25 mL of 0.1M HCl solution (i.e., 2.5 mmol of HCl) to 50 mL of 0.1M ammonia solution (i.e., 5 mmol of NH₃), 2.5 mmol of ammonia molecules are neutralized. The resulting 75 mL solution contains the remaining unneutralized 2.5 mmol of NH₃ molecules and 2.5 mmol of NH₃⁺.

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where, y = [OH^–] = [NH₄⁺]

The final 75 mL solution after neutralisation already contains 2.5 m mol NH₄⁺ ions (i.e. 0.033M), thus total concentration of NH₄⁺ ions is given as:

[NH₄⁺] = 0.033 + y

As y is small, [NH₄OH] = 0.033 M and [NH₄⁺] = 0.033M.

We know,

K_b = [NH₄⁺][OH^–] / [NH₄OH] = y(0.033)/(0.033) = 1.77 x 10^-5 M

Thus, y = 1.77 x 10^-5 = [OH^–]

[H⁺] = 10^-14 / 1.77 x 10^-5 = 0.56 x 10^-9

Hence, pH = 9.24

7.11.9 Hydrolysis of Salts and the pH of their Solutions

Salts formed by the reactions between acids and bases in definite proportions, undergo ionization in water. The cations/anions formed on ionization of salts either exist as hydrated ions in aqueous solutions or interact with water to reform corresponding acids/bases depending upon the nature of salts. The later process of interaction between water and cations/anions or both of salts is called hydrolysis. The pH of the solution gets affected by this interaction. The cations (e.g., Na⁺, K⁺, Ca²⁺, Ba²⁺, etc.) of strong bases and anions (e.g., Cl^–, Br^–, NO₃^–, ClO₄^– etc.) of strong acids simply get hydrated but do not hydrolyse, and therefore the solutions of salts formed from strong acids and bases are neutral i.e., their pH is 7. However, the other category of salts do undergo hydrolysis.

We now consider the hydrolysis of the salts of the following types :

(i) salts of weak acid and strong base e.g., CH₃COONa.

(ii) salts of strong acid and weak base e.g., NH₄Cl, and

(iii) salts of weak acid and weak base, e.g., CH₃COONH₄.

In the first case, CH₃COONa being a salt of weak acid, CH₃COOH and strong base, NaOH gets completely ionised in aqueous solution.

CH₃COONa(aq) → CH₃COO^– (aq)+ Na⁺(aq)

Acetate ion thus formed undergoes hydrolysis in water to give acetic acid and OH^– ions

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Without going into detailed calculation, it can be said that degree of hydrolysis is independent of concentration of solution, and pH of such solutions is determined by their pK values:

pH = 7 + (pK_a – pK_b) —————————————————————————————–(7.38)

The pH of solution can be greater than 7, if the difference is positive and it will be less than 7, if the difference is negative.

Problem 7.25

The pK_a of acetic acid and pK_b of ammonium hydroxide are 4.76 and 4.75 respectively. Calculate the pH of ammonium acetate solution.

Solution

pH = 7 + ½[pK_a – pK_b = 7 + ½[4.76 – 4.75] = 7 + ½[0.01] = 7 + 0.005 = 7.005

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Experiment related to pH

EXPERIMENT 5.1AimTo determine the pH of some fruit juices.TheorySeveral dyes show different colours at different pH. These act as acid-base indicators. Solution of a mixture of dyes can be used to obtain approximate pH value of a solution. A solution of a mixture of dyes can be obtained to measure pH values from zero to 14. It is called universal indicator. Some universal indicators can measure the pH change of even 0.5. In fact, dyes themselves are weak acids or bases. Colour change occurs as a result of change in the structure of dye due to acceptance or release of protons. Different forms of a dye have different colours and hence, colour change is observed when pH of the solution changes. A standard chart for the colour change of the universal indicator with pH is supplied with the indicator paper or solution and the comparison of observed colour change with the chart provides a good estimate of the pH of the solution.Material Required• Beakers (100 mL) : Four• Glass droppers : Four• Test tubes : Four• pH chart : One• Fruit juice: Lemon,orange,apple,pineapple• pH papers/universalindicator solutio:As per needProcedure(i)Procure fresh juices of lemon, orange, apple and pineapple in separate beakers of 100 mL capacity each.(ii)Transfer nearly 2 mL of the fresh juice (20 drops) with the help of a separate dropper for each juice in four different test tubes marked 1, 2, 3 and 4 respectively.(iii)Add two drops of the universal indicator in each test tube and mix the content of each test tube thoroughly by shaking.(iv)Match the colour appearing in each test tube with the standard pH chart.(v)Record your observations in Table 5.1.(vi)Repeat the experiment using pH papers to ascertain the pH of different juices and match the colour in each case with the one obtained with universal indicator.(vii)Arrange the pH value of the four juices in increasing order.

Name of the juice	Color with universal indicator	pH	Inference
Lemon
Orange
Apple
Pineapple

Precautions

(a) Add equal number of drops of universal indicator to equal volumes of solutions in each of the test tubes.

(b) Match the colour of the solution with pH chart carefully.

(d) Use only fresh juice for the experiment.

Discussion Questions

(i) Out of the four juices, which one is least acidic? Explain.

(ii) If we dilute each of the juices, what effect is likely to be observed on the pH values?

(iii) On mixing any two juices, would the pH alter or remain the same? Verify your answer experimentally.

(iv) How can you ascertain the pH of a soft drink ?

EXPERIMENT 5.2

Aim

To observe the variation in pH of acid/base with dilution.

Theory

Hydrogen ion concentration per unit volume decreases on dilution. Therefore, change in pH is expected on dilution of the solution.

Material Reqiured

• Boiling tubes:Eight

• Glass droppers:Four

• Test tubes:As per need

•0.1 M HCl solution: 20mL

•0.1 M NaOH solution: 20mL

•0.05 M H2SO 4 solution: 20mL

•pH paper/universal: As per need indicator

Procedure

(i)Take four boiling tubes and mark them as A, B, C and D. (Fig. 5.1).

(ii)Take 2mL of 0.1 M HCl in boiling tube A.

(iii) Take 2mL of 0.1 M HCl in boiling tube B and add 18 ml water to it and mix thoroughly.

(iv) Take 5mL of dilute HCl solution from boiling tube B in boiling tube C and add 15mL water to it.

(v) Take 5mL of diluted HCl from boiling tube C in boiling tube D and add 15 mL water to it.

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(vi) Cut a pH paper into small pieces and spread these on a clean glazed tile.

(vii) Take out some solution from boiling tube A with the help of a dropper and pour one drop on one of the pieces of pH paper kept on the glazed tile. Compare the colour of the pH glazed tile.

(vii) Take out some solution from boiling tube A with the help of a dropper and pour one drop on one of the pieces of pH paper kept on the glazed tile. Compare the colour of the pH paper with the standard chart.

(viii) Similarly test the pH of solutions of boiling tubes B, C, and D respectively and record your results as in Table 5.2.

(ix) Calculate the hydrogen ion concentration of solution B, C and D.

(x) Take out 1mL of solution from each boiling tube and transfer in separate test tubes. Add 2 drops of universal indicator to each of these test tubes. Shake the test tubes well and match the colour of these solutions with the standard pH chart to estimate the pH.

(xi) Similarly observe the change in pH of 0.05 M H₂SO₄ and 0.1M NaOH solution with dilution as detailed in steps (i) to (ix) above.

(xii) Record your observations in Table 5.2.

(xiii) Compare the result obtained by using universal indicator paper and that obtained by using universal indicator solution.

Boiling tube	HCl		H₂SO₄		NaOH
Boiling tube	Colour	pH	Colour	pH	Colour	pH
A
B
C
D

Result

(i)Concentration of solutions of test tube B, C and D are ____________.

(ii)Write your conclusion about the variation of pH with dilution.

Precautions

(a)Add equal number of drops of the universal indicator to equal amounts of solution in each of the boiling tubes.

(b)Match the colour of the solution with pH chart carefully.

Discussion Questions

(i) What trend is observed in the variation of pH with dilution for acidic as well as for basic solutions?

(ii) How do you explain the results of variation in pH with dilution?

(iii) If any two acidic solutions (say A and C) are mixed, what would happen to the pH of the mixture? Verify your answer experimentally.

(iv) For each acidic solution, whether we use HCl or H2SO 4, pH is same to a reasonably good extent, even though HCl is 0.1M, and H2SO4 is 0.05M. How do you explain this result?

(v) Will the pH of 0.1M acetic acid be the same as that of 0.1M hydrochloric acid? Verify your result and explain it?

EXPERIMENT5.3

Aim

To study the variation in pH by common ion effect in case of weak

acids and weak bases.

Theory

It is a known fact that the ionisation in the case of either a wtd/sup eak acid or a weak base is a reversible process. This can be represented as:

(1) HA = H⁺ + A^–(weak acid)

(2) BOH = B⁺ + OH^–(weak base)

The increase in concentration of A– ions in case (1) and that of B⁺ ions in case (2) would shift the equilibrium in the reverse direction thereby decreasing the concentration of H⁺ ions and OH^– ions in cases (1) and (2) respectively so as to maintain the constancy of equilibrium constant K. This change either in H⁺ ion concentration or OH^– ion concentration brings a change in the pH of the system, which can be judged either with the help of a pH paper or by using a universal indicator solution.

Material Required

• Beakers (100 mL) : Four

• Pipettes (25 mL) : Two

• Test tubes : Four

• pH chart : One

•Sodium ethanoate:2 g

•Ammonium chloride:2 g

•Ethanoic acid (1.0 M):50ml

•Ammonia solution (1.0 M):50ml

•pH paper and universal indicator: as per need

Procedure

(i) Take four 100 mL beakers and mark them as A, B, C and D.

(ii) Transfer 25 mL of 1M ethanoic acid in beaker ‘A’ and 25 mL of (1M) ammonia solution in beaker ‘B’.

(iii) Similarly transfer 25 mL of (1 M) ethanoic acid in beaker ‘C’ and 25 mL of (1.0 M) ammonia solution in beaker ‘D’. Now add 2 g sodium ethanoate in beaker ‘C’ and dissolve it. Likewise add 2 g of ammonium chloride in beaker ‘D’ and dissolve it by shaking the content of the beaker thoroughly.

(iv) Take approximately 2 mL (20 drops) of the solution from beakers A, B, C and D respectively into test tubes marked as 1, 2, 3 and 4.

(v) In each of the test tubes add 2 drops of universal indicator solution. Shake the content of the test tubes well and match the colour in each case with the standard pH chart.

(vi) Record your observations as given in Table 5.3.

(vii) Compare pH of the solution in test tubes 1 and 3 and record the change in pH.

(viii) Similarly compare pH of the solution in test tubes 2 and 4 and record the change in pH.

Sl. No. Test Tube	Composition of system	Color of pH paper	PH
1	CH₃COOH in water
2	NH₄OH(NH₃ in water
3	CH₃COOH+CH₃COONa
4	NH₄OH+NH₄

Result

(a)pH of acetic acid is _______

(b)pH of buffer of acetic acid and sodium acetate is ________ than acetic acid.

(c)pH of ammonia solution is _______

(d)pH of buffer of ammonia solution and ammonium chloride is _________ than the ammonia solution.

(e)Common ion effect ________ ionization of acid/base.

Precautions

(a)Try only weak acid/weak base and its salt for the study of the common ion effect.

(b)Handle the bottle of ammonium hydroxide with care.

(c)Add equal number of drops of the universal indicator in each of the test tubes.

(e)Store pH papers at a safe and dry place.

Discussion Questions

(i) The addition of sodium acetate to acetic acid increases the pH whereas, the addition of NH₄Cl to aqueous NH₃ solution (NH₄OH) decreases the pH of the system. How do you explain these observations?

(ii) Suggest suitable replacement for CH₃COONa for system 3 and NH₄Cl for system 4.

(iii) Suggest other pairs of weak acid and its salt and weak base and its salt to carry out the present investigations.

(iv) In salt analysis/mixture analysis, point out the situations where the variation in pH is carried out by common ion effect.

(v) How do buffer solutions resist change in the pH? Explain with a suitable example.

EXPERIMENT5.4

Aim

To study the change in pH during the titration of a strong acid with a strong base by using universal indicator.

Theory

It is assumed that strong acids and strong bases are completely dissociated in solution. During the process of neutralisation, H₊ ions obtained from the acid combine with the OH_– ions produced by base and form water. Therefore, when a solution of strong acid is added to a solution of strong base or vice versa, the pH of the solution changes. As the titration proceeds, initially there is slow

change in the pH but in the vicinity of the equivalence point there is very rapid change in the pH of the solution.

Material Required

•Burette:One

•Beakers (250 mL):Two

•Conical flask (100 mL):One

•Dropper:One

•pH chart:One

• Hydrochloric acid (0.1 M): 25 mL

• Sodium Hydroxide solution (0.1 M): 50 mL

• Universal indicator: As per requirement

Procedure

(i) Take 25 mL hydrochloric acid solution (0.1 M) in 100 mL Hydrochloric acid conical flask.

(ii) Add five drops of universal indicator solution to it.

(iii) Add (0.1 M) sodium hydroxide solution from the burette as

(iv) Shake the content of the flask well after each addition of sodium hydroxide solution. Note the colour of the solution

in the conical flask each time and find out the pH by comparing its colour with the pH chart.

(v) Note down your observations in Table 5.4.

(vi) Plot a graph of pH vs. total volume of NaOH added.

Table 5.4 : pH change during the neutralisation of25 mL of HCl (0.1 M) with NaOH (0.1 M) solution
Sl. No.	Volume of NaOH added in lots (ml)	Total volume of NaOH added to the solution in flask (ml)
1	0	0
2	12.5	12.5
3	10.0	22.5
4	2.3	24.8
5	0.1	24.9
6	0.1	25.0
7	0.1	25.1
8	0.1	25.2
9	0.1	25.3
10	0.1	25.4
11	0.1	25.9

Precautions

(a)To get good results perform the reaction with solutions of strong acid and strong base of same concentration.

(b)Handle the bottle of acid and base with care.

(c)Use small amount of indicator.

Result

Write down your result on the basis of data.

Discussion Questions

(i) What trend of pH change will you observe in the neutralisation of strong acid with strong base?

(ii) Do you expect the same trend of pH change for neutralisation of weak acid (acetic acid) with a strong base (sodium hydroxide)?

(iii) In which pH range should the indicator show colour change if the hydrochloric acid is to be neutralised by sodium hydroxide? Give answer after looking at the graph of the experiment.

(iv) Explain how does the study of pH change help in choosing the indicator for neutralisation reaction.

EXPERIMENT5.5

Aim

To study pH of solutions of sodium chloride, ferric chloride and sodium carbonate.

Theory

Salts of strong acid and strong base form neutral solutions while salts of weak acid/base and strong base/acid are basic and acidic respectively in nature. Salts of weak acid/base with strong base/ acid are hydrolysed in water while salts formed by neutralization of strong acid and strong base do not hydrolyse in solution. You have already learnt about this in your chemistry textbook.

