
Q1 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? Ans: Vapour pressure is the pressure exerted by the vapour in equilibrium with the liquid at a fixed temperature.
(a) If the volume of the container is suddenly increased, then the vapour pressure would decrease initially. This is because the amount of vapour remains the same, but the volume increases suddenly. As a result, the same amount of vapour is distributed in a larger volume.
(b) Since the temperature is constant, the rate of evaporation also remains constant. When the volume of the container is increased, the density of the vapour phase decreases. As a result, the rate of collisions of the vapour particles also decreases. Hence, the rate of condensation decreases initially.
(c) When equilibrium is restored finally, the rate of evaporation becomes equal to the rate of condensation. In this case, only the volume changes while the temperature remains constant. The vapour pressure depends on temperature and not on volume. Hence, the final vapour pressure will be equal to the original vapour pressure of the system.
Q2 What is Kc for the following equilibrium when the equilibrium concentration of each substance is: [SO2]= 0.60 M, [O2] = 0.82 M and [SO3] = 1.90 M ? 2SO2(g) + O2(g) ↔ 2SO3(g) Ans: The equilibrium constant (K_{c}) for the give reaction is:
K_{c }= [SO_{3}]2 / [SO_{2}]2 [O_{2}]
= (1.90)^{2} M^{2} / (0.60)^{2} (0.8321) M^{3}
= 12.239 M^{1 }Approx.
Hence, K_{c }for the equilibrium is 12.239 M^{1}
Q3 At a certain temperature and total pressure of 105 Pa, iodine vapour contains 40% by volume of I atoms I2(g) ↔ 2l (g) Calculate Kp for the equilibrium. Ans: Partial pressure of I atoms,
p_{I }_{ }= p_{total} x 40/100
= 10^{5} x 40/100
= 4 x 10^{4} Pa
Partial pressure of I_{2} molecules,
p_{I }_{2 } = _{ }p_{total} x 60/100
= 10^{5} x 60/100
= 6 x 10^{4} Pa
Now, for the given reaction,
Kp = (pI)^{2} / pI_{2}
= (4 x 10^{4} )^{ 2} Pa^{2} / 6 x 10^{4} Pa
= 2.67 x 10^{4} Pa
Q4 Write the expression for the equilibrium constant, Kc for each of the following reactions: (i) 2NOCl (g) ↔ 2NO (g) + Cl2 (g) (ii) 2Cu(NO3)2 (s) ↔ 2CuO (s) + 4NO2 (g) + O2 (g) (iii) CH3COOC2H5(aq) + H2O(l) ↔ CH3COOH (aq) + C2H5OH (aq) (iv) Fe3+ (aq) + 3OH– (aq) ↔ Fe(OH)3 (s) (v) I2 (s) + 5F2 ↔ 2IF5 Ans: Q5 Find out the value of Kc for each of the following equilibria from the value of Kp: (i) 2NOCl (g) ↔ 2NO (g) + Cl2 (g); Kp = 1.8 × 10–2 at 500 K (ii) CaCO3 (s) ↔ CaO(s) + CO2(g); Kp = 167 at 1073 K Ans: (a) The relation between K_{p} and K_{c} is given as:
K_{p} = K_{c} (RT)^{Δn}
(a) Here, Δn = 3  2 = 1
R = 0.0831 barLmol^{1}K^{1}
T = 500 K
Kp = 1.8 x 10^{2}
Now, K_{p} = K_{c} (RT)^{Δn}
⇒ K_{c} = 1.8 x 10^{2 }/ (0.0831x500)
= 4.33 x 10^{4 }(approx.)
(b) Here, Δn = 2  1 = 1
R = 0.0831 barLmol^{1}K^{1}
T = 1073 K
K^{p }= 167
Now, K_{p} = K_{c} (RT)^{Δn}
⇒ K_{c} = 167 / (0.0831x1073)
= 1.87 (approx.)
Q6 For the following equilibrium, Kc = 6.3 × 1014 at 1000 K NO (g) + O3 (g) ↔ NO2 (g) + O2 (g) Both the forward and reverse reactions in the equilibrium are elementary bimolecular reactions. What is Kc, for the reverse reaction? Ans: It is given that K_{c }for the forward reaction is 6.3 × 10^{14}
Then, K_{c } for the reverse reaction will be,
K^{’}_{c } = 1 / K_{c}
= 1 / 6.3 × 10^{14}
= 1.59 x 10^{15}
Q7 Explain why pure liquids and solids can be ignored while writing the equilibrium constant expression? Ans: : For a pure substance (both solids and liquids),
Pure Substance = Number of moles / Volume
= Mass/Molecular Mass / Volume
= Mass / Volume x Molecular Mass
= Density / Molecular Mass
Now, the molecular mass and density (at a particular temperature) of a pure substance is always fixed and is accounted for the equilibrium constant. Therefore, the values of pure substances are not mentioned in the equilibrium constant expression.
