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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 pKa of the acid or pKb 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:

The equilibrium constant is given by the equation:

K = {[Ba2+][SO42-]} / [BaSO4]

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

Ksp = K[BaSO4] = [Ba2+][SO42-] ————————————————————————————————(7.39)

We call Ksp the solubility product constant or simply solubility product. The experimental value of Ksp 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) = Sor 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(PO43-)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 [PO43-] = 4S

and Ksp = (3S)3 (4S)4 = 6912 (S)7

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

A solid salt of the general formula Mxp+ Xyq- with molar solubility S in equilibrium with its saturated solution may be represented by the equation:

(where X x p+ = y x q)

And its solubility product constant is given by:

Ksp = [Mp+]x[Xq- ]y = (xS)x(yS)y ————————————————————————-(7.40)

= xx . yy . S(x + y)

S(x + y) = Ksp / xx . yy

S = (Ksp / xx . y y)1 / x + y ———————————————————————————(7.41)

The term Ksp in equation is given by Qsp (section 7.6.2) when the concentration of one or more species is not the concentration under equilibrium. Obviously under equilibrium conditions Ksp = Qsp 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, Ksp of Some Common Ionic Salts at 298K.

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, Ksp = 1.1 x 10-23.

Solution

A2X3 → 2A3+ + 3X2-

Ksp = [A3+]2 [X2-]3 = 1.1 x 10-23

If S = solubility of A2X3, then

[A3+] = 2S; [X2] = 3S

therefore, Ksp = (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 Ksp 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

Ksp = [Ag+][CN] = 6 x 10-17

Ksp = [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 Ksp = Qsp. 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 Ksp = Qsp. 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 Qsp. 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.

Ksp = 2.0 x 10-15 = [Ni2+] [OH]2

= (S) (0.10 + 2S)2

As Ksp 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 Ksp = Qsp. We have to satisfy two equilibria simultaneously i.e.,

Ksp = [M+] [X],

Ka = [H+(aq)][X(aq)]/[HX(aq)]

Taking inverse of both side and adding 1 we get

[HX]/[X] + 1 = [H+]/Ka + 1

([HX]+[H])/[X] = ([H+] + Ka)/Ka

Now, again taking inverse, we get

[X] / {[X] + [HX]} = f = Ka / (Ka + [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

Ksp = [S] [f S] = S2 {Ka / (Ka + [H+])} and

S = {Ksp ([H+] + Ka ) /Ka }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, Kc is expressed as the concentration of products divided by reactants, each term raised to the stoichiometric coefficient. For reaction,

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 Kp and is written by replacing concentration terms by partial pressures in Kc expression. The direction of reaction can be predicted by reaction quotient Qc which is equal to Kc 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 (Ka) and weak bases (Kb) 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]) ; pKa = -log[Ka] ; pKb = -log[Kb]; and pKw = -log[Kw] etc.). The ionization of water has been considered and we note that the equation: pH + pOH = pKw 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 (Ksp). 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 Kc for the following equilibrium when the equilibrium concentration of each substance is: [SO2]= 0.60M, [O2] = 0.82M and [SO3] = 1.90M ?

7.3 At a certain temperature and total pressure of 105Pa, iodine vapour contains 40% by volume of I atoms

Calculate Kp for the equilibrium.

7.4 Write the expression for the equilibrium constant, Kc for each of the following reactions:

7.5 Find out the value of Kc for each of the following equilibria from the value of Kp:

7.6 For the following equilibrium, Kc= 6.3 x 1014 at 1000 K

Both the forward and reverse reactions in the equilibrium are elementary bimolecular reactions. What is Kc, 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 O2- takes place as follows:

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 Kc= 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:

When 0.087 mol of NO and 0.0437 mol of Br2 are mixed in a closed container at constant temperature, 0.0518 mol of NOBr is obtained at equilibrium. Calculate equilibrium amount of NO and Br2 .

7.10 At 450K, Kp= 2.0 x 1010/bar for the given reaction at equilibrium.

What is Kc at this temperature ?

7.11 A sample of HI(g) is placed in flask at a pressure of 0.2 atm. At equilibrium the partial pressure of HI(g) is 0.04 atm. What is Kp for the given equilibrium ? below:

7.12 A mixture of 1.57 mol of N2, 1.92 mol of H2 and 8.13 mol of NH3 is introduced into a 20 L reaction vessel at 500 K. At this temperature, the equilibrium constant, Kc for the reaction  is 1.7 x 102. Is the reaction mixture at equilibrium? If not, what is the direction of the net reaction?