Material Required

• Boiling tubes: Three

• Test tubes: Three

• Glass droppers: Three

•pH paper/universal indicator:As per need

•0.1 M NaCl solution:As per need

•0.1 M FeCl₃ solution:As per need

•0.1 M Na₂CO₃ solution:As per need

Procedure

(i) Take three boiling tubes and mark them as A, B and C.

(ii) Take 20 mL of 0.1 M solution(s) of NaCl, FeCl₃ and Na₂CO₃ in boiling tubes A, B and C respectively.

(iii) Cut the pH paper in small pieces and spread the pieces on a clean glazed tile.

(iv) Test the pH of the solution in boiling tubes A, B and C as in the experiment 5.1.

(v) Arrange three clean test tubes in a test tube stand.

(vi) Number the test tubes as 1, 2 and 3.

(vii) Pour 4 mL solution from boiling tube A in each of the test tubes.

(viii) Add 5 mL, 10 mL and 15 mL water in the test tubes 1, 2 and 3 respectively.

(ix) Note the pH of the solutions of test tubes 1, 2 and 3 with the help of pH paper and universal indicator.

(x) Repeat the experiment with the solutions of boiling tubes B and C.

(xi) Note your results in tabular form as in Table 5.5.

Table 5.5 : pH of NaCl, FeCl₃ and Na₂CO₃ solutions of different concentrations
Solution	pH of solution
Solution	Test tube 1	Test tube 2	Test tube 3
NaCl
FeCl₃
Na₂CO₃

Result

Write down the result on the basis of your observations.

Precautions

(a)Use freshly prepared solutions.

(b)Do not leave the bottles open after taking out salt.

(c)Use separate and clean test tube for each solution.

(d)Store the pH paper at a safe and dry place.

Discussion Questions

(i) Why are FeCl₃ and Na₂CO₃ solutions not neutral?

(ii) Why are the salts of strong acid and strong base not hydrolysed? Explain.

(iii) How is the phenomenon of hydrolysis useful in salt analysis?

(iv) What is the effect of dilution on pH of salt solution? Verify and explain your results.

7.12 BUFFER SOLUTIONS

Many body fluids e.g., blood or urine have definite pH and any deviation in their pH indicates malfunctioning of the body. The control of pH is also very important in many chemical and biochemical processes. Many medical and cosmetic formulations require that these be kept and administered at a particular pH. The solutions which resist change in pH on dilution or with the addition of small amounts of acid or alkali are called Buffer Solutions. Buffer solutions of known pH can be prepared from the knowledge of pK_a of the acid or pK_b of base and by controlling the ratio of the salt and acid or salt and base. A mixture of acetic acid and sodium acetate acts as buffer solution around pH 4.75 and a mixture of ammonium chloride and ammonium hydroxide acts as a buffer around pH 9.25. You will learn more about buffer solutions in higher classes.

7.13 SOLUBILITY EQUILIBRIA OF SPARINGLY SOLUBLE SALTS

We have already known that the solubility of ionic solids in water varies a great deal. Some of these (like calcium chloride) are so soluble that they are hygroscopic in nature and even absorb water vapour from atmosphere. Others (such as lithium fluoride) have so little solubility that they are commonly termed as insoluble. The solubility depends on a number of factors important amongst which are the lattice enthalpy of the salt and the solvation enthalpy of the ions in a solution. For a salt to dissolve in a solvent the strong forces of attraction between its ions (lattice enthalpy) must be overcome by the ion-solvent interactions. The solvation enthalpy of ions is referred to in terms of solvation which is always negative i.e. energy is released in the process of solvation. The amount of solvation enthalpy depends on the nature of the solvent. In case of a non-polar (covalent) solvent, solvation enthalpy is small and hence, not sufficient to overcome lattice enthalpy of the salt. Consequently, the salt does not dissolve in non-polar solvent. As a general rule , for a salt to be able to dissolve in a particular solvent its solvation enthalpy must be greater than its lattice enthalpy so that the latter may be overcome by former. Each salt has its characteristic solubility which depends on temperature. We classify salts on the basis of their solubility in the following three categories.

Category I	Soluble	Solubility > 0.1M
Category II	Slightly Soluble	0.01M<Solubility< 0.1M
Category III	Sparingly Soluble	Solubility < 0.01M

We shall now consider the equilibrium between the sparingly soluble ionic salt and its saturated aqueous solution.

7.13.1 Solubility Product Constant

Let us now have a solid like barium sulphate in contact with its saturated aqueous solution. The equilibrium between the undisolved solid and the ions in a saturated solution can be represented by the equation:

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The equilibrium constant is given by the equation:

K = {[Ba²⁺][SO₄^2-]} / [BaSO₄]

For a pure solid substance the concentration remains constant and we can write

K_sp = K[BaSO₄] = [Ba²⁺][SO₄^2-] ————————————————————————————————(7.39)

We call K_sp the solubility product constant or simply solubility product. The experimental value of K_sp in above equation at 298K is 1.1 x 10^-10. This means that for solid barium sulphate in equilibrium with its saturated solution, the product of the concentrations of barium and sulphate ions is equal to its solubility product constant. The concentrations of the two ions will be equal to the molar solubility of the barium sulphate. If molar solubility is S, then

1.1 x 10^-10 = (S)(S) = S²or S = 1.05 x 10^-5.

Thus, molar solubility of barium sulphate will be equal to 1.05 x 10^-5 mol L^-1.

A salt may give on dissociation two or more than two anions and cations carrying different charges. For example, consider a salt like zirconium phosphate of molecular formula (Zr⁴⁺)₃(PO₄^3-)₄. It dissociates into 3 zirconium cations of charge +4 and 4 phosphate anions of charge -3. If the molar solubility of zirconium phosphate is S, then it can be seen from the stoichiometry of the compound that

[Zr⁴⁺] = 3S and [PO₄^3-] = 4S

and K_sp = (3S)³ (4S)⁴ = 6912 (S)⁷

or S = {K_sp / (3³ x 4⁴)}^1/7 = (K_sp / 6912)^1/7

A solid salt of the general formula M_x^p+ X_y^q- with molar solubility S in equilibrium with its saturated solution may be represented by the equation:

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(where X x p⁺ = y x q^–)

And its solubility product constant is given by:

K_sp = [M^p+]^x[X^q- ]^y = (xS)^x(yS)^y ————————————————————————-(7.40)

= x^x . y^y . S^{(x + y)}

S^{(x + y)} = K_sp / x^x . y^y

S = (K_sp / x^x . y ^y)^{1 / x + y} ———————————————————————————(7.41)

The term K_sp in equation is given by Q_sp (section 7.6.2) when the concentration of one or more species is not the concentration under equilibrium. Obviously under equilibrium conditions K_sp = Q_sp but otherwise it gives the direction of the processes of precipitation or dissolution. The solubility product constants of a number of common salts at 298K are given in Table 7.9.

Table 7.9 The Solubility Product Constants, K_sp of Some Common Ionic Salts at 298K.

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Problem 7.26

Calculate the solubility of A₂X₃ in pure water, assuming that neither kind of ion reacts with water. The solubility product of A₂X₃, K_sp = 1.1 x 10^-23.

Solution

A₂X₃ → 2A³⁺ + 3X^2-

K_sp = [A³⁺]² [X^2-]³ = 1.1 x 10^-23

If S = solubility of A₂X₃, then

[A³⁺] = 2S; [X²] = 3S

therefore, K_sp = (2S)²(3S)³ = 108S⁵ = 1.1 x 10^-23

thus, S⁵ = 1 x 10^-25

S = 1.0 x 10^-5 mol/L.

Problem 7.27

The values of K_sp of two sparingly soluble salts Ni(OH)₂ and AgCN are 2.0 x 10^-15 and 6 x 0^-17 respectively. Which salt is more soluble? Explain.

Solution

AgCN = Ag+ + CN-

K_sp = [Ag⁺][CN^–] = 6 x 10^-17

Ni (OH)2 = Ni2+ + 2OH-

K_sp = [Ni²⁺][OH^–]²= 2 x 10^-15

Let [Ag⁺] = S₁, then [CN^–] = S₁

Let [Ni²⁺] = S₂, then [OH^–] = 2S₂

S₁² = 6 x 10^-17 , S₁ = 7.8 x 10^-9

(S₂)(2S₂)² = 2 x 10^-15, S₂ = 0.58 x10^-4

Ni(OH)₂ is more soluble than AgCN.

7.13.2 Common Ion Effect on Solubility of Ionic Salts

It is expected from Le Chatelier’s principle that if we increase the concentration of any one of the ions, it should combine with the ion of its opposite charge and some of the salt will be precipitated till once again K_sp = Q_sp. Similarly, if the concentration of one of the ions is decreased, more salt will dissolve to increase the concentration of both the ions till once again K_sp = Q_sp. This is applicable even to soluble salts like sodium chloride except that due to higher concentrations of the ions, we use their activities instead of their molarities in the expression for Q_sp. Thus if we take a saturated solution of sodium chloride and pass HCl gas through it, then sodium chloride is precipitated due to increased concentration (activity) of chloride ion available from the dissociation of HCl. Sodium chloride thus obtained is of very high purity and we can get rid of impurities like sodium and magnesium sulphates. The common ion effect is also used for almost complete precipitation of a particular ion as its sparingly soluble salt, with very low value of solubility product for gravimetric estimation. Thus we can precipitate silver ion as silver chloride, ferric ion as its hydroxide (or hydrated ferric oxide) and barium ion as its sulphate for quantitative estimations.

Problem 7.28

Calculate the molar solubility of Ni(OH)₂ in 0.10 M NaOH. The ionic product of Ni(OH)₂ is 2.0 x 10^-15.

Solution

Let the solubility of Ni(OH)₂ be equal to S. Dissolution of S mol/L of Ni(OH)₂ provides S mol/L of Ni²⁺ and 2S mol/L of OH^–, but the total concentration of OH^– = (0.10 + 2S) mol/L because the solution already contains 0.10 mol/L of OH^– from NaOH.

K_sp = 2.0 x 10^-15 = [Ni²⁺] [OH^–]²

= (S) (0.10 + 2S)²

As K_sp is small, 2S << 0.10, thus, (0.10 + 2S) ≈ 0.10

Hence,

2.0 x 10^-15 = S (0.10)²

S = 2.0 x 10^-13 M = [Ni²⁺]

The solubility of salts of weak acids like phosphates increases at lower pH. This is because at lower pH the concentration of the anion decreases due to its protonation. This in turn increase the solubility of the salt so that K_sp = Q_sp. We have to satisfy two equilibria simultaneously i.e.,

K_sp = [M⁺] [X^–],

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HX ( aq ) = H+ ( aq ) + X- ( aq )

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K_a = [H⁺(aq)][X^–(aq)]/[HX(aq)]

Taking inverse of both side and adding 1 we get

[HX]/[X^–] + 1 = [H⁺]/K_a + 1

([HX]+[H^–])/[X^–] = ([H⁺] + K_a)/K_a

Now, again taking inverse, we get

[X^–] / {[X^–] + [HX]} = f = K_a / (K_a + [H⁺]) and it can be seen that ‘f ’ decreases as pH decreases. If S is the solubility of the salt at a given pH then

K_sp = [S] [f S] = S² {K_a / (K_a + [H⁺])} and

S = {K_sp ([H⁺] + K_a ) /K_a }^1/2 —————————————————————–(7.42)

Thus solubility S increases with increase in [H⁺] or decrease in pH.

SUMMARY

When the number of molecules leaving the liquid to vapour equals the number of molecules returning to the liquid from vapour, equilibrium is said to be attained and is dynamic in nature. Equilibrium can be established for both physical and chemical processes and at this stage rate of forward and reverse reactions are equal. Equilibrium constant, K_c is expressed as the concentration of products divided by reactants, each term raised to the stoichiometric coefficient. For reaction,

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Equilibrium constant has constant value at a fixed temperature and at this stage all the macroscopic properties such as concentration, pressure, etc. become constant. For a gaseous reaction equilibrium constant is expressed as K_p and is written by replacing concentration terms by partial pressures in K_c expression. The direction of reaction can be predicted by reaction quotient Q_c which is equal to K_c at equilibrium. Le Chatelier‘s principle states that the change in any factor such as temperature, pressure, concentration, etc. will cause the equilibrium to shift in such a direction so as to reduce or counteract the effect of the change. It can be used to study the effect of various factors such as temperature, concentration, pressure, catalyst and inert gases on the direction of equilibrium and to control the yield of products by controlling these factors.Catalyst does not effect the equilibrium composition of a reaction mixture but increases the rate of chemical reaction by making available a new lower energy pathway for conversion of reactants to products and vice-versa.

All substances that conduct electricity in aqueous solutions are called electrolytes. Acids, bases and salts are electrolytes and the conduction of electricity by their aqueous solutions is due to anions and cations produced by the dissociation or ionization of electrolytes in aqueous solution. The strong electrolytes are completely dissociated. In weak electrolytes there is equilibrium between the ions and the unionized electrolyte molecules. According to Arrhenius, acids give hydrogen ions while bases produce hydroxyl ions in their aqueous solutions. Brönsted-Lowry on the other hand, defined an acid as a proton donor and a base as a proton acceptor. When a Brönsted-Lowry acid reacts with a base, it produces its conjugate base and a conjugate acid corresponding to the base with which it reacts. Thus a conjugate pair of acid-base differs only by one proton. Lewis further generalised the definition of an acid as an electron pair acceptor and a base as an electron pair donor. The expressions for ionization (equilibrium) constants of weak acids (K_a) and weak bases (K_b) are developed using Arrhenius definition. The degree of ionization and its dependence on concentration and common ion are discussed. The pH scale (pH = -log[H⁺]) for the hydrogen ion concentration (activity) has been introduced and extended to other quantities (pOH = – log[OH^–]) ; pK_a = -log[K_a] ; pK_b = -log[K_b]; and pK_w = -log[K_w] etc.). The ionization of water has been considered and we note that the equation: pH + pOH = pK_w is always satisfied. The salts of strong acid and weak base, weak acid and strong base, and weak acid and weak base undergo hydrolysis in aqueous solution.The definition of buffer solutions, and their importance are discussed briefly. The solubility equilibrium of sparingly soluble salts is discussed and the equilibrium constant is introduced as solubility product constant (K_sp). Its relationship with solubility of the salt is established. The conditions of precipitation of the salt from their solutions or their dissolution in water are worked out. The role of common ion and the solubility of sparingly soluble salts is also discussed.

SUGGESTED ACTIVITIES FOR STUDENTS REGARDING THIS UNIT

(a) The student may use pH paper in determining the pH of fresh juices of various vegetables and fruits, soft drinks, body fluids and also that of water samples available.

(b) The pH paper may also be used to determine the pH of different salt solutions and from that he/she may determine if these are formed from strong/weak acids and bases.

(c) They may prepare some buffer solutions by mixing the solutions of sodium acetate and acetic acid and determine their pH using pH paper.

(d) They may be provided with different indicators to observe their colours in solutions of varying pH.

(e) They may perform some acid-base titrations using indicators.

(f) They may observe common ion effect on the solubility of sparingly soluble salts.

(g) If pH meter is available in their school, they may measure the pH with it and compare the results obtained with that of the pH paper.