Q8 Reaction between N2 and O2– 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 Kc = 2.0 × 10–37, determine the composition of equilibrium mixture. Ans: Let the concentration of N_{2}O at equilibrium be x. The given reaction is:
2N_{2} (g) + O_{2} (g) ↔ 2N_{2}O (g)
Intial Concentration: 0.482 mol 0.933 mol 0 mol
At equilibrium (0.482x) mol (0.933 x) mol x mol
Therefore, at equilibrium, in the 10 L vessel:
N_{2} = 0.482x / 10
O_{2} = 0.933x/2 / 10
N_{2}O = x / 10
The value of equilibrium constant i.e.K_{c} = 2.0 × 10^{37} is very small. Therefore, the amount of N_{2} and O_{2} reacted is also very small. Thus, x can be neglected from the expressions of molar concentrations of N_{2} and O_{2}.
Then,
N2 = 0.482/10 = 0.0482 molL1 and O2 = 0.933/10 = 0.0933 molL1
Now,
Q9 Nitric oxide reacts with Br2 and gives nitrosyl bromide as per reaction given below: 2NO (g) + Br2 (g) ↔ 2NOBr (g) 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 . Ans: The given reaction is:
2NO (g) + Br_{2} (g) ↔ 2NOBr (g)
2 mol 1 mol 2 mol
Now, 2 mol of NOBr are formed from 2 mol of NO. Therefore, 0.0518 mol of NOBr are formed from 0.0518 mol of NO.
Again, 2 mol of NOBr are formed from 1 mol of Br.
Therefore, 0.0518 mol of NOBr are formed from 0.0518/2 mol of Br, or 0.0259 mol of NO.
The amount of NO and Br present initially is as follows:
[NO] = 0.087 mol [Br_{2}] = 0.0437 mol
Therefore, the amount of NO present at equilibrium is: [NO] = 0.087  0.0518 = 0.0352 mol
And, the amount of Br present at equilibrium is: [Br_{2}] = 0.04370.0259 = 0.0178 mol
Q10 At 450K, Kp= 2.0 × 1010/bar for the given reaction at equilibrium. 2SO2(g) + O2(g) ↔ 2SO3 (g) What is Kc at this temperature ? Ans: For the given reaction,
Δn = 2  3 =1
T = 450 K
R = 0.0831 bar L bar K^{1} mol^{1}
K_{p} = 2.0 × 10^{10} bar^{ 1}
We know that,
K_{p} = K_{c} (RT)Δn
Q11 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 ? 2HI (g) ↔ H2 (g) + I2 (g) Ans: The initial concentration of HI is 0.2 atm. At equilibrium, it has a partial pressure of 0.04 atm. Therefore, a decrease in the pressure of HI is 0.2  0.04 = 0.16. The given reaction is:
2HI (g) ↔ H_{2} (g) + I_{2} (g)
Intial concentration 0.2 atm 0 0
At equilibrium 0.04 atm 0.16/2 2.15/2
Therefore,
Therefore, Hence, the value of K_{p} for the given equilibrium is 4.0.
Q12 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 N2 (g) + 3H2 (g) ↔ 2NH3 (g) is 1.7 × 102. Is the reaction mixture at equilibrium? If not, what is the direction of the net reaction? Ans: The given reaction is:
N_{2 }(g) + 3H_{2} (g) ↔ 2NH_{3} (g)
The given concetration of various species is:
N_{2} = 1.57/20 molL^{1}
H_{2} = 1.92/20 molL^{1}
NH_{3 }= 8.13/20 molL^{1}
Now, reaction quotient Qc is:
Qc = (8.13/20)^{2} / (1.57/20) (1.92/20)^{3}
^{ } = 2.4 x 10^{3}
Since,Qc ≠ Kc the reaction mixture is not at equilibrium.
Again, Qc > Kc. Hence, the reaction will proceed in the reverse direction.