7.13 The equilibrium constant expression for a gas reaction is,

Kc = [ 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,

Calculate the equilibrium constant for the reaction.

7.15 At 700 K, equilibrium constant for the reaction:

is 54.8. If 0.5 mol L-1 of HI(g) is present at equilibrium at 700 K, what are the concentration of H2(g) and I2(g) assuming that we initially started with HI(g) and allowed it to reach equilibrium at 700K?

7.16 What is the equilibrium concentration of each of the substances in the

equilibrium when the initial concentration of ICl was 0.78 M ?

7.17 Kp = 0.04 atm at 899 K for the equilibrium shown below. What is the equilibrium concentration of C2H6 when it is placed in a flask at 4.0 atm pressure and allowed to come to equilibrium?

7.18 Ethyl acetate is formed by the reaction between ethanol and acetic acid and the equilibrium is represented as:

(i) Write the concentration ratio (reaction quotient), Qc, 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 Kc is 8.3 x 10-3, what are the concentrations of PCl3 and Cl2 at equilibrium?

7.20 One of the reaction that takes place in producing steel from iron ore is the reduction of iron(II) oxide by carbon monoxide to give iron metal and CO2.

What are the equilibrium partial pressures of CO and CO2 at 1050 K if the initial partial pressures are: pCO= 1.4 atm and 2 CO p = 0.80 atm?

7.21 Equilibrium constant, Kc for the reaction

 at 500 K is 0.061

At a particular time, the analysis shows that composition of the reaction mixture is 3.0 mol L-1 N2, 2.0 mol L-1 H2 and 0.5 mol L-1 NH3. Is the reaction at equilibrium? If not in which direction does the reaction tend to proceed to reach equilibrium?

7.22 Bromine monochloride, BrCl decomposes into bromine and chlorine and reaches the equilibrium:

for which Kc= 32 at 500 K. If initially pure BrCl is present at a concentration of 3.3 x 10-3mol L-1, what is its molar concentration in the mixture at equilibrium?

7.23 At 1127 K and 1 atm pressure, a gaseous mixture of CO and CO2 in equilibrium with soild carbon has 90.55% CO by mass

Calculate Kc for this reaction at the above temperature.

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

where ΔfGθ (NO2) = 52.0 kJ/mol

ΔfGθ (NO) = 87.0 kJ/mol

ΔfGθ (O2) = 0 kJ/mol

7.25 Does the number of moles of reaction products increase, decrease or remain same when each of the following equilibria is subjected to a decrease in pressure by increasing the volume?

7.26 Which of the following reactions will get affected by increasing the pressure? Also, mention whether change will cause the reaction to go into forward or backward direction.

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

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:

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

(b) How will the values of Kp 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:

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

a) write an expression for Kc for the reaction.

b) what is the value of Kc for the reverse reaction at the same temperature ?

c) what would be the effect on Kc if (i) more PCl5 is added (ii) pressure is increased (iii) the temperature is increased ?

7.31 Dihydrogen gas used in Haber‘s process is produced by reacting methane from natural gas with high temperature steam. The first stage of two stage reaction involves the formation of CO and H2. In second stage, CO formed in first stage is reacted with more steam in water gas shift reaction,

If a reaction vessel at 400°C is charged with an equimolar mixture of CO and steam such that pCO = pH2O = 4.0 bar, what will be the partial pressure of H2 at equilibrium? Kp= 10.1 at 400°C

7.32 Predict which of the following reaction will have appreciable concentration of reactants and products:

7.33 The value of Kc for the reaction  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,  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, Kc 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, CO32-, 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 10-4and 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 HSion 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 S2- 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 pKa.

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 pKa of bromoacetic acid.

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

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 pKa 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

(c) Human blood, 7.38 (d) Human saliva, 6.4.

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 10-13respectively. 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 Ksp = 7.4 x 10-8 ).