EXERCISES

7.1 A liquid is in equilibrium with its vapour in a sealed container at a fixed temperature. The volume of the container is suddenly increased.

a) What is the initial effect of the change on vapour pressure?

b) How do rates of evaporation and condensation change initially?

c) What happens when equilibrium is restored finally and what will be the final vapour pressure?

7.2 What is K_c for the following equilibrium when the equilibrium concentration of each substance is: [SO₂]= 0.60M, [O₂] = 0.82M and [SO₃] = 1.90M ?

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Both the forward and reverse reactions in the equilibrium are elementary bimolecular reactions. What is K_c, for the reverse reaction?

7.7 Explain why pure liquids and solids can be ignored while writing the equilibrium constant expression?

7.8 Reaction between N₂ and O_2- takes place as follows:

2N2 ( g ) + O2 ( g ) = 2N2O ( g )

If a mixture of 0.482 mol N₂ and 0.933 mol of O₂ is placed in a 10 L reaction vessel and allowed to form N₂O at a temperature for which K_c= 2.0 x 10^-37, determine the composition of equilibrium mixture.

7.9 Nitric oxide reacts with Br₂ and gives nitrosyl bromide as per reaction given below:

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7.13 The equilibrium constant expression for a gas reaction is,

K_c = [ NH₃ ]⁴[ O₂]⁵/[ NO]⁴ [H₂O]⁶

Write the balanced chemical equation corresponding to this expression.

7.14 One mole of H₂O and one mole of CO are taken in 10 L vessel and heated to 725 K. At equilibrium 40% of water (by mass) reacts with CO according to the equation,

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(i) Write the concentration ratio (reaction quotient), Q_c, for this reaction (note: water is not in excess and is not a solvent in this reaction)

(ii) At 293 K, if one starts with 1.00 mol of acetic acid and 0.18 mol of ethanol, there is 0.171 mol of ethyl acetate in the final equilibrium mixture. Calculate the equilibrium constant.

(iii) Starting with 0.5 mol of ethanol and 1.0 mol of acetic acid and maintaining it at 293 K, 0.214 mol of ethyl acetate is found after sometime. Has equilibrium been reached?

7.19 A sample of pure PCl5 was introduced into an evacuated vessel at 473 K. After equilibrium was attained, concentration of PCl₅ was found to be 0.5 x 10^-1 mol L^-1. If value of K_c is 8.3 x 10^-3, what are the concentrations of PCl₃ and Cl₂ at equilibrium?

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Calculate K_c for this reaction at the above temperature.

7.24 Calculate a) ΔG^θ and b) the equilibrium constant for the formation of NO₂ from NO and O₂ at 298K

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7.27 The equilibrium constant for the following reaction is 1.6 x 10⁵ at 1024K

H2 ( g ) + Br2 ( g ) = 2HBr ( g )

Find the equilibrium pressure of all gases if 10.0 bar of HBr is introduced into a sealed container at 1024K.

7.28 Dihydrogen gas is obtained from natural gas by partial oxidation with steam as per following endothermic reaction:

CH4 ( g ) + H2O ( g ) = CO ( g ) + 3 H2 ( g )

(a) Write as expression for K_p for the above reaction.

(b) How will the values of K_p and composition of equilibrium mixture be affected by

(i) increasing the pressure

(ii) increasing the temperature

(iii) using a catalyst ?

7.29 Describe the effect of :

a) addition of H₂

b) addition of CH₃OH

c) removal of CO

d) removal of CH₃OH

on the equilibrium of the reaction:

2H2 ( g ) + CO ( g ) = CH3OH ( g )

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7.30 At 473 K, equilibrium constant K^c for decomposition of phosphorus pentachloride, PCl₅ is 8.3 x 10^-3. If decomposition is depicted as,

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7.33 The value of K_c for the reaction 3O2 ( g ) = 2O3 ( g ) is 2.0 x 10^-50 at 25°C. If the equilibrium concentration of O₂ in air at 25°C is 1.6 x 10^-2, what is the concentration of O₃?

7.34 The reaction, CO ( g ) + 3 H2 ( g ) = CH4 ( g ) + H2O ( g ) is at equilibrium at 1300 K in a 1L flask. It also contain 0.30 mol of CO, 0.10 mol of H₂ and 0.02 mol of H₂O and an unknown amount of CH₄ in the flask. Determine the concentration of CH₄ in the mixture. The equilibrium constant, K_c for the reaction at the given temperature is 3.90.

7.35 What is meant by the conjugate acid-base pair? Find the conjugate acid/base for the following species:

HNO₂, CN^–, HClO₄, F^–, OH^–, CO₃^2-, and S₂

7.36 Which of the followings are Lewis acids? H₂O, BF₃, H⁺, and NH₄⁺

7.37 What will be the conjugate bases for the Brönsted acids: HF, H₂SO₄ and HCO₃^–?

7.38 Write the conjugate acids for the following Brönsted bases: NH₂^–, NH₃ and HCOO^–.

7.39 The species: H₂O, HCO₃^–, HSO₄^– and NH₃ can act both as Brönsted acids and bases. For each case give the corresponding conjugate acid and base.

7.40 Classify the following species into Lewis acids and Lewis bases and show how these act as Lewis acid/base: (a) OH^– (b) F^– (c) H⁺ (d) BCl₃ .

7.41 The concentration of hydrogen ion in a sample of soft drink is 3.8 x 10^-3 M. what is its pH?

7.42 The pH of a sample of vinegar is 3.76. Calculate the concentration of hydrogen ion in it.

7.43 The ionization constant of HF, HCOOH and HCN at 298K are 6.8 x 10^-4 ,1.8 x 10^-4 and 4.8 x 10^-9 respectively. Calculate the ionization constants of the corresponding conjugate base.

7.44 The ionization constant of phenol is 1.0 x 10^-10. What is the concentration of phenolate ion in 0.05 M solution of phenol? What will be its degree of ionization if the solution is also 0.01M in sodium phenolate?

7.45 The first ionization constant of H₂S is 9.1 x 10^-8. Calculate the concentration of HS^– ion in its 0.1M solution. How will this concentration be affected if the solution is 0.1M in HCl also ? If the second dissociation constant of H₂S is 1.2 x 10^-13, calculate the concentration of S^2- under both conditions.

7.46 The ionization constant of acetic acid is 1.74 x 10^-5. Calculate the degree of dissociation of acetic acid in its 0.05 M solution. Calculate the concentration of acetate ion in the solution and its pH.

7.47 It has been found that the pH of a 0.01M solution of an organic acid is 4.15. Calculate the concentration of the anion, the ionization constant of the acid and its pK_a.

7.48 Assuming complete dissociation, calculate the pH of the following solutions:

(a) 0.003 M HCl (b) 0.005 M NaOH (c) 0.002 M HBr (d) 0.002 M KOH

7.49 Calculate the pH of the following solutions:

a) 2 g of TlOH dissolved in water to give 2 litre of solution.

b) 0.3 g of Ca(OH)₂ dissolved in water to give 500 mL of solution.

c) 0.3 g of NaOH dissolved in water to give 200 mL of solution.

d) 1mL of 13.6 M HCl is diluted with water to give 1 litre of solution.

7.50 The degree of ionization of a 0.1M bromoacetic acid solution is 0.132. Calculate the pH of the solution and the pK_a of bromoacetic acid.

7.51 The pH of 0.005M codeine (C₁₈H₂₁NO₃) solution is 9.95. Calculate its ionization constant and pK_b.

7.52 What is the pH of 0.001M aniline solution ? The ionization constant of aniline can be taken from Table 7.7. Calculate the degree of ionization of aniline in the solution. Also calculate the ionization constant of the conjugate acid of aniline.

7.53 Calculate the degree of ionization of 0.05M acetic acid if its pK_a value is 4.74. How is the degree of dissociation affected when its solution also contains

(a) 0.01M (b) 0.1M in HCl ?

7.54 The ionization constant of dimethylamine is 5.4 x 10^-4. Calculate its degree of ionization in its 0.02M solution. What percentage of dimethylamine is ionized if the solution is also 0.1M in NaOH?

7.55 Calculate the hydrogen ion concentration in the following biological fluids whose pH are given below:

(a) Human muscle-fluid, 6.83 (b) Human stomach fluid, 1.2

7.56 The pH of milk, black coffee, tomato juice, lemon juice and egg white are 6.8, 5.0, 4.2, 2.2 and 7.8 respectively. Calculate corresponding hydrogen ion concentration in each.

7.57 If 0.561 g of KOH is dissolved in water to give 200 mL of solution at 298 K. Calculate the concentrations of potassium, hydrogen and hydroxyl ions. What is its pH?

7.58 The solubility of Sr(OH)₂ at 298 K is 19.23 g/L of solution. Calculate the concentrations of strontium and hydroxyl ions and the pH of the solution.

7.59 The ionization constant of propanoic acid is 1.32 x 10^-5. Calculate the degree of ionization of the acid in its 0.05M solution and also its pH. What will be its degree of ionization if the solution is 0.01M in HCl also?

7.60 The pH of 0.1M solution of cyanic acid (HCNO) is 2.34. Calculate the ionization constant of the acid and its degree of ionization in the solution.

7.61 The ionization constant of nitrous acid is 4.5 x 10^-4. Calculate the pH of 0.04 M sodium nitrite solution and also its degree of hydrolysis.

7.62 A 0.02M solution of pyridinium hydrochloride has pH = 3.44. Calculate the ionization constant of pyridine.

7.63 Predict if the solutions of the following salts are neutral, acidic or basic:

NaCl, KBr, NaCN, NH4NO₃, NaNO₂ and KF

7.64 The ionization constant of chloroacetic acid is 1.35 x 10^-3. What will be the pH of 0.1M acid and its 0.1M sodium salt solution?

7.65 Ionic product of water at 310 K is 2.7 x 10^-14. What is the pH of neutral water at this temperature?

7.66 Calculate the pH of the resultant mixtures:

a) 10 mL of 0.2M Ca(OH)₂ + 25 mL of 0.1M HCl

b) 10 mL of 0.01M H₂SO₄ + 10 mL of 0.01M Ca(OH)₂

c) 10 mL of 0.1M H₂SO₄ + 10 mL of 0.1M KOH

7.67 Determine the solubilities of silver chromate, barium chromate, ferric hydroxide, lead chloride and mercurous iodide at 298K from their solubility product constants given in Table 7.9. Determine also the molarities of individual ions.

7.68 The solubility product constant of Ag₂CrO₄ and AgBr are 1.1 x 10^-12 and 5.0 x 10^-13 respectively. Calculate the ratio of the molarities of their saturated solutions.

7.69 Equal volumes of 0.002 M solutions of sodium iodate and cupric chlorate are mixed together. Will it lead to precipitation of copper iodate? (For cupric iodate K_sp = 7.4 x 10^-8 ).

7.70 The ionization constant f benzoic acid is 6.46 x 10^-5 and K_sp for silver benzoate is 2.5 x 10^-13. How many times is silver benzoate more soluble in a buffer of pH 3.19 compared to its solubility in pure water?

7.71 What is the maximum concentration of equimolar solutions of ferrous sulphate and sodium sulphide so that when mixed in equal volumes, there is no precipitation of iron sulphide? (For iron sulphide, K_sp = 6.3 x 10^-18).

7.72 What is the minimum volume of water required to dissolve 1g of calcium sulphate at 298 K? (For calcium sulphate, K_sp is 9.1 x 10^-6).

7.73 The concentration of sulphide ion in 0.1M HCl solution saturated with hydrogen sulphide is 1.0 x 10^-19 M. If 10 mL of this is added to 5 mL of 0.04 M solution of the following: FeSO₄, MnCl₂, ZnCl₂ and CdCl₂. in which of these solutions precipitation will take place?

Answer to Some Selected Problems

7.2 12.237 molL^–1

7.3 2.67 x 10⁴

7.5 (i) 4.4 x 10^–4 (ii) 1.90

7.6 1.59 x 10^–15

7.8 [N₂] = 0.0482 molL^–1, [O₂] = 0.0933 molL^–1, [N₂O] = 6.6 x 10^–21 molL^–1

7.9 0.0352mol of NO and 0.0178mol of Br₂

7.10 7.47 x 10¹¹ M^–1

7.11 4.0

7.12 Q_c = 2.379 x 10³. No, reaction is not at equilibrium.

7.14 0.44

7.15 0.068 molL^–1 each of H₂ and I₂

7.16 [I₂] = [Cl₂] = 0.21 M, [ICl] = 0.36 M

7.17 [C₂H₆]eq = 3.62 atm

7.18 (i) [CH₃COOC₂H₅][H₂O] / [CH₃COOH][C₂H₅OH]

(ii) 22.9 (iii) value of Q_c is less than K_c therefore equilibrium is not attained.

7.19 0.02molL^–1 for both.

7.20 [PCO] = 1.739atm, [PCO₂] = 0.461atm.

7.21 No, the reaction proceeds to form more products.

7.22 3 x 10^–4 molL^–1

7.23 6.46

7.24 a) – 35.0kJ, b) 1.365 x 10⁶

7.27 [PH₂]eq = [PBr₂]eq = 2.5 x 10^–2bar, [PHBr] = 10.0 bar

7.30 b) 120.48

7.31 [H₂]eq = 0.96 bar

7.33 2.86 x 10^–28 M

7.34 5.85 x 10^–2

7.35 NO₂^–, HCN, ClO₄, HF, H₂O, HCO₃^–, HS^–

7.36 BF₃, H⁺, NH₄⁺

7.37 F^–, HSO₄^–, CO₃^2–

7.38 NH₃, NH₄⁺, HCOOH

7.41 2.42

7.42 1.7 x 10^–4M

7.43 F^–= 1.5 x 10^–11, HCOO^–= 5.6 x 10^–11, CN^–= 2.08 x 10^–6

7.44 [phenolate ion]= 2.2 x 10^–6, pH= 5.65, α= 4.47 x 10^–5. pH of 0.01M sodium phenolate solution= 9.30.