Q13 The equilibrium constant expression for a gas reaction is, Write the balanced chemical equation corresponding to this expression. Ans: The balanced chemical equation corresponding to the given expression can be written as:
4NO(g) + 6H_{2}O(g) ↔ 4NH_{3(g)} + 5O_{2(g)}
Q14 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, H2O (g) + CO (g) ↔ H2 (g) + CO2 (g) Calculate the equilibrium constant for the reaction. Ans: The given reaction is:
H_{2}O (g) + CO (g) ↔ H_{2} (g) + CO_{2} (g)
Initial Concentration: 1/10 M 1/10 M 0 0
At equilibrium 10.4/10 M 10.4/10 M 0.4/10M 0.4/10M
= 0.06M 0.06M 0.04M 0.04M
Therefore, the equilibrium constant for the reaction,
Q15 At 700 K, equilibrium constant for the reaction: H2 (g) + I2 (g) ↔ 2HI (g) 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? Ans: It is given that equilibrium constant K_{c }for the reaction
H_{2} (g) + I_{2} (g) ↔ 2HI (g) is 54.8.
Therefore, at equilibrium, the equilibrium constant K'_{c}for the reaction
2HI (g) ↔ H_{2} (g) + I_{2} (g) will be 1/54.8
HI = 0.5 molL^{1}
Let the concentrations of hydrogen and iodine at equilibrium be x molL^{1} .
Hence, at equilibrium,
Q16 What is the equilibrium concentration of each of the substances in the equilibrium when the initial concentration of ICl was 0.78 M ? 2ICl (g) ↔ I2 (g) + Cl2 (g); Kc = 0.14 Ans: The given reaction is:
2ICl (g) ↔ I_{2} (g) + Cl_{2} (g)
Initial conc. 0.78 M 0 0
At equilibrium (0.78  2x) M x M x MNow we can write, Kc = I_{2 } x_{ } Cl_{2 }/ (ICl)^{2} _{ }
Q17 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? C2H6 (g) ↔ C2H4 (g) + H2 (g) Ans: Let p be the pressure exerted by ethene and hydrogen gas (each) at equilibrium.
Now, according to the reaction,
C_{2}H_{6} (g) ↔ C_{2}H_{4} (g) + H_{2} (g)
Initial conc. 4.0atm 0 0
At equilibrium 4.0p p pWe can write,
Hence, at equilibrium,
C2H6  4  p = 4 .038
= 3.62 atm
Q18 Ethyl acetate is formed by the reaction between ethanol and acetic acid and the equilibrium is represented as: CH3COOH (l) + C2H5OH (l) ↔ CH3COOC2H5 (l) + H2O (l) (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? Ans: (i) Reaction quotient,
(ii) Let the volume of the reaction mixture be V. Also, here we will consider that water is a solvent and is present in excess. The given reaction is:
Therefore, equilibrium constant for the given reaction is:
Q19 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 × 10–1 mol L–1. If value of Kc is 8.3 × 10–3, what are the concentrations of PCl3 and Cl2 at equilibrium? PCl5 (g) ↔ PCl3 (g) + Cl2(g) Ans: Let the concentrations of both PCl_{3} and Cl_{2} at equilibrium be x molL^{1}. The given reaction is:
PCl_{5} (g) ↔ PCl_{3} (g) + Cl_{2}(g)
at equilibrium 0.5x10^{1}molL^{1} xmolL^{1} xmolL^{1}
it is given that the value of equilibrium constant, Kc is 8.3x10^{–3}
Now we can write the expression for equilibrium as:
Q20 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. FeO (s) + CO (g) ↔ Fe (s) + CO2 (g); Kp = 0.265 atm at 1050K What are the equilibrium partial pressures of CO and CO2 at 1050 K if the initial partial pressures are: pCO= 1.4 atm and Pco2 = 0.80 atm Ans: Q21 Equilibrium constant, Kc for the reaction N2 (g) + 3H2 (g) ↔ 2NH3 (g) 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? Ans: The given reaction is:
N_{2} (g) + 3H_{2} (g) ↔ 2NH_{3} (g)
at a particular time: 3.0molL^{1} 2.0 molL^{1} 0.5molL^{1}
Now, we know that,
Qc = [NH_{3}]^{2} / [N_{2}][H_{2}]^{3}
^{ } = (0.5)^{2} / (3.0)(2.0)^{3}
^{ }= 0.0104
It is given that K_{c} = 0.061
Since,Qc ≠ Kc, the reaction mixture is not at equilibrium.
Again, Qc < Kc, the reaction will proceed in the forward direction to reach equilibrium.