7.70 The ionization constant f benzoic acid is 6.46 x 10-5 and Ksp 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, Ksp = 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, Ksp 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 Qc = 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 Qc is less than Kc 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, CO32–

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, [S2–]= 1.2 x 10–13M, in 0.1M HCl [S2–]= 1.09 x 10–19M

7.46 [Ac]= 0.00093, pH= 3.03

7.47 [A] = 7.08 x10–5M, Ka= 5.08 x 10–7, pKa= 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, pKa = 2.70

7.51 Kb = 1.6 x 10–6, pKb = 5.8

7.52 α = 6.53 x 10–4, Ka = 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 Ka = 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 Kb = 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 CrO42– = 0.65 x 10–4M; Barium Chromate S = 1.1 x 10–5M;
Molarity of Ba2+ and CrO42– 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 Kc and Kp is

Kp = Kc(RT)Δn

What would be the value of Δn for the reaction

(i) 1

(ii) 0.5

(iii) 1.5

(iv) 2

2. For the reaction  , the standard free energy is ΔGΘ > 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–1respectively. The value of Kc for the reaction  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.

(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, Ka, is the measure of strength of an acid. The Kavalues 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. Ka1 , Ka2 and Ka3 are the respective ionisation constants for the following reactions.

The correct relationship between Ka1 , Ka2 and Ka3 is

(i) Ka3 = Ka1 × Ka2

(ii) Ka3 = Ka1 + Ka2

(iii) Ka3 = Ka1 – Ka2

(iv) Ka3 = Ka1 / Ka2

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 (Ka = 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 .

(i) ΔGΘ = 0

(ii) ΔGΘ > 0

(iii) ΔGΘ < 0

(iv) ΔGΘ = –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.

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, Kc , for the following reaction is 5.

What would be the equilibrium constant Kc for the reaction

(i) 0.04

(ii) 0.4

(iii) 25

(iv) 2.5

19. In which of the following reactions, the equilibrium remains unaffected on

addition of small amount of argon at constant volume?


(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 , 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:

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 Kc for the reaction  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 (Qsp) 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. (Ksp 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)

ΔfHΘ[CaO(s)] = – 635.1 kJ mol–1

ΔfHΘ[CO2(g)] = – 393.5 kJ mol–1

ΔfHΘ[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

39. For the reaction : 

Equilibrium constant Kc = [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) ΔGΘ > 0 (a) K > 1
(ii) ΔGΘ < 0 (b) K = 1
(iii) ΔGΘ = 0 (c) K = 0
(d) K < 1

41. Match the following species with the corresponding conjugate acid

Species Conjugate acid
(i) NH3 (a) CO32–
(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) ΔGΘ = 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 same 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 Ka and Kb 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 Kcand Qc?

(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.

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 Axp+Byq− 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 :

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) ΔGΘ = 0
Justification : ΔGΘ = – RT lnK
At the stage of half completion of reaction [A] = [B], Therefore, K = 1.
Thus, ΔGΘ = 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 :
H3: → 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.

Ksp for BaSO4 in water = [Ba2+] [SO42–] = (S) (S) = S2
But S = 8 × 10–4 mol dm–3
∴ Ksp = (8×10–4)2 = 64 × 10–8 ………………………………………………….. (1)
The expression for Ksp in the presence of sulphuric acid will be as follows :
Ksp = (S) (S + 0.01) ………………………………………………………………………………………………. (2)
Since value of Ksp 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 = [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.

Ksp = [Al3+] [OH]3 = (S) (3S)3 = 27 S4
S4 = Ksp/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. Ksp of PbCl2 = 3.2 × 10–8
Let S be the solubility of PbCl2.

Ksp = [Pb2+] [Cl]2 = (S) (2S)2 = 4S3
Ksp = 4S3
S3 = Ksp/4 = 3.2 x 10–8/4 mol L–1 = 8 × 10–9 mol L–1


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. ΔrHΘ = ΔfHΘ[CaO(s)] + ΔfHΘ [CO2(g)] – ΔfHΘ [CaCO3(s)]
∴ ΔrHΘ = 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) Qc < Kc
(ii) Qc > Kc
(iii) Qc = Kc

where, Qc is reaction quotient in terms of concentration and Kc is equilibrium constant.

53.


S moles of AxBy dissolve to give xS moles of Ap+ and y S moles of Bq–.]

54. ΔG = ΔGΘ + RTlnQ
ΔGΘ = Change in free energy as the reaction proceeds
ΔG = Standard free energy change
Q = Reaction quotient
R = Gas constant
T = Absolute temperature
Since ΔGΘ = – 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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