7.45 [HS^–]= 9.54 x 10^–5, in 0.1M HCl [HS^–] = 9.1 x 10^–8M, [S^2–]= 1.2 x 10^–13M, in 0.1M HCl [S^2–]= 1.09 x 10^–19M

7.46 [Ac^–]= 0.00093, pH= 3.03

7.47 [A^–] = 7.08 x10^–5M, K_a= 5.08 x 10^–7, pK_a= 6.29

7.48 a) 2.52 b) 11.70 c) 2.70 d) 11.30

7.49 a) 11.65 b) 12.21 c) 12.57 c) 1.87

7.50 pH = 1.88, pK_a = 2.70

7.51 K_b = 1.6 x 10^–6, pK_b = 5.8

7.52 α = 6.53 x 10^–4, K_a = 2.34 x 10^–5

7.53 a) 0.0018 b) 0.00018

7.54 α = 0.0054

7.55 a) 1.48 x 10^–7M, b) 0.063 c) 4.17 x 10^–8M d) 3.98 x 10^–7

7.56 a) 1.5 x 10^–7M, b) 10^–5M, c) 6.31 x 10^–5M d) 6.31 x 10^–3M

7.57 [K⁺] = [OH^–] = 0.05M, [H⁺] = 2.0 x 10^–13M

7.58 [Sr²⁺] = 0.1581M, [OH^–] = 0.3162M , pH = 13.50

7.59 α = 1.63 x 10^–2, pH = 3.09. In presence of 0.01M HCl, α = 1.32 x 10^–3

7.60 K_a = 2.09 x 10^–4 and degree of ionization = 0.0457

7.61 pH = 7.97. Degree of hydrolysis = 2.36 x 10^–5

7.62 K_b = 1.5 x 10^–9

7.63 NaCl, KBr solutions are neutral, NaCN, NaNO₂ and KF solutions are basic and NH₄NO₃ solution is acidic.

7.64 (a) pH of acid solution= 1.94 (b) its salt solution= 2.87

7.65 pH = 6.78

7.66 a) 11.2 b) 7.00 c) 3.00

7.67 Silver chromate S= 0.65 x 10^–4M; Molarity of Ag⁺ = 1.30 x 10^–4M
Molarity of CrO₄^2– = 0.65 x 10^–4M; Barium Chromate S = 1.1 x 10^–5M;
Molarity of Ba²⁺ and CrO₄^2– each is 1.1 x 10^–5M; Ferric Hydroxide S = 1.39 x 10^–10M;
Molarity of Fe³⁺ = 1.39 x 10^–10M; Molarity of [OH^–] = 4.17 x 10^–10M
Lead Chloride S = 1.59 x 10^–2M; Molarity of Pb²⁺ = 1.59 x 10^–2M
Molarity of Cl^– = 3.18 x 10^–2M; Mercurous Iodide S = 2.24 x 10^–10M;
Molarity of Hg₂²⁺ = 2.24 x 10^–10M and molarity of I^– = 4.48 x 10^–10M

7.68 Silver chromate is more soluble and the ratio of their molarities = 91.9

7.69 No precipitate

7.70 Silver benzoate is 3.317 times more soluble at lower pH

7.71 The highest molarity for the solution is 2.5 x 10^–9M

7.72 2.46 litre of water

7.73 Precipitation will take place in cadmium chloride solution

I. Multiple Choice Questions (Type-I)

1. We know that the relationship between K_c and K_p is

K_p = K_c(RT)^Δn

What would be the value of Δn for the reaction

NH4Cl ( s ) = NH3 ( g ) + HCl ( g )

(i) 1

(ii) 0.5

(iii) 1.5

(iv) 2

2. For the reaction H2 ( g ) + I2 ( g ) = 2HI ( g ) , the standard free energy is ΔG^Θ > 0. The equilibrium constant (K ) would be __________.

(i) K = 0

(ii) K > 1

(iii) K = 1

(iv) K < 1

3. Which of the following is not a general characteristic of equilibria involving

physical processes?

(i) Equilibrium is possible only in a closed system at a given temperature.

(ii) All measurable properties of the system remain constant.

(iii) All the physical processes stop at equilibrium.

(iv) The opposing processes occur at the same rate and there is dynamic but stable condition.

4. PCl₅, PCl₃ and Cl₂ are at equilibrium at 500K in a closed container and their concentrations are 0.8 × 10^–3 mol L^–1, 1.2 × 10^–3 mol L^–1 and 1.2 × 10^–3 mol L^–1 respectively. The value of Kc for the reaction

PCl5 ( g ) = PCl3 ( g ) + Cl2

will be

(i) 1.8 × 10³ mol L^–1

(ii) 1.8 × 10^–3

(iii) 1.8 × 10^–3 L mol^–1

(iv) 0.55 × 10⁴

5. Which of the following statements is incorrect?

(i) In equilibrium mixture of ice and water kept in perfectly insulated flask mass of ice and water does not change with time.

(ii) The intensity of red colour increases when oxalic acid is added to a solution containing iron (III) nitrate and potassium thiocyanate.

(iii) On addition of catalyst the equilibrium constant value is not affected.

(iv) Equilibrium constant for a reaction with negative ΔH value decreases

as the temperature increases.

6. When hydrochloric acid is added to cobalt nitrate solution at room temperature, the following reaction takes place and the reaction mixture becomes blue. On cooling the mixture it becomes pink. On the basis of this information mark the correct answer.

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(i) ΔH > 0 for the reaction

(ii) ΔH < 0 for the reaction

(iii) ΔH = 0 for the reaction

(iv) The sign of ΔH cannot be predicted on the basis of this information.

7. The pH of neutral water at 25°C is 7.0. As the temperature increases, ionisation

of water increases, however, the concentration of H⁺ ions and OH^– ions are equal. What will be the pH of pure water at 60°C?

(i) Equal to 7.0

(ii) Greater than 7.0

(iii) Less than 7.0

(iv) Equal to zero

8. The ionisation constant of an acid, K_a, is the measure of strength of an acid. The K_a values of acetic acid, hypochlorous acid and formic acid are 1.74 × 10^–5, 3.0 × 10^–8 and 1.8 × 10^–4 respectively. Which of the following orders of pH of 0.1 mol dm^–3 solutions of these acids is correct?

(i) acetic acid > hypochlorous acid > formic acid

(ii) hypochlorous acid > acetic acid > formic acid

(iii) formic acid > hypochlorous acid > acetic acid

(iv) formic acid > acetic acid > hypochlorous acid

9. K_a₁ , K_a₂ and K_a₃ are the respective ionisation constants for the following reactions.

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The correct relationship between K_a₁ , K_a₂ and K_a₃ is

(i) K_a₃ = K_a₁ × K_a₂

(ii) K_a₃ = K_a₁ + K_a₂

(iii) K_a₃ = K_a₁ – K_a₂

(iv) K_a₃ = K_a₁ / K_a₂

10. Acidity of BF₃ can be explained on the basis of which of the following concepts?

(i) Arrhenius concept

(ii) Bronsted Lowry concept

(iii) Lewis concept

(iv) Bronsted Lowry as well as Lewis concept.

11. Which of the following will produce a buffer solution when mixed in equal

volumes?

(i) 0.1 mol dm^–3 NH₄OH and 0.1 mol dm^–3 HCl

(ii) 0.05 mol dm^–3 NH₄OH and 0.1 mol dm^–3 HCl

(iii) 0.1 mol dm^–3 NH₄OH and 0.05 mol dm^–3 HCl

(iv) 0.1 mol dm^–3 CH₄COONa and 0.1 mol dm^–3 NaOH

12. In which of the following solvents is silver chloride most soluble?

(i) 0.1 mol dm^–3 AgNO₃ solution

(ii) 0.1 mol dm^–3 HCl solution

(iii) H₂O

(iv) Aqueous ammonia

13. What will be the value of pH of 0.01 mol dm^–3 CH₃COOH (K_a = 1.74 × 10^–5 )?

(i) 3.4

(ii) 3.6

(iii) 3.9

(iv) 3.0

14. Ka for CH₃COOH is 1.8 × 10^–5 and K b for NH₄OH is 1.8 × 10^–5 . The pH of ammonium acetate will be

(i) 7.005

(ii) 4.75

(iii) 7.0

(iv) Between 6 and 7

15. Which of the following options will be correct for the stage of half completion

of the reaction A = B

(i) ΔG^Θ = 0

(ii) ΔG^Θ > 0

(iii) ΔG^Θ < 0

(iv) ΔG^Θ = –RT ln2

16. On increasing the pressure, in which direction will the gas phase reaction

proceed to re-establish equilibrium, is predicted by applying the Le Chatelier’s

principle. Consider the reaction.

N2 ( g ) + 3H2 ( g ) = 2NH3 ( g )

Which of the following is correct, if the total pressure at which the equilibrium

is established, is increased without changing the temperature?

(i) K will remain same

(ii) K will decrease

(iii) K will increase

(iv) K will increase initially and decrease when pressure is very high

17. What will be the correct order of vapour pressure of water, acetone and ether

at 30°C. Given that among these compounds, water has maximum boiling point and ether has minimum boiling point?

(i) Water < ether < acetone

(ii) Water < acetone < ether

(iii) Ether < acetone < water

(iv) Acetone < ether < water

18. At 500 K, equilibrium constant, K_c , for the following reaction is 5.

🙂

(iv) The equilibrium will remain unaffected in all the three cases.

II. Multiple Choice Questions (Type-II)

In the following questions two or more options may be correct.

20. For the reaction N2O4 ( g ) = 2 NO2 ( g ), the value of K is 50 at 400 K and 1700 at 500 K. Which of the following options is correct?

(i) The reaction is endothermic

(ii) The reaction is exothermic

(iii) If NO₂ (g) and N₂O₄ (g) are mixed at 400 K at partial pressures 20 bar and 2 bar respectively, more N2O4 (g) will be formed.

(iv) The entropy of the system increases.

21. At a particular temperature and atmospheric pressure, the solid and liquid

phases of a pure substance can exist in equilibrium. Which of the following term defines this temperature?

(i) Normal melting point

(ii) Equilibrium temperature

(iii) Boiling point

(iv) Freezing point

III. Short Answer Type

22. The ionisation of hydrochloric in water is given below:

HCl ( aq ) + H2O ( l ) = H3O+ ( aq ) + Cl- ( aq )

Label two conjugate acid-base pairs in this ionisation.

23. The aqueous solution of sugar does not conduct electricity. However, when

sodium chloride is added to water, it conducts electricity. How will you explain

this statement on the basis of ionisation and how is it affected by concentration

of sodium chloride?

24. BF₃ does not have proton but still acts as an acid and reacts with NH₃. Why is it so? What type of bond is formed between the two?

25. Ionisation constant of a weak base MOH, is given by the expression

🙂

Values of ionisation constant of some weak bases at a particular temperature are given below:

Base Dimethylamine Urea Pyridine Ammonia

K b 5.4 × 10–4 1.3 × 10–14 1.77 × 10–9 1.77 × 10–5

Arrange the bases in decreasing order of the extent of their ionisation at equilibrium. Which of the above base is the strongest?

26. Conjugate acid of a weak base is always stronger. What will be the decreasing

order of basic strength of the following conjugate bases?

OH^–, RO^– , CH₃COO^– , Cl^–

27. Arrange the following in increasing order of pH.

KNO₃(aq), CH₃COONa (aq), NH₃Cl (aq), C₆H₅COONH₄(aq)

28. The value of K_c for the reaction 2HI ( g ) = H2 ( g ) + I2 ( g ) is 1 × 10^–4

At a given time, the composition of reaction mixture is

[HI] = 2 × 10^–5 mol, [H₂] = 1 × 10^–5 mol and [I₂] = 1 × 10^–5 mol

In which direction will the reaction proceed?

29. On the basis of the equation pH = –log [H⁺], the pH of 10^–8 mol dm^–3 solution of HCl should be 8. However, it is observed to be less than 7.0. Explain the reason.

30. pH of a solution of a strong acid is 5.0. What will be the pH of the solution

obtained after diluting the given solution a 100 times?

31. A sparingly soluble salt gets precipitated only when the product of

concentration of its ions in the solution (Q_sp) becomes greater than its solubility product. If the solubility of BaSO4 in water is 8 × 10^–4 mol dm^–3. Calculate its solubility in 0.01 mol dm^–3 of H₂SO₄.

32. pH of 0.08 mol dm^–3 HOCl solution is 2.85. Calculate its ionisation constant.

33. Calculate the pH of a solution formed by mixing equal volumes of two solutions

A and B of a strong acid having pH = 6 and pH = 4 respectively.

34. The solubility product of Al (OH)₃ is 2.7 × 10^–11. Calculate its solubility in gL^–1 and also find out pH of this solution. (Atomic mass of Al = 27 u).

35. Calculate the volume of water required to dissolve 0.1 g lead (II) chloride to

get a saturated solution. (K_sp of PbCl₂ = 3.2 × 10^–8 , atomic mass of Pb = 207 u).

36. A reaction between ammonia and boron trifluoride is given below:

: NH₃ + BF₃ → H₃N:BF₃

Identify the acid and base in this reaction. Which theory explains it? What is

the hybridisation of B and N in the reactants?

37. Following data is given for the reaction: CaCO₃ (s) &rightarrow; CaO (s) + CO₂(g)

Δ_fH^Θ[CaO(s)] = – 635.1 kJ mol^–1

Δ_fH^Θ[CO₂(g)] = – 393.5 kJ mol^–1

Δ_fH^Θ[CaCO₃(s)] = – 1206.9 kJ mol^–1

Predict the effect of temperature on the equilibrium constant of the above reaction.

IV. Matching Type

38. Match the following equilibria with the corresponding condition

🙂

Equilibrium constant K_c = [NH₃]²/[N₂][H₂]³

Some reactions are written below in Column I and their equilibrium constants in terms of Kc are written in Column II. Match the following reactions with the

corresponding equilibrium constant

🙂

40. Match standard free energy of the reaction with the corresponding equilibrium constant

(i)	ΔG^Θ > 0	(a)	K > 1
(ii)	ΔG^Θ < 0	(b)	K = 1
(iii)	ΔG^Θ = 0	(c)	K = 0
		(d)	K < 1

41. Match the following species with the corresponding conjugate acid

	Species		Conjugate acid
(i)	NH₃	(a)	CO₃^2–
(ii)	HCO₃^–	(b)	NH₄⁺
(iii)	H₂O	(c)	H₃O⁺
(iv)	HSO₄^–	(d)	H₂SO₄
		(e)	H₂CO₃

42. Match the following graphical variation with their description

🙂

43. Match Column (I) with Column (II).

	Column I		Column II
(i)	Equilibrium	(a)	ΔG > 0, K < 1
(ii)	Spontaneous reaction	(b)	ΔG = 0
(iii)	Non spontaneous reaction	(c)	ΔG^Θ = 0
		(d)	ΔG < 0, K > 1

V. Assertion and Reason Type

In the following questions a statement of Assertion (A) followed by a statement

of Reason (R) is given. Choose the correct option out of the choices given

below each question.

44. Assertion (A) : Increasing order of acidity of hydrogen halides is HF < HCl < HBr < HI

Reason (R) : While comparing acids formed by the elements belonging to the subsame group of periodic table, H–A bond strength is a more important factor in determining acidity of an acid than the polar nature of the bond.

(i) Both A and R are true and R is the correct explanation of A.

(ii) Both A and R are true but R is not the correct explanation of A.

(iii) A is true but R is false.

(iv) Both A and R are false.

45. Assertion (A) : A solution containing a mixture of acetic acid and sodium

acetate maintains a constant value of pH on addition of small amounts of acid or alkali.

Reason (R) : A solution containing a mixture of acetic acid and sodium acetate acts as a buffer solution around pH 4.75.

(i) Both A and R are true and R is correct explanation of A.

(ii) Both A and R are true but R is not the correct explanation of A.

(iii) A is true but R is false.

(iv) Both A and R are false.

46. Assertion (A): The ionisation of hydrogen sulphide in water is low in the

presence of hydrochloric acid.

Reason (R) : Hydrogen sulphide is a weak acid.

(i) Both A and R are true and R is correct explanation of A.

(ii) Both A and R are true but R is not correct explanation of A.

(iii) A is true but R is false

(iv) Both A and R are false

47. Assertion (A): For any chemical reaction at a particular temperature, the equilibrium constant is fixed and is a characteristic property.