Q22 Bromine monochloride, BrCl decomposes into bromine and chlorine and reaches the equilibrium: 2BrCl (g) ↔ Br2 (g) + Cl2 (g) for which Kc= 32 at 500 K. If initially pure BrCl is present at a concentration of 3.3 × 10–3 mol L–1, what is its molar concentration in the mixture at equilibrium? Ans: Let the amount of bromine and chlorine formed at equilibrium be x. The given reaction is:
2BrCl (g) ↔ Br_{2} (g) + Cl_{2} (g)
Initial Conc. 3.3x10^{3 } 0 0
at equilibrium 3.3x10^{3} 2x x x
Now, we can write,
Kc = [Br_{2}][Cl_{2}] / [BrCl]^{2}
⇒ (x) x (x) / (3.3x10^{3} 2x)^{2 }= 32
⇒ x / (3.3x10^{3} 2x) = 5.66
⇒ x = 18.678x10^{3}  11.32x
⇒ x + 11.32x = 18.678x10^{3}
⇒ 12.32x = 18.678x10^{3}
⇒ x = 1.5 x 10^{3}
Therefore, at equilibrium,
[BrCl] = 3.3x10^{3 } (2 x 1.5 x 10^{3})
= 3.3x10^{3}  3.0x10^{3}
^{ }= 0.3 x 10^{3}
^{ } = 3.0 x 10^{4} mol L^{1}
Q23 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 C (s) + CO2 (g) ↔ 2CO (g) Calculate Kc for this reaction at the above temperature. Ans: Let the total mass of the gaseous mixture be 100 g.
Mass of CO = 90.55 g
And, mass of CO_{2} = (100  90.55) = 9.45 g
Now, number of moles of CO, n_{co} = 90.55/28 = 3.234 mol
Number of moles of CO_{2}, n_{co2} = 9.45/44 = 0.215 mol
Partial pressure of CO,
For the given reaction,
Δn = 2  1 = 1
we know that,
Kp = Kc (RT)^{Δn}
⇒ 14.19 = Kc (0.082 x 1127)^{1}
⇒ Kc = 14.19 / 0.082 x 1127
= 0.154 (arrpox.)
Q24 Calculate a) ΔG0 and b) the equilibrium constant for the formation of NO2 from NO and O2 at 298K NO (g) + ½ O2 (g) ↔ NO2 (g) where ΔfG0 (NO2) = 52.0 kJ/mol ΔfG0 (NO) = 87.0 kJ/mol ΔfG0 (O2) = 0 kJ/mol Ans: (a) For the given reaction,
ΔG° = ΔG°( Products)  ΔG°( Reactants)
ΔG° = 52.0 {87.0 + 0} = 35.0 kJ mol^{1 }
(b) We know that,
ΔG° = RT log K_{c}
ΔG° = 2.303 RT log K_{c}
Hence, the equilibrium constant for the given reaction K_{c} is 1.36 × 106
Q25 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? (a) PCl5 (g) ↔ PCl3 (g) + Cl2 (g) (b) CaO (s) + CO2 (g) ↔ CaCO3 (s) (c) 3Fe (s) + 4H2O (g) ↔ Fe3O4 (s) + 4H2 (g) Ans: (a) The number of moles of reaction products will increase. According to Le Chatelier's principle, if pressure is decreased, then the equilibrium shifts in the direction in which the number of moles of gases is more. In the given reaction, the number of moles of gaseous products is more than that of gaseous reactants. Thus, the reaction will proceed in the forward direction. As a result, the number of moles of reaction products will increase.
(b) The number of moles of reaction products will decrease.
(c) The number of moles of reaction products remains the same.
Q26 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. (i) COCl2 (g) ↔ CO (g) + Cl2 (g) (ii) CH4 (g) + 2S2 (g) ↔ CS2 (g) + 2H2S (g) (iii) CO2 (g) + C (s) ↔ 2CO (g) (iv) 2H2 (g) + CO (g) ↔ CH3OH (g) (v) CaCO3 (s) ↔ CaO (s) + CO2 (g) (vi) 4 NH3 (g) + 5O2 (g) ↔ 4NO (g) + 6H2O(g) Ans: The reactions given in (i), (iii), (iv), (v), and (vi) will get affected by increasing the pressure.
The reaction given in (iv) will proceed in the forward direction because the number of moles of gaseous reactants is more than that of gaseous products.
The reactions given in (i), (iii), (v), and (vi) will shift in the backward direction because the number of moles of gaseous reactants is less than that of gaseous products.
Q27 The equilibrium constant for the following reaction is 1.6 ×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. Ans: Given,k_{p} for the reaction i.e., H_{2}(g) + Br_{2}(g) ↔ 2HBr(g) is 1.6 ×10^{5}
Therefore, for the reaction
2HBr(g) ↔ H_{2}(g) + Br_{2}(g)
the equilibrium constant will be,
K'_{p} = 1/K_{p}
= 1/1.6 ×10^{5}
^{ }= 6.25x10^{6}
Now, let p be the pressure of both H_{2} and Br_{2} at equilibrium.
2HBr(g) ↔ H_{2}(g) + Br_{2}(g)
Initial Conc. 10 0 0
at equilibrium 102p p p
Now, we can write,
Q28 Dihydrogen gas is obtained from natural gas by partial oxidation with steam as per following endothermic reaction: CH4 (g) + H2O (g) ↔ CO (g) + 3H2 (g) (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 ? Ans: (a) For the given reaction,
(b) (i) According to Le Chatelier's principle, the equilibrium will shift in the backward direction.
(ii) According to Le Chatelier's principle, as the reaction is endothermic, the equilibrium will shift in the forward direction.
(iii) The equilibrium of the reaction is not affected by the presence of a catalyst. A catalyst only increases the rate of a reaction. Thus, equilibrium will be attained quickly.
Q29 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) Ans: (a) According to Le Chatelier's principle, on addition of H_{2}, the equilibrium of the given reaction will shift in the forward direction.
(b) On addition of CH_{3}OH, the equilibrium will shift in the backward direction.
(c) On removing CO, the equilibrium will shift in the backward direction.
(d) On removing CH_{3}OH, the equilibrium will shift in the forward direction.
Q30 At 473 K, equilibrium constant Kc for decomposition of phosphorus pentachloride, PCl5 is 8.3 ×103. If decomposition is depicted as, PCl5 (g) ↔ PCl3 (g) + Cl2 (g) ΔrH0 = 124.0 kJ mol–1 (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 ? Ans: (c) (i) Kc would remain the same because in this case, the temperature remains the same.
(ii) Kc is constant at constant temperature. Thus, in this case, Kc would not change.
(iii) In an endothermic reaction, the value of Kc increases with an increase in temperature. Since the given reaction in an endothermic reaction, the value of Kc will increase if the temperature is increased.
Q31 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, CO (g) + H2O (g) ↔ CO2 (g) + H2 (g) 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. Ans: Let the partial pressure of both carbon dioxide and hydrogen gas be p. The given reaction is:
CO (g) + H_{2}O (g) ↔ CO_{2} (g) + H_{2} (g)
Initial Conc. 4.0 bar 4.0 bar 0 0
At equilibrium 4.0p 4.0p p p
It is given that K_{p} = 10.1
Now,
Hence, at equilibrium, the partial pressure of H_{2} will be 3.04 bar.
Q32 Predict which of the following reaction will have appreciable concentration of reactants and products: a) Cl2 (g) ↔ 2Cl (g) Kc = 5 ×10–39 b) Cl2 (g) + 2NO (g) ↔ 2NOCl (g) Kc = 3.7 × 108 c) Cl2 (g) + 2NO2 (g) ↔ 2NO2Cl (g) Kc = 1.8 Ans: If the value of K_{c} lies between 10^{3} and 10^{3}, a reaction has appreciable concentration of reactants and products.
Thus, the reaction given in (c) will have appreciable concentration of reactants and products.
Q33 The value of Kc for the reaction 3O2 (g) ↔ 2O3 (g) is 2.0 ×10–50 at 25°C. If the equilibrium concentration of O2 in air at 25°C is 1.6 ×10–2, what is the concentration of O3? Ans: The given reaction is:
3O_{2} (g) ↔ 2O_{3} (g)
Then, Kc = [O_{3} (g)]^{2} / [O_{2} (g)]^{3}
It is given that Kc = 2.0 ×10^{–50 }and O_{2} (g) = 1.6 ×10^{–2}
Then we have,
2.0 ×10^{–50} = [O_{3} (g)]^{2 }/ [1.6 ×10^{–2 }]^{3}
⇒ [O_{3} (g)]^{2} = [2.0 ×10^{–50}] x [1.6 ×10^{–2 }]^{3}
⇒ [O_{3} (g)]^{2 }= 8.192 x 10^{56}
⇒ O_{3} (g) = 2.86x10^{28 }M
Hence, the concentration of O_{2} (g) = 2.86x10^{28 }M
Q34 The reaction, CO(g) + 3H2(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, Kc for the reaction at the given temperature is 3.90. Ans: Let the concentration of methane at equilibrium be x.