Reason (R) : Equilibrium constant is independent of temperature.

(i) Both A and R are true and R is correct explanation of A.

(ii) Both A and R are true but R is not correct explanation of A.

(iii) A is true but R is false.

(iv) Both A and R are false.

48. Assertion (A) : Aqueous solution of ammonium carbonate is basic.

Reason (R) : Acidic/basic nature of a salt solution of a salt of weak acid and weak base depends on K_a and K_b value of the acid and the base forming it.

(i) Both A and R are true and R is correct explanation of A.

(ii) Both A and R are true but R is not correct explanation of A.

(iii) A is true but R is false.

(iv) Both A and R are false.

49. Assertion (A): An aqueous solution of ammonium acetate can act as a buffer.

Reason (R) : Acetic acid is a weak acid and NH₄OH is a weak base.

(i) Both A and R are true and R is correct explanation of A.

(ii) Both A and R are true but R is not correct explanation of A.

(iii) A is false but R is true.

(iv) Both A and R are false.

50. Assertion (A): In the dissociation of PCl₅ at constant pressure and temperature addition of helium at equilibrium increases the dissociation of PCl₅.

Reason (R) : Helium removes Cl₂ from the field of action.

(i) Both A and R are true and R is correct explanation of A.

(ii) Both A and R are true but R is not correct explanation of A.

(iii) A is true but R is false.

(iv) Both A and R are false.

VI. Long Answer Type

51. How can you predict the following stages of a reaction by comparing the value of K_c and Q_c?

(i) Net reaction proceeds in the forward direction.

(ii) Net reaction proceeds in the backward direction.

(iii) No net reaction occurs.

52. On the basis of Le Chatelier principle explain how temperature and pressure can be adjusted to increase the yield of ammonia in the following reaction.

N2 ( g ) + 3H2 ( g ) = 2NH3 ( g ) ΔH = -92.38 kJ per mole

What will be the effect of addition of argon to the above reaction mixture at constant volume?

53. A sparingly soluble salt having general formula p qx A_x^p+B_y^q− and molar solubility S is in equilibrium with its saturated solution. Derive a relationship between the solubility and solubility product for such salt.

54. Write a relation between ΔG and Q and define the meaning of each term and answer the following :

(a) Why a reaction proceeds forward when Q < K and no net reaction occurs when Q = K.

(b) Explain the effect of increase in pressure in terms of reaction quotient Q.for the reaction :

CO ( g ) + 3H2 ( g ) = CH4 ( g ) + H2O ( g )

ANSWERS

I. Multiple Choice Questions (Type-I)

1. (iv) 2. (iv) 3. (iii) 4. (ii) 5. (ii) 6. (i) 7. (iii) 8. (iv) 9. (i) 10. (iii) 11. (iii) 12. (iv) 13. (i) 14. (iii)
15. (i) ΔG^Θ = 0
Justification : ΔG^Θ = – RT lnK
At the stage of half completion of reaction [A] = [B], Therefore, K = 1.
Thus, ΔG^Θ = 0

16. (i), Justification: According to Le-Chatelier’s principle, at constant temperature, the equilibrium composition will change but K will remain same.
17. (ii) 18. (i) 19. (iv)

II. Multiple Choice Questions (Type-II)

20. (i), (iii) and (iv)
Justification :
(i) K increases with increase in temperature.
(iii) Q > K, Therefore, reaction proceeds in the backward direction.
(iv) Δ n > 0, Therefore, Δ S > 0.

21. (i) and (iv)

III. Short Answer Type

22.
HCl           Cl^–
acid         conjugate base
H₂O           H₃O⁺
base         conjugate acid

23.
• Sugar does not ionise in water but NaCl ionises completely in water and produces Na⁺ and Cl^– ions.
• Conductance increases with increase in concentration of salt due to release of more ions.

24. BF₃ acts as a Lewis acid as it is electron deficient compound and coordinate bond is formed as given below :
H₃N : → BF₃

25. • Order of extent of ionisation at equilibrium is as follows :
Dimethylamine > Ammonia > Pyridine > Urea
• Since dimethylamine will ionise to the maximum extent it is the strongest base out of the four given bases.

26. RO^– > OH^– > CH₃COO^– > Cl^–
27. NH₄Cl < C₆H₅COONH₄ < KNO₃ < CH₃COONa

28. At a given time the reaction quotient Q for the reaction will be given by the expression.

Q = [H₂][I₂]/[HI]²
= 1 × 10^–5 × 1 × 10^–5/(2 x 10^–5)² = 1/4
= 0.25 = 2.5 × 10^–1

As the value of reaction quotient is greater than the value of Kc i.e. 1× 10^–4 the reaction will proceed in the reverse direction.

29. Concentration of 10^–8 mol dm^–3 indicates that the solution is very dilute. Hence, the contribution of H₃O⁺ concentration from water is significant and should also be included for the calculation of pH.

30. (i) pH = 5
[H⁺] = 10^–5 mol L^–1
On 100 times dilution
[H⁺] = 10^–7 mol L^–1
On calculating the pH using the equation pH = – log [H⁺], value of pH comes out to be 7. It is not possible. This indicates that solution is very
dilute. Hence,
Total hydrogen ion concentration = [H⁺]
= [Contribution of H₃O⁺ ion concentration of acid ] + [ Contribution of H₃O⁺ ion concentration of water ]
= 10^–7 + 10^–7.

pH = 2 × 10^–7 = 7 – log 2 = 7 – 0.3010 = 6.6990

31.

🙂

K_sp for BaSO₄ in water = [Ba²⁺] [SO₄^2–] = (S) (S) = S²
But S = 8 × 10–4 mol dm–3
∴ K_sp = (8×10^–4)2 = 64 × 10^–8 ………………………………………………….. (1)
The expression for K_sp in the presence of sulphuric acid will be as follows :
K_sp = (S) (S + 0.01) ………………………………………………………………………………………………. (2)
Since value of K_sp will not change in the presence of sulphuric acid, therefore from (1) and (2)
(S) (S + 0.01) = 64 × 10^–8
⇒ S² + 0.01 S = 64 × 10^–8
⇒ S² + 0.01 S – 64 × 10^–8 = 0

🙂

32. pH of HOCl = 2.85
But, – pH = log [H⁺]
∴ – 2.85 = log [H⁺]

🙂

🙂
⇒ [H⁺] = 1.413 × 10^–3

For weak mono basic acid [H⁺] = √(Ka X C )
⇒ K_a = [H⁺]²/C = (1.413 x 10^–3)²/0.08
= 24.957 × 10^–6 = 2.4957 × 10^–5

33.
pH of Solution A = 6
Therefore, concentration of [H+] ion in solution A = 10^–6 mol L^–1
pH of Solution B = 4
Therefore, Concentration of [H⁺] ion concentration of solution B = 10^–4 mol L^–1
On mixing one litre of each solution, total volume = 1L + 1L = 2L
Amount of H⁺ ions in 1L of Solution A= Concentration × volume V
= 10^–6 mol × 1L
Amount of H⁺ ions in 1L of solution B = 10^–4 mol × 1L
∴ Total amount of H⁺ ions in the solution formed by mixing solutions A and B is (10^–6 mol + 10^–4 mol)
This amount is present in 2L solution.
∴ Total [H⁺] = 10^–4(1 + 0.01)/2 = 1 .01 x 10^–4/2 mol L^–1 = 1 .01 x 10^–4/2 mol L^–1
= 0.5 × 10^–4 mol L^–1
= 5 × 10^–5 mol L^–1

pH = – log [H⁺] = – log (5 × 10^–5)
= – [log 5 + (– 5 log 10)]
= – log 5 + 5
= 5 – log 5
= 5 – 0.6990
= 4.3010 = 4.3

34. Let S be the solubility of Al(OH)₃.

🙂

K_sp = [Al³⁺] [OH^–]³ = (S) (3S)³ = 27 S⁴
S⁴ = K_sp/27 = 27 × 10^–11/27 x 10 = 1 × 10^–12
S = 1× 10^–3 mol L^–1

(i) Solubility of Al(OH)₃
Molar mass of Al (OH)₃ is 78 g. Therefore,
Solubility of Al (OH)₃ in g L^–1 = 1 × 10^–3 × 78 g L^–1 = 78 × 10^–3 g L^–1 = 7.8 × 10^–2 g L^–1

(ii) pH of the solution
S = 1×10^–3 mol L^–1
[OH^– ] = 3S = 3×1×10^–3 = 3 × 10^–3
pOH = 3 – log 3
pH = 14 – pOH = 11 + log 3 = 11.4771

35. K_sp of PbCl₂ = 3.2 × 10^–8
Let S be the solubility of PbCl₂.

🙂

Molar mass of PbCl₂ = 278
∴ Solubility of PbCl₂ in g L^–1= 2 × 10^–3 × 278 g L^–1
= 556 × 10^–3 g L^–1
= 0.556 g L^–1
To get saturated solution, 0.556 g of PbCl₂ is dissolved in 1 L water.
0.1 g PbCl₂ is dissolved in 0 .1/0 .556 L = 0.1798 L water.

To make a saturated solution, dissolution of 0.1 g PbCl₂ in 0.1798 L ≈ 0.2 L of water will be required.

37. Δ_rH^Θ = Δ_fH^Θ[CaO(s)] + Δ_fH^Θ [CO₂(g)] – Δ_fH^Θ [CaCO₃(s)]
∴ Δ_rH^Θ = 178.3 kJ mol^–1

The reaction is endothermic. Hence, according to Le-Chatelier’s principle, reaction will proceed in forward direction on increasing temperature.

IV. Matching Type

38. (i) → (b) (ii) → (d) (iii) → (c) (iv) → (a)
39. (i) → (d) (ii) → (c) (iii) → (b)
40. (i) → (d) (ii) → (a) (iii) → (b)
41. (i) → (b) (ii) → (e) (iii) → (c) (iv) → (d)
42. (i) → (c) (ii) → (a) (iii) → (b)
43. (i) → (b) and (c) (ii) → (d) (iii) → (a)

V. Assertion and Reason Type

44. (i) 45. (i) 46. (ii) 47.(iii) 48. (i) 49. (iii) 50. (iv)

VI. Long Answer Type

51. (i) Q_c < K_c
(ii) Q_c > K_c
(iii) Q_c = K_c

where, Q_c is reaction quotient in terms of concentration and K_c is equilibrium constant.

53.

🙂

🙂
S moles of A_xB_y dissolve to give xS moles of A^p+ and y S moles of B^q–.]

54. ΔG = ΔG^Θ + RTlnQ
ΔG^Θ = Change in free energy as the reaction proceeds
ΔG = Standard free energy change
Q = Reaction quotient
R = Gas constant
T = Absolute temperature
Since ΔG^Θ = – RT lnK
∴ ΔG = – RT lnK + RT lnQ = RT ln Q/K

If Q < K, ΔG will be negative. Reaction proceeds in the forward direction.
If Q = K, ΔG = 0, no net reaction.
[Hint: Next relate Q with concentration of CO, H₂, CH₄ and H₂O in view of reduced volume (increased pressure). Show that Q < K and hence the reaction proceeds in forward direction.]

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🙂

4 ) How can I get all your lectures ?

Ans : – Apart from my lectures there are approx 700 GB of PCM ( Phy, Chem, Math ) lectures. It takes approx 3 years of continuous download from scattered sources. I have ( 20,000 )Thousands of these. You can take ALL of them from me in an external 1 TB hard disk, instead of spending so much money and time again for downloading. These cover ( by Various Professors ) everything of Chemistry, Physics, Maths… Lot of this is from outside India … as foreigners have much wider heart than Indians ( as most of GNU / open source software have been developed by Non-Indians ). I observed the gaps in these videos, and thus I am solving IIT, APhO, Roorkey, IPhO Numericals. Videos made by me along with these videos gives a complete preparation.

Send me a mail at mokshya@gmail.com to contact me.

search for videos in http://www.skmclasses.kinja.com
You will get most videos. I say most because I do not upload all videos that I make. I have many more videos which are not in the net.

🙂

5 ) How do you get benefited out of this ?

Ans :- If anyone learns we all will have better people in this world. I will have better “ YOU “.
🙂

6 ) Why do you call yourself a Zookeeper ?

Ans :- This is very nicely explained at https://zookeepersblog.wordpress.com/z00keeper-why-do-i-call-myself-a-zoookeeper/

🙂

7 ) Where do you stay ?

Ans :- Presently I am in Bangalore.

🙂

8 ) If I need videos in a few topics can you make them for me ?

Ans :- We actively answers doubts at doubtpoint.
see http://skmclasses.weebly.com/doubtpoint.html
In case you appreciate our time and efforts involved in answering complicated Questions, then get Quality answers at doubtpoint.

🙂

9 ) Why did you write an article saying there are No Poor students ?

Ans :- There are lots of NGOs and others working for rural / poor children education at lower classes. While very less effort is on for std 9 till 12. Also see the answer in question number ( 3 ) above. In more than 2 decades of teaching I never met a Poor child who was seriously interested in ( higher ) studies. As I have a mind / thinking of a ” Physicist “, I go by ” Experimental Observation “.

It is not about what is being said about poor in media / TV etc, or ” what it should be ” ( ? ) …. It is about what I see happening. Also to add ( confuse ? you more )…. You must be knowing that in several states over many years now girl students have better ( by marks as well as by pass percentage ) result in std 10 / Board Exams….. well but NEVER a girl student came FIRST in IIT JEE … why ? [ The best rank by a Girl student is mostly in 2 digits, very rarely in single digit ] ????? So ????

🙂

10 ) How much do I have to study to make it to IIT ?

Ans :- My experience of Teaching for IIT JEE since 1989, tells me, Total 200 hours per subject ( PCM ) is sufficient. If you see my Maths and Physics videos, each subject is more than 200 hours. So if someone sees all the videos diligently, takes notes and remembers, …… Done.

🙂

11 ) What is EAMCET ?

Ans :- Engineering Agriculture and Medicine Common Entrance Test is conducted by JNT University Hyderabad on behalf of APSCHE. This examination is the gateway for entry into various professional courses offered in Government/Private Colleges in Andhra Pradesh.

12 ) In your videos are you covering other Exams apart from IIT ?

Ans : – Yes. See many videos made by solving problems of MPPET, Rajasthan / J&K CET, UPSEAT ( UPES Engineering Aptitude Test ), MHCET, BCECE ( Bihar Combined Entrance Competitive Examination Board ), WB JEE etc

🙂

13 ) What is SCRA ?

Ans : – Special Class Railway Apprentice (SCRA) exam is conducted by Union Public Service Commission (UPSC) board, for about 10 seats.That translates into an astonishing ratio of 1 selection per 10,000 applicants. The SCRA scheme was started in 1927 by the British, to select a handful of most intelligent Indians to assist them in their Railway Operations, after training at their Railway’s largest workshop, i.e. Jamalpur Workshop, and for one year in United Kingdom. The selected candidates were required to appear in the Mechanical Engineering Degree Examination held by Engineering Council (London).