CO(g) + 3H_{2}(g) ↔ CH_{4}(g) + H_{2}O(g)
At equilibrium 0.3/1 M 0.1/1 M x 0.02/1 M
It is given that K_{c}= 3.90.
Therefore,
Kc = [CH_{4}(g)] [ H_{2}O(g)] / [ CO(g) ] [H_{2}(g)]^{3}
⇒ [x] [0.02] / (0.3) (0.1)^{3} = 3.90
⇒ x = (3.90) (0.3) (0.1)^{3 }/ [0.02]
⇒ x = 0.00117 / 0.02
= 0.0585 M
= 5.85 x 10^{2}
Hence, the concentration of CH_{4} at equilibrium is 5.85 × 10^{2} M.
Q35 What is meant by the conjugate acidbase pair? Find the conjugate acid/base for the following species: HNO2, CN– , HClO4, F –, OH–, CO2–3 and S Ans: A conjugate acidbase pair is a pair that differs only by one proton. The conjugate acidbase for the given species is mentioned in the table below.
Species Conjugate acidbase HNO_{2} NO^{}_{2 }(base) CN^{–} HCN (Acid) HClO_{4} ClO^{}_{4} (base) F^{ –} HF (Acid) OH^{–} H_{2}O(Acid) / O^{2} (base) CO^{2–}_{3} HCO^{–}_{3 }(Acid) S^{} HS^{ }(Acid) Q36 Which of the followings are Lewis acids? H2O, BF3, H+ and NH+4 Ans: Lewis acids are those acids which can accept a pair of electrons. For example, BF_{3}, H^{+}, and NH^{+}_{4} are Lewis acids.
Q37 What will be the conjugate bases for the Brönsted acids: HF, H2SO4 and HCO3? Ans: The table below lists the conjugate bases for the given Bronsted acids.
Bronsted acid Conjugate base HF F^{} H_{2}SO_{4} HSO^{}_{4} HCO_{3} CO^{2}_{3} Q38 Write the conjugate acids for the following Brönsted bases: NH–2 , NH3 and HCOO–. Ans: The table below lists the conjugate acids for the given Brönsted bases.
Bronsted base Conjugate acid NH^{–}_{2} NH_{3} NH_{3 } NH^{+}_{4} HCOO^{–} HCOOH Q39 The species: H2O, HCO–3, HSO4 and NH3 can act both as Brönsted acids and bases. For each case give the corresponding conjugate acid and base. Ans: The table below lists the conjugate acids and conjugate bases for the given species.
Species Conjugate acid Conjugate base H_{2}O H_{3}O^{+} OH^{} HCO^{–}_{3} H_{2}CO_{3} CO^{2}_{3} HSO^{}_{4} H_{2}SO_{4} SO^{2}_{4} NH_{3} NH^{+}_{4} NH^{}_{2} Q40 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 Ans: (a) OH^{ }is a Lewis base since it can donate its lone pair of electrons.
(b) F^{ i}s a Lewis base since it can donate a pair of electrons.
(c) H^{+} is a Lewis acid since it can accept a pair of electrons.
(d) BCl_{3} is a Lewis acid since it can accept a pair of electrons.
Q41 The concentration of hydrogen ion in a sample of soft drink is 3.8 × 10–3 M. what is its pH? Ans: Given that:
[H^{+}] = 3.8 × 10^{–3}
∴ pH value of soft drink
= log [H^{+}]
= log [3.8 × 10^{–3}]
= log [3.8]  log[10^{–3}]
= log [3.8] + 3
= 0.58 + 3
= 2.42
Q42 The pH of a sample of vinegar is 3.76. Calculate the concentration of hydrogen ion in it. Ans: Given that:
pH = 3.76
It is known that,
pH = log [H^{+}]
⇒ log [H^{+}] = pH
⇒ [H^{+}] = antilog (pH)
= antilog(3.76)
= 1.74x10^{4} M
Hence, the concentration of hydrogen ion in the given sample of vinegar is 1.74 × 10^{4} M.