Thanks for your time. To become my friend in google+ ( search me as mokshya@gmail.com and send friend request )

Read http://edge.org/responses/what-scientific-concept-would-improve-everybodys-cognitive-toolkit
🙂
The following video is a must see for full CO2 cycle, plates of Earth, Geological activities, stability of weather

🙂
Article in Nature says CO2 increase is good for the trees
http://thegwpf.org/science-news/6086-co2-is-greening-the-planet-savannahs-soon-to-be-covered-by-forests.html
🙂
http://climaterealists.com/index.php?id=9752

BBC documentary Crescent and Cross shows the 1000 years of fight between Christians and Muslims. Millions have been killed in the name of Religion. To decided whose GOD is better, and which GOD to follow. The fight continues.
–

Summary of Women

🙂
The Virus of Faith

🙂
The God delusion

🙂
cassiopeia facts about evolution

–

Intermediate Fossil records shown and explained nicely Fossils, Genes, and Embryos http://www.youtube.com/watch?v=fdpMrE7BdHQ

The Rise Of Narcissism In Women

🙂
13 type of women whom you should never court
http://timesofindia.indiatimes.com/life-style/relationships/man-woman/13-Women-you-should-never-court/articleshow/14637014.cms

🙂
Media teaching Misandry in India

http://www.youtube.com/watch?v=-M2txSbOPIo

Summary of problems with women
http://problemwithwomentoday.blogspot.in/2009/12/problem-with-women-today-what-in-hell.html

🙂
Eyeopener men ? women only exists
http://www.youtube.com/watch?v=6ZAuqkqxk9A

🙂
Each of you is an Activist in some way or other. You are trying to propagate those thoughts, ideas that you feel concerned / excited about.
–

Did you analyze your effectiveness ?

Culturomics can help you

😀
see how biased women are. Experimental proof. Women are happy when they see another woman is beating a man ( see how women misbehave with men )

🙂
http://www.youtube.com/watch?v=LlFAd4YdQks

–

see detailed statistics at

An eye opener in Misandry

My sincere advice would be to be EXTREMELY careful ( and preferably away ) of girls. As girls age; statistically certain behavior in them has been observed. Most Male can NOT manage those behaviors… Domestic violence, divorce etc are rising very fast. Almost in all cases boys / males are HUGE loosers. Be extremely choosy ( and think from several angles ) before even talking to a girl.
🙂
https://zookeepersblog.wordpress.com/save-the-male/

🙂

How women manipulate men
http://www.angryharry.com/esWomenManipulateMen.htm

Gender Biased Laws in India
https://zookeepersblog.wordpress.com/biased-laws/

🙂

Only men are victimised

Men are BETTER than women
http://www.menarebetterthanwomen.com/
🙂

see http://www.youtube.com/watch?&v=T0xoKiH8JJM#!
🙂

Male Psychology http://www.youtube.com/watch?v=uwxgavf2xWE

Women are more violent than men
http://www.independent.co.uk/news/uk/home-news/women-are-more-violent-says-study-622388.html

🙂

In the year 2010, 168 men ended their lives everyday ( on average ). More husbands committed suicide than wives.
🙂

http://www.rediff.com/news/report/ncrb-stats-show-more-married-men-committing-suicide/20111028.htm

It is EXTREMELY unfortunate that media projects men as fools, women as superiors, Husbands as servants, and replaceable morons. In ad after ad worldwide from so many companies, similar msg to disintegrate the world is being bombarded. It is highly unacceptable misandry

🙂

It is NOT at all funny that media shows violence against MEN. Some advertisers are trying to create a new ” Socially acceptable culture ” of slapping Men ( by modern city women ). We ( all men ) take objection to these advertisements.
We oppose this Misandry bad culture. Please share to increase awareness against Men bashing

🙂

Think what are you doing … why are you doing ?