Q43 The ionization constant of HF, HCOOH and HCN at 298K are 6.8 × 10–4, 1.8 × 10–4 and 4.8 × 10–9 respectively. Calculate the ionization constants of the corresponding conjugate base. Ans: It is known that,
K_{b} = K_{w} / K_{a}
Given
K_{a} of HF = 6.8 × 10^{4}
Hence, K_{b} of its conjugate base F = K_{w} / K_{a}
= 10^{14} / 6.8 × 10^{4}
= 1.5 x 10^{11}
Given, K_{a} of HCOOH = 1.8 × 10^{4}
Hence, K_{b} of its conjugate base HCOO^{ }= K_{w} / K_{a}
= 10^{14} / 1.8 × 10^{4}
= 5.6x10^{11}
Given, K_{a} of HCN = 4.8 × 10^{9}
Hence, Kb of its conjugate base CN^{ }= K_{w} / K_{a}
= 10^{14} / 4.8 × 10^{9}
= 2.8 x 10^{6}
Q44 The ionization constant of phenol is 1.0 × 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? Ans: Ionization of phenol:
C_{6}H_{5}OH + H_{2}O ↔ C_{6}H_{5}O^{} + H_{3}O^{+}
Initial Conc. 0.05 0 0
At equilibrium 0.05x x x
Ka = [C_{6}H_{5}O^{} x H_{3}O^{+}] / C_{6}H_{5}OH
Ka = x x / 0.05x
Ka = x^{2} / 0.05x
As the value of the ionization constant is very less, x will be very small. Thus,
we can ignore x in the denominator
Now, let α be the degree of ionization of phenol in the presence of 0.01 M C_{6}H_{5}ONa.
Q45 The first ionization constant of H2S is 9.1 × 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 × 10–13, calculate the concentration of S2– under both conditions. Ans: (i) To calculate the concentration of HS^{} ion:
Case I (in the absence of HCl):
Let the concentration of HS^{} be x M.
H_{2}S ↔ H^{+} + HS^{}
C_{i} 0.1 0 0
C_{f } 0.1x x x
Then K_{a1} = [H^{+ }] [ HS^{}] / H_{2}S
9.1 × 10^{–8 }= xx / 0.1x
(9.1 × 10^{–8}) (0.1x) = x^{2}
Taking 0.1  x M ; 0.1M, we have
(9.1 × 10^{–8}) (0.1) = x^{2}
9.1 x 10^{9 }= x^{2}
Case II (in the presence of HCl):
In the presence of 0.1 M of HCl, let [HS^{}] be y M.
Then, H_{2}S ↔ H^{+} + HS^{}
C_{i} 0.1 0 0
C_{f } 0.1y y y
Also, HCI ↔ H^{+} + CI^{}
0.1 0.1
Now, K_{a1} = [H^{+ }] [ HS^{}] / H_{2}S
K_{a1} = [y] [0.1+y] / [0.1y]
9.1 × 10^{–8 }= y x 0.1 / 0.1 (∵ 0.1y; 0.1M) (and 0.1+y; 0.1M)
9.1 × 10^{–8 }= y
⇒ [ HS^{}] = 9.1 × 10^{–8}
(ii) To calculate the concentration of [S^{2}]:
Case I (in the absence of 0.1 M HCl):
HS^{ } ↔ H^{+} + S^{2}
HS^{ } = 9.54 x 10^{5} M (From first ionization, case I)
Let S^{2 }be X.
Also, [H^{+}] = 9.54 x 10^{5} M (From first ionization, case I)
K_{a2 } = (9.54 x 10^{5}) (X) / (9.54 x 10^{5})
1.2x1013 = X = S^{2}
Case II (in the presence of 0.1 M HCl):
Again, let the concentration of HS^{} be X' M.
Q46 The ionization constant of acetic acid is 1.74 x 105. 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. Ans: 1) CH_{3}COOH ↔ CH_{3}COO^{} + H^{+} K_{a} = 1.74x10^{5}
2) H_{2}O + H_{2}O ↔ H_{3}O^{+} + OH^{} K_{w} = 1.0x10^{14}
Since K_{a} > K_{w}
CH_{3}COOH + H_{2}O ↔ CH_{3}COO^{} + H_{3}O^{+}
Ci = 0.05 0 0
0.05 0.05α 0.05α 0.05α
k_{a} = 0.05α x 0.05α / 0.05 0.05α
=
0.05α x 0.05α /0.05(1α)= 0.05α^{2} / (1α)
⇒ 1.74x10^{5} = 0.05α^{2} / (1α)
⇒ 1.74x10^{5}  1.74x10^{5} α = 0.05α^{2}
⇒ 0.05α^{2} + 1.74x10^{5} α  1.74x10^{5} =0
D = b^{2 } 4ac
= (1.74x10^{5 })^{2}  4 (0.05) (1.74x10^{5})
= 3.02x10^{25} + 0.348x10^{5}
Method 2:
Degree of Dissociation,
CH_{3}COOH ↔ CH_{3}COO^{} + H^{+}
Thus, concentration of CH_{3}COO^{} = c.α
= 0.05x1.86x10^{2}
= 0.93x10^{2}
=.00093M
Since [oAc] = [H^{+}]
[H^{+}] = .00093 = 0.093x102
pH = log[H^{+}]
= log (0.93x10^{2})
∴ pH = 3.03
Hence, the concentration of acetate ion in the solution is 0.00093 M and its Ph is 3.03.