Every Man must know this …

🙂
Manginas, White Knights, & Other Chivalrous Dogs

!
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key words

_________________________________________________________________________
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receptors reduction regioselectivity retro reaction rhodium ring closure ring contraction ring expansion ring opening ruthenium samarium scandium Schiff bases selenium self-assembly silicon sodium solid-phase synthesis solvent effects spectroscopy sphingolipids spiro compounds stereoselective synthesis stereoselectivity steric hindrance steroids Stille reaction substituent effects sulfates sulfonamides sulfones sulfoxides sulfur supported catalysis supramolecular tandem reaction tautomerism terpenoids thioacetals thiols tin titanium total synthesis transesterification transition metals transition states tungsten Umpolung vinylidene complexes vitamins Wacker reaction Wittig reaction ylides zeolites zinc BRST Quantization Effective field theories Field Theories Higher Dimensions Field Theories Lower Dimensions Large Extra Dimensions Lattice Quantum Field Theory Nonperturbative Effects Renormalization Group Renormalization Regularization skmclasses.weebly.com Renormalons Sigma Models 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SKMClasses.weebly.com remove IITJEE SKMClasses.weebly.com electron Chemistry energy second ionisation energy energy IITJEE SKMClasses.weebly.com one electron IITJEE SKMClasses.weebly.com IITJEE SKMClasses.weebly.com ion one mole gaseous 1+ ions IITJEE SKMClasses.weebly.com one mole gaseous 2+ ions isotopes Atoms skmclasses.weebly.com element IITJEE SKMClasses.weebly.com different numbers neutrons different masses le Chatelier’s principle system dynamic equilibrium subjected change position equilibrium will shift minimise change limiting reagent substance chemical reaction IITJEE SKMClasses.weebly.com runs out first lone pair outer shell pair electrons IITJEE SKMClasses.weebly.com involved chemical bonding mass nucleon number particles protons aneutrons) nucleus mechanism sequence steps showing path taken electrons reaction metallic bond electrostatic attraction IITJEE SKMClasses.weebly.com positive metal ions adelocalised electrons molar mass substance units molar mass IITJEE SKMClasses.weebly.com molar volume IITJEE SKMClasses.weebly.com mole gas. units molar volume IITJEE SKMClasses.weebly.com dm room temperature skmclasses.weebly.com pressure molar volume approximately 24.0 substance containing IITJEE skmclasses.weebly.com many particles thereIITJEE SKMClasses.weebly.com carbon atoms exactly 12 g carbon isotope molecular formula number atoms IITJEE SKMClasses.weebly.com element molecule molecular ion M positive ion formed mass spectrometry IITJEE SKMClasses.weebly.com molecule loses electron molecule small group atoms held together covalent bonds monomer small molecule IITJEE SKMClasses.weebly.com combines IITJEE SKMClasses.weebly.com monomers polymer nomenclature system naming compounds nucleophile atom group atoms attracted electron deficient centre atom donates pair electrons covalent bond nucleophilic substitution type substitution reaction IITJEE SKMClasses.weebly.com nucleophile attracted electron deficient centre atom, IITJEE SKMClasses.weebly.com donates pair electrons IITJEE SKMClasses.weebly.com new covalent bond oxidation Loss electrons IITJEE SKMClasses.weebly.com increase oxidation number oxidation number measure number electrons IITJEE SKMClasses.weebly.com IITJEE SKMClasses.weebly.com atom uses bond IITJEE SKMClasses.weebly.com atoms another element. Oxidation numbers IITJEE SKMClasses.weebly.com derive d rules oxidising agent reagent IITJEE SKMClasses.weebly.com oxidises (takes electrons from) another species percentage yield period horizontal row elements Periodic Table Elements show trends properties across period periodicity regular periodic variation properties elements IITJEE SKMClasses.weebly.com atomic number position Periodic Table permanent dipole small charge difference across bond resulting IITJEE SKMClasses.weebly.com difference electronegativities bonded atoms permanent dipole dipole force attractive force IITJEE SKMClasses.weebly.com permanent dipoles neighbouring polar molecules pi bond (p bond reactive part double bond formed above skmclasses.weebly.com below plane bonded atoms sideways overlap p orbitalspolar covalent bond bond IITJEE SKMClasses.weebly.com permanent dipole polar molecule molecule IITJEE SKMClasses.weebly.com IITJEE SKMClasses.weebly.com overall dipole skmclasses account dipoles across bonds polymer long molecular chain built monomer units precipitation reaction formation solid solution during chemical reaction Precipitates IITJEE SKMClasses.weebly.com formed IITJEE SKMClasses.weebly.com two aqueous solutions IITJEE SKMClasses.weebly.com mixed together principal quantum number n number representing relative overall energy orbital IITJEE SKMClasses.weebly.com increases distance nucleus sets orbitals IITJEE SKMClasses.weebly.com value IITJEE skmclasses.weebly.com electron shells energy levels propagation two repeated radical substitution IITJEE SKMClasses.weebly.com build up products chain reaction radical species unpaired electron rate reaction change concentration reactant product redox reaction reaction IITJEE SKMClasses.weebly.com reduction skmclasses.weebly.com oxidation take IITJEE SKMClasses.weebly.com reducing agent reagent IITJEE SKMClasses.weebly.com reduces (adds electron to) species reduction Gain electrons decrease oxidation number yield actual amount mol product theoretical amount mol product Chemistry reflux continual boiling skmclasses.weebly.com condensing reaction mixture ensure IITJEE SKMClasses.weebly.com reaction IITJEE SKMClasses.weebly.com without contents flask boiling dry relative atomic mass weighted mean mass atom element compared one twelfth mass IITJEE SKMClasses.weebly.com atom carbon relative formula mass weighted mean mass formula unit compared IITJEE SKMClasses.weebly.com one twelfth mass atom carbon relative isotopic mass mass atom isotope compared IITJEE SKMClasses.weebly.com one twelfth mass atom carbon relative molecular mass weighted mean mass molecule compared twelfth mass atom carbon 12 repeat unit specific arrangement atom s IITJEE SKMClasses.weebly.com occurs structure over over again. Repeat units IITJEE SKMClasses.weebly.com included brackets outside IITJEE SKMClasses.weebly.com symbol n Salt chemical compound formed IITJEE SKMClasses.weebly.com IITJEE SKMClasses.weebly.com acid IITJEE SKMClasses.weebly.com H+ ion acid IITJEE skmclasses.weebly.com been replaced metal ion another positive ion such IITJEE skmclasses.weebly.com ammonium ion, NH saturated hydrocarbon IITJEE SKMClasses.weebly.com single bonds only shell group atomic orbitals IITJEE SKMClasses.weebly.com skmclasses.weebly.com principal quantum number known main energy level simple molecular lattice three dimensional structure molecules, bonded together weak intermolecular forces skeletal formula simplified organic formula, IITJEE SKMClasses.weebly.com hydrogen atoms removed alkyl chains, leaving carbon skeleton skmclasses.weebly.com associated functional groups species particle IITJEE SKMClasses.weebly.com part chemical reaction specific heat capacity, c energy IITJEE SKMClasses.weebly.com raise temperature 1 g substance 1 C spectator ions Ions present part chemical reaction standard conditions pressure 100 kPa 1 atmosphere stated temperature usually 298 K (25 °C), skmclasses.weebly.com concentration 1 mol dm reactions aqueous solutions standard enthalpies enthalpystandard solution solution known concentration Standard solutions normally IITJEE SKMClasses.weebly.com titrations IITJEE SKMClasses.weebly.com determine unknown information another substance Chemistry standard state physical state substance under standard conditions 100 kPa 1 atmosphere) skmclasses.weebly.com 298 K 25 C stereoisomers Compounds skmclasses.weebly.com structural formula IITJEE SKMClasses.weebly.com different arrangement atoms space stoichiometry molar relationship IITJEE SKMClasses.weebly.com relative quantities substances part reaction stratosphere second layer Earth’s atmosphere, containing ‘ozone layer’, about 10 km IITJEE SKMClasses.weebly.com 50 km above Earth’s surface structural formula formula showing minimal detail skmclasses.weebly.com arrangement atoms molecule structural isomers Molecules IITJEE SKMClasses.weebly.com skmclasses.weebly.com molecular formula different structural arrangements atoms subshell group skmclasses.weebly.com type atomic orbitals s, p, d f within shell substitution reaction reaction IITJEE SKMClasses.weebly.com atom group atoms replaced different atom group atoms termination step end radical substitution IITJEE SKMClasses.weebly.com two radicals combine IITJEE SKMClasses.weebly.com molecule thermal decomposition breaking chemical substance IITJEE SKMClasses.weebly.com heat skmclasses least two chemical substances troposphere lowest layer Earth’s atmosphere extending Earth’s surface about 7 km (above poles) about 20 km above tropics unsaturated hydrocarbon hydrocarbon containing carbon carbon multiple bonds van der Waals’ forces Very weak attractive forces IITJEE SKMClasses.weebly.com induced dipoles neighbouring molecules volatility ease IITJEE 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actinium (89) skmclasses.weebly.com lawrencium (103 activated complex – structure IITJEE SKMClasses.weebly.com forms because collisionIITJEE SKMClasses.weebly.com molecules new bondsvIITJEE SKMClasses.weebly.com formed activation energy – minimum energy IITJEE SKMClasses.weebly.com must be inputIITJEE SKMClasses.weebly.com chemical system activity series actual yield addition reaction – within organic chemistry, IITJEE SKMClasses.weebly.com two IITJEE SKMClasses.weebly.com molecules combineIITJEE SKMClasses.weebly.com IITJEE SKMClasses.weebly.com larger aeration mixing air skmclasses liquid solid alkali metals metals Group 1 on periodic table alkaline earth metals – metals Group 2 on periodic table allomer substance IITJEE SKMClasses.weebly.com hIITJEE skmclasses.weebly.comdifferent composition another skmclasses.weebly.comcrystalline structure allotropy elements IITJEE SKMClasses.weebly.com different structures skmclasses.weebly.com therefore different forms IITJEE 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See IITJEE SKMClasses.weebly.com benzene atom – chemical element IITJEE SKMClasses.weebly.com smallest form, skmclasses.weebly.com made up neutrons skmclasses.weebly.comprotons within nucleus skmclasses.weebly.comelectrons circling nucleus atomic mass unit atomic number number representing IITJEE SKMClasses.weebly.com element IITJEE SKMClasses.weebly.com corresponds IITJEE SKMClasses.weebly.com number protons within nucleus atomic orbital region IITJEE SKMClasses.weebly.com electron atom may be found atomic radius average atomic mass Avogadro’s law Avogadro’s number number particles mole substance ( 6.02×10^23 ) barometer deviceIITJEE SKMClasses.weebly.comIITJEE SKMClasses.weebly.com measure pressure atmosphere base substance IITJEE SKMClasses.weebly.com accepts proton skmclasses.weebly.com high pH; common example sodium hydroxide (NaOH biochemistry chemistry organisms boiling phase transition liquid vaporizing boiling point temperature IITJEE SKMClasses.weebly.com substance 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Rao Academy in Bull Temple Road light Standard conditions skmclasses.weebly.com temperature skmclasses.weebly.compressure SATP standardisationIITJEE SKMClasses.weebly.com order compare experimental results (25 °C skmclasses.weebly.com 100.000 kPa state matter matter having homogeneous, macroscopic phase; gas, plasma Ria Institute Technology in Marathahalli liquid solidIITJEE SKM Classes.weebly.com well known increasing concentration sublimation – phase transitionIITJEE SKMClasses.weebly.com solidIITJEE SKMClasses.weebly.com limewater fuel gas subatomic particles – particles IITJEE SKMClasses.weebly.comIITJEE SKMClasses.weebly.com smaller atom; examplesIITJEE SKMClasses.weebly.com protons neutrons skmclasses.weebly.comelectrons substance – material IITJEE SKMClasses.weebly.com definite chemical composition Phase diagram showing triple skmclasses.weebly.comcritical points substance talc mineral representing one on Mohs Scale skmclasses.weebly.comcomposed hydrated magnesium silicate IITJEE SKMClasses.weebly.com chemical formula H2Mg3(SiO3)4 Mg3Si4O10(OH)2 temperature – average energy microscopic motions particles theoretical yield yield theory model describing nature phenomenon thermal conductivity property material Communication skmclasses.weebly.com Careers R.M.V. Extn. 2nd Stage conduct heat (often noted IITJEE skmclasses.weebly.com k thermochemistry study absorption release heat within chemical reaction thermodynamics study effects changing temperature, volume pressure work, heat, skmclasses.weebly.com energy on macroscopic scale I-Bas Consulting Pvt. Ltd., Ulsoor thermodynamic stability IITJEE SKMClasses.weebly.com system IITJEE SKMClasses.weebly.com lowest energy state IITJEE SKMClasses.weebly.com environment equilibrium thermometer device measures average energy system titration – process titrating one solution IITJEE SKMClasses.weebly.com another Cavalier India, Kalyan Nagar called volumetric analysis torr unit measure pressure (1 Torr equivalentIITJEE SKMClasses.weebly.com 133.322 Pa 1.3158×10-3 atm transition metal elements IITJEE SKMClasses.weebly.com incomplete d sub-shells IITJEE SKMClasses.weebly.com may referredIITJEE SKMClasses.weebly.com IITJEE skmclasses.weebly.com d-block elements transuranic element – element IITJEE SKMClasses.weebly.com atomic number greater 92; none transuranic elementsIITJEE SKMClasses.weebly.com stable triple bond – sharing three pairs electrons within covalent bond example N2 triple point temperature skmclasses.weebly.compressure three phasesIITJEE SKMClasses.weebly.com skmclasses.weebly.com Water special National IAS Academy, Raja Rajeshwari Nagar phase diagram Tyndall effect effect light scattering colloidal mixture IITJEE SKMClasses.weebly.com one substance dispersed evenly through another suspended particles UN number four digit codeIITJEE SKMClasses.weebly.comIITJEE SKMClasses.weebly.com note hazardous skmclasses.weebly.com flammable substances uncertainty characteristic IITJEE SKMClasses.weebly.com measurement IITJEE SKMClasses.weebly.com involves estimation any amount cannot be exactly reproducible Uncertainty principle knowing Shaping Lives Education Pvt. Ltd., Rajaji Nagar location particle makes momentum uncertain knowing momentum particle makes location uncertain unit cell smallest repeating unit lattice unit factor statements Manhattan Review, Jaya Nagar convertingIITJEE SKMClasses.weebly.com units universal ideal gas constant proportionality constant ideal gas law (0.08206 L·atm/(K·mol)) valence electron outermost electrons IITJEE SKMClasses.weebly.com atom IITJEE SKMClasses.weebly.comIITJEE SKMClasses.weebly.com located electron shells Valence bond theory theory explaining chemical bonding within molecules discussing valencies number chemical bonds formed IITJEE SKMClasses.weebly.com atom van der Waals force – one forces (attraction/repulsion)IITJEE SKMClasses.weebly.com molecules van ‘t Hoff factor – ratio moles particles solutionIITJEE SKMClasses.weebly.com moles solute dissolved vapor IITJEE SKMClasses.weebly.com substance below critical temperature gas phase vapour pressure – pressure vapour over liquid at equilibrium vaporization phase changeIITJEE SKMClasses.weebly.com liquidIITJEE SKMClasses.weebly.com gas viscosity – resistance liquidIITJEE SKMClasses.weebly.com flow (oil) volt one joule workIITJEE SKMClasses.weebly.com coulomb unit electrical potential transferred voltmeter – instrument IITJEE SKMClasses.weebly.com measures cell potential volumetric analysis Endeavor, Jaya Nagar 5th Block titration water – H2O – chemical substance, major part cells skmclasses.weebly.com Earth, skmclasses.weebly.com covalently bonded wave function function describing electron’s position three-dimensional space worknamount force over distance skmclasses.weebly.com terms joules energy X-ray ionizing, electromagnetic radiation gamma skmclasses.weebly.comUV rays X-ray diffraction – method skmclasses.weebly.com establishing structures crystalline solids using singe wavelength X-rays skmclasses.weebly.com looking diffraction pattern X-ray photoelectron spectroscopy spectroscopic technique IITJEE SKMClasses.weebly.com measure composition material yield amount product produced during chemical reaction zone melting way remove impuritiesIITJEE SKMClasses.weebly.com IITJEE SKMClasses.weebly.com element melting skmclasses.weebly.com slowly travel IITJEE SKMClasses.weebly.com ingot (cast) Zwitterion chemical compound whose net charge zero skmclasses.weebly.comhence electrically neutral IITJEE SKMClasses.weebly.com positive skmclasses.weebly.com negative charges due formal charge, owing partial charges IITJEE SKMClasses.weebly.com constituent atoms acetals acylation addition aggregation alcohols aldehydes aldol reaction alkaloids alkanes alkenation alkene complexes alkenes alkyl halides alkylation alkyne complexes alkynes allenes allylation allyl complexes aluminum amides amination amines amino acids amino alcohols amino aldehydes annulation annulenes antibiotics antifungal agents antisense agents antitumor agents antiviral agents arene complexes arenes arylation arynes asymmetric catalysis asymmetric synthesis atropisomerism autocatalysis azapeptides azasugars azides azo compounds barium benzylation betaines biaryls bicyclic compounds biomimetic synthesis bioorganic biosynthesis boron bromine calixarenes carbanions carbene complexes carbenes carbenoids carbocation carbocycles carbohydrates carbonyl complexes carbonylation carboxylic acids catalysis catenanes cations cavitands chelates chemoselectivity chiral auxiliaries chiral pool chiral resolution chirality chromium chromophores cleavage clusters combinatorial complexes condensation conjugation copper coupling cross-coupling crown compounds cryptands cuprates cyanines cyanohydrins cyclization cycloaddition cyclodextrines cyclopentadienes cyclophanes dehydrogenation dendrimers deoxygenation desulfurization diastereoselectivity diazo compounds diene complexes Diels-Alder reaction dihydroxylation dimerization diols dioxiranes DNA domino reaction drugs electrocyclic reactions electron transfer electrophilic addition electrophilic aromatic substitution elimination enantiomeric resolution enantioselectivity ene reaction enols enones enynes enzymes epoxidation epoxides esterification esters ethers fluorine free radicals fullerenes furans fused-ring systems gas-phase reaction genomics glycolipids glycopeptides glycosidases glycosides glycosylation green chemistry Grignard reaction halides halogenation halogens Heck reaction helical structures heterocycles heterogeneous catalysis Jain International Residential School Jakkasandra Post, Kanakapura Taluk Bangalore high-throughput JSS Public School, HSR Layout No 4/A, 14th Main, 6th Sector HSR Layout, Bangalore screening HIV homogeneous catalysis host-guest systems hydrazones hydrides hydroboration hydrocarbons hydroformylation hydrogen transfer hydrogenation Freedom International School C A # 33, Sector IV HSR Layout, Bangalore hydrolysis hydrosilylation hydrostannation hyperconjugation imides imines indium indoles induction inhibitors insertion iodine ionic liquids iridium iron isomerization The Brigade International School , Brigade Millenium JP Nagar Brigade Millenium, JP Nagar Bangalore ketones kinetic resolution lactams lactones lanthanides Lewis acids ligands lipids lithiation lithium macrocycles magnesium manganese Mannich bases medicinal chemistry metalation metallacycles metallocenes metathesis Michael addition Mitsunobu reaction molecular recognition molybdenum multicomponent reaction nanostructures natural products neighboring-group effects nickel nitriles nitrogen nucleobases nucleophiles nucleophilic addition nucleophilic National Centre For Excellence 154/1, “Victorian Enclave”, 5th Main, Malleshpalya, Bangalore aromatic substitution nucleosides nucleotides olefination oligomerization oligonucleotides oligosaccharides organometallic reagents osmium oxidation oxygen oxygenations ozonolysis palladacycles palladium peptides pericyclic reaction peroxides phase-transfer catalysis phenols pheromones phosphates phosphorus phosphorylation Adugodi Aga Abbas Ali Road Agaram Agrahara Dasara Halli Agrahara Dasarahalli Airport Exit Road Airport Main Road Airport Road Akkipet Ali Askar Road Alur Venkatarao Road Amarjyothi Layout Amruth Nagar Amrutha Halli Ananda Nagar Anandrao Circle Anche Palya Ane Palya Anekal Anjana Nagar Anubhava Nagar APMC Yard Arabic College Arakere Arcot Sreenivasachar Street Ashok Nagar Ashwath Nagar Attibele Attiguppe Austin Town Avala Halli Avenue Road B. Narayanapura Babusahib Palya Bagalagunte Bagalur Balaji Nagar Balepet Banashankari Banashankari 1st Stage Banashankari 2nd Stage Banashankari 3rd Stage Banaswadi Banaswadi Ring Road Bangalore G.P.O Bannerghatta Bannerghatta Road Bapuji Nagar Basappa Circle Basava Nagar Basavanagudi Basaveshwara Nagar Basaveshwara Nagar 2nd Stage Basaveshwara Nagar 3rd Block Basaveshwara Nagar 3rd Stage Basaveshwara Road Bazaar Street Begur BEL Road Bellandur Bellandur Outer Ring Road Bellary Road BEML Layout Benagana Halli Bendre Nagar Benson Town Bharati Nagar Bhattara Halli Bhoopasandra Bhuvaneshwari Nagar Bidadi Bileka Halli Bilekahalli Binny Mill Road Bismillah Nagar Bommana Halli Bommanahalli Kendriya Vidyalaya Malleswaram 18th Cross Malleswaram Bangalore Bommasandra Bommasandra Industrial Area Brigade Road Brindavan Nagar Brookefield Brunton Road BTM 1st Stage BTM 2nd Stage Bull Temple Road Palace Orchards/Sadashivnagar area located north city centre IITJEE SKMClasses.weebly.com property prices higher brackets possibly IITJEE SKMClasses.weebly.com up-market residential area in Bangalore M.G. Road/Brigade Road M.G. Road skmclasses.weebly.comBrigade Road main commercial areas Bangalore. Residential areas nearbyIITJEE SKMClasses.weebly.com Brunton Road Rest House Road, St. Mark’s Road skmclasses.weebly.comLavelle Road Airport Road/Indiranagar eastern suburb, Indiranagar is easily accessible IITJEE city centre skmclasses.weebly.com Airport Koramangala Located south Indiranagar, Koramangala quite favourite IITJEE SKMClasses.weebly.com IT professionals Despite 7 kmsIITJEE SKMClasses.weebly.com city centre, property values Ulsoor scenic man-made lake Ulsoor seen a spurt building activity last few years.IITJEE SKMClasses.weebly.com proximityIITJEE SKMClasses.weebly.com M.G Road jacked up property prices here Jayanagar/J.P. Nagar/Banashankari proximity areas Electronic City main reason skmclasses.weebly.comtheir growth recent past Jayanagar largest colonies Asia skmclasses.weebly.comthese areas popular areas Bangalore. Jayanagara originally namedIITJEE SKMClasses.weebly.com Sri Jayachamarajendra wodeyar last king Mysore. Later Sri Kumaran Children’s Home Survey No 44 – 50, Mallasandra Village Uttarahalli Hobli, Off Kanakapura Main Road, Bangalore skmclasseslocality namedIITJEE SKMClasses.weebly.com current DD kendra is situated known IITJEE skmclasses.weebly.com JC Nagar or Jayachamarajendra Nagar Delhi Public School, North Campus Survey No. 35/A, Sathanur Village Jala Hobli, Bangalore Jayanagar IITJEE SKMClasses.weebly.com literally Victory City Jayanagar IITJEE skmclasses.weebly.com traditionally regarded IITJEE skmclasses.weebly.com southern end Bangalore South End Circle “, wherein six roadsIITJEE SKMClasses.weebly.com different areas meet skmclasses.weebly.com historic Ashoka Pillar mark southern end city bear this fact. newer extensions IITJEE SKMClasses.weebly.com taken away this distinctionIITJEE SKMClasses.weebly.com Jayanagar still remains one IITJEE SKMClasses.weebly.com southern parts city Malleshwaram Basavanagudi Malleshwaram north Bangalore, Basavanagudi south IITJEE SKMClasses.weebly.com areas oldest Bangalore skmclasses.weebly.com residents IITJEE SKMClasses.weebly.com original inhabitants City. Malleswaram PSBB Learning Leadership Academy
# 52, Sahasra Deepika Road, Laxmipura Village, Off Bannerghatta Main Road Bangalore located actually north-west Bangalore derives IITJEE SKMClasses.weebly.com name IITJEE SKMClasses.weebly.com famous Kaadu Malleshwara temple 8th Cross in Malleshwaram, skmclasses.weebly.comGandhibazar/ DVG Road in Basavanagudi IITJEE SKMClasses.weebly.com popular areas in Bangalore skmclasses.weebly.comshopping during festival times. Malleswaram been homeIITJEE SKMClasses.weebly.com several important personalities skmclasses.weebly.cominstitutions. Bangalore’s own Nobel laureate, C.V. Raman, late Veena Doreswamy Iyengar skmclasses.weebly.com M.Chinnaswamy cricket stadium is named, academician M.P.L. Sastry, poet G.P. Rajaratnam skmclasses.weebly.com Dewan Seshadri Iyer institutions IITJEE SKMClasses.weebly.com Canara Union club Konkani-speaking people in 1930 IITJEE SKMClasses.weebly.comIITJEE SKMClasses.weebly.com this day hosts a variety cultural activities Malleswaram Association, hub area’s sporting activity since 1929 skmclasses.weebly.com Chowdaiah Memorial hosting great names music skmclasses.weebly.comtheatre. AccordingIITJEE SKMClasses.weebly.com recent figures available IITJEE SKMClasses.weebly.com Bangalore Development Authority BDA Malleswaram’s net population density is 521 personsIITJEE SKMClasses.weebly.com hectare, Bangalore City Corporation standard is 352IITJEE SKMClasses.weebly.com hectare Sadhashivnagar Sadashivanagar arguably IITJEE SKMClasses.weebly.com elite skmclasses.weebly.comexpensive neighborhood in Bangalore India fashionable among politicians, movie starsIITJEE SKMClasses.weebly.com millionaires afford homes “Beverly Hills Bangalore,” having IITJEE SKMClasses.weebly.com address in Sadashivanagar connotes high level prestige success fame Vijayanagar derivesIITJEE SKMClasses.weebly.com nameIITJEE SKMClasses.weebly.com Vijayanagara empire IITJEE SKMClasses.weebly.com flourished in south India during 15th skmclasses.weebly.com16th centuries.Vijayanag ar East is popularly known IITJEE base skmclasses.weebly.com RPC Layout (Railway Parallel Colony Layout), since this layout is along railway track. IITJEE skmclasses.weebly.com recently renamed Hampi Nagar Hampi capital Vijayanagar Empire Vijayanagar houses a large Public Library, IITJEE SKMClasses.weebly.com is one largest in Karnataka Halasuru Halasuru formerly known IITJEE skmclasses.weebly.com Ulsoor oldest neighbourhoods Indian city Bangalore predominant Tamil speaking population renowned skmclasses.weebly.com numerous temples skmclasses.weebly.comrather narrow streets skmclassesprominant areas CityIITJEE SKMClasses.weebly.com Sanjay Nagar skmclasses.weebly.com RT Nagar, Hebbal, Vyalikaval, Yeshwanthpur, Sriramapura, Rajajinagar, Rajarajeshwarinagar, Chickpet, Chamarajpet, V V Puram, Mavalli, Hanumanthanagar, Padmanabhanagar Hosakerehalli Sarakki, BTM Layout, Domlur, Gandhinagar, Vasanthanagar, Vivek Nagar, Cox Town, Frazer Town Benson Town Bangalore Roads Many roads Bangalore had European names South Parade Road, Albert Victor Road, Hardinge Road, Grant Road several roads Bangalore derived Delhi Public School Sarjapur, Bangalore East Survey No.43/1B & 45, Sulikunte Village, Dommasandra Post, Bangalore IITJEE SKMClasses.weebly.com military nomenclature Mahatma Gandhi Road MG Raod called IITJEE skmclasses.weebly.com South Parade Roadskmclasses.weebly.com nomenclature Independence Edify School Electronic City
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Hardinge Road old name Pampa Mahakavi Road. sometime, Cunningham Road crowded bazaar being called Sampangi Ramaswamy Temple Road Race Course Road became Devraj Urs Road National Public School, Rajajinagar 1036-A, Purandarapura, V Block, Rajajinagar, Bangalore skmclasses.weebly.comGrant Road became Vittal Mallya Road IITJEE SKMClasses.weebly.com two Vittal Mallya Roads skmclasses bund Sampangi Tank Kanteerava Stadium Gear Innovative International School GEAR Road, Doddakannelli, Off Sarjapur Road & Outer Ring Road, Bangalore IITJEE SKMClasses.weebly.com built MacIver Town Shantala Nagar Assayee Road Meanee Road those names commemoration wars fought Madras New Horizon Gurukul Ring Road Marathalli, Behind New Horizon College of Engineering, Bangalore , Bangalore IITJEE skmclasses.weebly.com Sappers, BGS National Public School Ramalingeshwara Cave Temple Hulimavu, Bangalore IITJEE SKMClasses.weebly.com Presidency School (Bangalore – East) CA Site 7P1A, 2nd A Main, 3rd A cross, East of NGEF Layout, Kasturinagar, Bangalore British Army against Marathas first decade 19th century Basavanagudi, meaning temple Basava skmclasses.weebly.com big bull situated area reason behind naming area Basavanagudi extension skmclassesformed around 1900. Gandhi Bazar, earlier known merely Angadi Beedhi School Of India Anekal Road, Bannerghatta, Bangalore skmclasses formed Kumarapark came skmclasses existence 1947, year Indian Independence, whereas Jayanagar skmclasses.weebly.comRajajinagarIITJEE SKMClasses.weebly.com thought year later 1948 skmclasses.weebly.com orchards Bangalore Palace skmclasses developed housing colony skmclasses.weebly.comnamed Sadashivanagar 1960,IITJEE SKMClasses.weebly.com Orchids The International School Jalahalli, Nagarbavi, Mysore Road, Sarjapur Road, BTM, Bangalore well-known freedom fighter Dakshina Kannada Karnad Sadashiva Rao BVK Iyengar Road Byappana Halli Byatarayanapura Byrasandra C.V Raman Nagar Cambridge Layout Cambridge Road Cantonment Carmelaram Castle Street Central Street Chamarajapet Shanthi Theatre South End Circle INOX Shree Garuda Swagath Mall, 4th Floor, Tilak Nagar Main Road INOX Bangalore Central-2, 5th Floor, 45th Cross Maheshwari Theater Bannerghatta Main Road Gopalan Cinemas Gopalan Innovation Mall, JP Nagar 3rd Phase Chandapura Chandra Layout Global Academy For Learning Sri Chowdeshwari Farm, Near Global Village IT Park, National Public School, HSR Layout P2/32, Sector 4, HSR Layout Bangalore Pattanagere Main Road, Rajarajeshwarinagar, Bangalore Chickpet Chikkabanavara Chikkadugodi Chikkallasandra Chikkamavalli Cholara Palya Chowdeshwari Temple Street Chunchagatta Church Street Clevelskmclasses.weebly.com Town CMH Road Coles Park Commercial Street Commissariat Road Cooke Town Corporation Circle Cottonpet Cox Town Crescent Road Cubbon Park Cubbon Road Cubbonpet Cunningham Road Dairy Circle Dasara Halli Dasarahalli Devaiah Park Devana Halli Devanahalli Devara Chikkana Halli Devara Jeevana Halli Devasandra Dharmaram College Dickenson Road Dispensary Road Dodda Banaswadi Dodda Bommasandra Dodda Kallasandra Dodda Kanna Hally Dodda Mavalli Doddaballapur Road Doddaballapura Doddana Kundi Dollars Colony Domlur Domlur 2nd Stage Domlur Ring Road Dooravani Nagar Dr. Ambedkar Veedhi Dr. DVG Road Delhi Public School, South 11 K.M., kanakapura Road Konanakunte Post, Bangalore Dr. Raj Kumar Road Dr. TCM Royan Road Ejipura Electronic City Field Marshal Cariappa Road Frazer Town Ganapathi Nagar Gandhi Bazaar Gandhi Nagar Ganga Nagar Gangadhar Chetty Road Ganigarpet Garvebhavi Palya Gavipuram Extension Gayathri Nagar Geddala Halli Geddalahalli Giri Nagar Giri Nagar 1st Phase Giri Nagar 2nd Phase GM Palya Gokula Golf Course Road Gorgunte Palya Govindaraj Nagar Green Park Extension, Guddada Halli Gundopanth Street National Public School, Indiranagar 12 A Main HAL II Stage, Bangalore H.Siddaiah Road Haines Road HAL HAL 2nd Stage HAL 3rd Stage HAL Airport Road Hampi Nagar Hanumantha Nagar Hayes Road HBR Layout Hebbal Kempapura Hebbal Ring Road Hegde Nagar Heggana Halli Hennur Hesaraghatta HKP Road HMT Layout Hongasandra Hoody Horamavu Hosakere Halli photochemistry photooxidation piperidines polyanions polycations polycycles polymers Porphyrins prostaglandins 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Training iPhone Development Training Mobile Application Testing Training Mobile Gaming Training Mobile Application Development Training Oakridge International School Oakridge International School, Sarjapur Road, , Bangalore School of India, Bannerghatta, Bangalore Delhi Public School DPS North Campus, Yelahanka, Bangalore Jain International Residential School (JIRS), Jakkasandra Post, Bangalore Delhi Public School (DPS East), Sarjapur, Bangalore TREAMIS World School, Electronics City, Bangalore South Delhi Public School (South), Kanakapura Road, Bangalore The Deen’s Academy, Whitefield, Bangalore National Public School (NPS), Koramangala, Bangalore Royale Concorde International School, Kalyan Nagar, Bangalore Freedom International School, HSR Layout, Bangalore Air Force School Army Public School Bangalore Military School BGS International School Cambridge Public School Delhi Public School Deva Matha Central School Jain International Residential School Kendriya Vidyalaya A M C School 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Colin English Elementary UNITED STATES Jose de Escandon Elementary, UNITED STATES Lincoln Elementary School, UNITED STATES Qingdao Amerasia International School CHINA Roland Park K-8 Magnet School for International Studies, UNITED STATES Theodore Roosevelt Elementary School, UNITED STATES Woodrow Wilson Elementary UNITED STATES Middle Years Programme Cache La Poudre Middle School, UNITED STATES Carl Hankey K-8 School, UNITED STATES Cedar Shoals High School UNITED STATES Concord High School, UNITED STATES Harry Stone Montessori Academy, UNITED STATES International School of Monterey, UNITED STATES Johnnie R. Carr Middle School, UNITED STATES Prairie Seeds Academy, UNITED STATES Roland Park K-8 Magnet School, UNITED STATES Sterling Middle School UNITED STATES The Aga Khan Academy, Hyderabad, INDIA Diploma Programme Ausangate Bilingual School PERU Author’s School Istochnik RUSSIAN FEDERATION Colegio Fiscal Técnico El Chaco ECUADOR Colegio Juan Bautista Montini ECUADOR Colegio Nacional Ciudad de Cuenca ECUADOR Colegio Nacional Experimental Salcedo, ECUADOR Colegio Nacional Machachi, ECUADOR Colegio Nacional Mixto El Playon, ECUADOR Colegio Técnico Cascales, ECUADOR Dar al Marefa Private School, UNITED ARAB EMIRATES Escola Internacional del Camp SPAIN Gymnasium Jovan Jovanovic Zmaj SERBIA ISTEK Private Acibadem Schools TURKEY Instituto Superior Tecnológico Carlos Cisneros ECUADOR Instituto Superior Tecnológico Daniel Alvarez Burneo ECUADOR Instituto Técnico Superior Isabel de Godin ECUADOR King Abdulaziz Saudi School Rome ITALY Riga State Gymnasium Nr. 2 LATVIA Saudi School Vienna AUSTRIA State IS Seeheim Jugenheim/Schuldorf Bergstrasse GERMANY Unidad Educativa Bolívar, ECUADOR Unidad Educativa Abelardo Moncayo, ECUADOR Unidad Educativa Fiscomisional Verbo Divino, ECUADOR Unidad Educativa Mayor ECUADOR Unidad Educativa Nueva Semilla, ECUADOR Unidad Educativa Temporal Juan Bautista Vásquez, ECUADOR Primary Years Programme Ajman Academy UNITED ARAB EMIRATES British International School Kiev UKRAINE Cache La Poudre Elementary School, UNITED STATES Dr. Thomas S. O Connell Elementary School UNITED STATES Gems World Academy Abu Dhabi UNITED ARAB EMIRATES Hebron-Harman Elementary School, UNITED STATES International School of Solothurn, SWITZERLAND Lisa-Junior Primary School AUSTRIA Madison Richard Simis Elementary School, UNITED STATES Miina Härma Gümnaasium, ESTONIA Riffenburgh Elementary School UNITED STATES Roscoe Wilson Elementary School, UNITED STATES Singapore International School INDIA William H. Wharton K-8 Dual Language Academy UNITED STATES World Academy of Tirana, ALBANIA École Centrale, CANADA École Micheline-Brodeur CANADA École Saint-Édouard, CANADA École élémentaire catholique Jean-Paul II CANADA Özel Istanbul Coskun Koleji Anaokulu & Ilkokulu TURKEY Middle Years Programme Abraham Lincoln Middle School UNITED STATES Beijing Huijia Private School CHINA Cakir Middle School TURKEY Durango High School UNITED STATES Emirates IS Meadows, UNITED ARAB EMIRATES Madison International School, MEXICO Meadow Park Middle School, UNITED STATES North Central High School, UNITED STATES Phuket International Academy Day School THAILAND Ray Wiltsey Middle School UNITED STATES Rockridge Secondary School CANADA School Lane Charter School UNITED STATES Strothoff International School Rhein-Main Campus Dreieich GERMANY Tsukuba International School JAPAN École Père-Marquette, CANADA École secondaire Saint-Luc, CANADA Diploma Programme Anania Shirakatsy Armenian National Lyceum Ed’l Complex-CJSC, ARMENIA COLEGIO ALAUDA SPAIN Colegio Británico, MEXICO Colegio Nacional Camilo Gallegos Toledo ECUADOR Colegio Nacional Experimental Amazonas, ECUADOR Colegio Nacional Experimental María Angélica Idrobo, ECUADOR Colegio Nacional San José, ECUADOR Eastern Mediterranean International School, ISRAEL Emirates National School UNITED ARAB EMIRATES GEMS American Academy Abu Dhabi, UNITED ARAB EMIRATES German International School Sharjah UNITED ARAB EMIRATES Instituto Tecnológico Superior Angel Polibio Chaves, ECUADOR International School Moshi Arusha Campus TANZANIA, UNITED REPUBLIC OF International School of Bydgoszcz POLAND Ludoteca Elementary & High School, Padre Víctor Grados, ECUADOR Léman International School Chengdu CHINA Metropolitan School of Panama, PANAMA Munic. Atms. Educ. Institution Kogalym Secondary School ?8, RUSSIAN FEDERATION Phorms Bilingual Gymnasium, GERMANY Royal High School, UNITED STATES SIS Swiss International School Stuttgart-Fellbach, GERMANY Seedling Public School INDIA The British School of Beijing CHINA Unidad Educativa Fiscal Experimental del Milenio, ECUADOR Unidad Educativa Juan de Velasco ECUADOR Unidad Educativa Tumbaco, ECUADOR École secondaire Gaétan Gervais, CANADA École secondaire Hanmer CANADA Stonehill International School American School of Bombay Mumbai Day school offering PYP MYP DP Dhirubhai Ambani International School Mumbai Day school offering DP Ecole Mondiale World School, Mumbai Day school offering DP Jamnabai Narsee School Mumbai Day school offering DP Ahmedabad International School Ahmedabad Day School offering PYP Mahatma Gandhi International School Ahmedabad Day school offering MYP Mahindra United World College of India Pune Boarding school offering DP Mercedes-Benz International School Pune American Embassy School Delhi Day school offering DP The British School, Delhi Day school offering DP Pathways World School, Gurgaon Boarding school offering PYP DP SelaQui World School, Dehra Dun Boarding school offering DP Canadian International School, Bangalore Mixed Boarding Day school offering DP International School of Bangalore, Bangalore Mixed Boarding Day school offering DP Oakridge International School Hyderabad Day school offering PYP Chinmaya International Residential School Coimbatore Boarding school offering DP Good Shepherd International School Ooty Boarding school offering DP Kodaikanal International School, Kodaikanal Boarding school offering DP Home Tuition Group teachers available small groupsstudents IB International Baccalaureate Programme, IGCSE, ISc, ICSE, CBSE Schools offering IB ( International Baccalaureate ) Programme Bangalore International School Geddalahalli Hennur Bagalur Road Kothanur Post Bengaluru India 560 077 Stonehill International School, 1st Floor, Embassy Point #150, Infantry Road Bengaluru 560 001 Stonehill International School 259/333/334/335 Tarahunise Post Jala Hobli, Bengaluru North 562157 Candor International School Begur Koppa Road, Hullahalli Off Bannerghatta Road, Near Electronic City Bangalore 560105 Greenwood High International School Bengaluru, No.8-14, Chickkawadayarapura, Near Heggondahalli Gunjur Post, Varthur Sarjapur Road, Bangalore 560087 Sarla Birla Academy, Bannerghatta, Bangalore, Canadian International School, Yelahanka, Bangalore Indus International School Billapura Cross Sarjapur Bangalore

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SKM Classes Bangalore

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