Q47 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. Ans: Let the organic acid be HA.
⇒ HA ↔ H^{+ } + A^{}
Concentration of HA = 0.01 M
pH = 4.15
log[H^{+}] = 4.15
[H^{+}] = 7.08x10^{5}
Now, K_{a} = [H^{+}] [A^{}] / [HA]
[H^{+}] = [A^{}] = 7.08x10^{5}
[HA] = 0.01
Then,
K_{a} = 7.08x10^{5 }x 7.08x10^{5 }/ 0.01
K_{a} = 5.01x10^{7}
pK_{a }= log K_{a}
= log(5.01x10^{7})
pK_{a} = 6.3001
Q48 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 Ans: (i) 0.003MHCl:
H_{2}O + HCl ↔ H3O^{+} + Cl^{}
Since HCl is completely ionized,
[H_{3}O^{+}] = [ HCl]
⇒ [H_{3}O^{+}] = = 0.003
Now
pH = log [H_{3}O^{+}] = log (0.003)
= 2.52
Hence, the pH of the solution is 2.52.
(b) 0.005 M NaOH
NaOH_{(aq)} ↔ Na^{+}_{(aq)} + HO^{}_{(aq)}
[NaOH] = [ HO^{}]
⇒ [ HO^{}] = 0.05
pOH = log[ HO^{}] = log (0.05)
= 2.30
∴ pH = 14  2.30 = 11.70
Hence, the pH of the solution is 11.70.
(c) 0.002 M HBr
HBr + H_{2}O ↔ H_{3}O^{+} + Br^{}
[HBr] = [H_{3}O^{+}]
⇒ [H_{3}O^{+}] = 0.002
∴ pH = log [H_{3}O^{+}] = log (0.002)
= 2.69
Hence, the pH of the solution is 2.69.
(d) 0.002 M KOH
KOH_{(aq)} ↔ K^{+}_{(aq) } + OH^{}_{(aq)}
[OH^{}] = [KOH]
⇒ [OH^{}] = 0.002
Now pOH = log[OH^{}] = log (0.002)
= 2.69
∴ pH = 142.69 = 11.31
Hence, the pH of the solution is 11.31.
Q49 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. Ans: For 2g of TlOH dissolved in water to give 2 L of solution:
[TIOH(aq)] = 2/2 g/L
= 2/2 x 1/221 M
= 1/221 M
TIOH_{(aq)} → TI^{+}_{(aq)} + OH^{}_{(aq)}
OH^{}_{(aq) }= TIOH_{(aq) }= 1/221M
Kw = [H^{+}] [OH^{}]
10^{14 }= [H^{+}] [1/221]
[H^{+}] = 221x10^{14}
⇒ pH = log [H^{+}] = log ( 221x10^{14})
= 11.65
(b) For 0.3 g of Ca(OH)_{2} dissolved in water to give 500 mL of solution:
Ca(OH)_{2} → Ca^{2+} + 2OH^{}
[Ca(OH)_{2}] = 0.3x1000/500 = 0.6M
OH^{}_{(aq)} = 2 x [Ca(OH)_{2(aq)}] = 2 x 0.6 = 1.2M
[H+] = Kw / OH^{}_{(aq) }
= 10^{14}/1.2 M
= 0.833 x 10^{14}
pH = log(0.833 x 10^{14})
= log(8.33 x 10^{13})
= (0.902 + 13)
= 12.098
(c) For 0.3 g of NaOH dissolved in water to give 200 mL of solution:
NaOH → Na^{ +}_{(aq)} + OH^{}_{(aq)}
[NaOH] = 0.3 x 1000/200 = 1.5M
[OH^{}_{(aq)}] = 1.5M
Then [H+] = 10^{14 }/ 1.5
= 6.66 x 10^{13}
pH = log ( 6.66 x 10^{13})
= 12.18
(d) For 1mL of 13.6 M HCl diluted with water to give 1 L of solution:
13.6 x 1 mL = M_{2 }x 1000 mL
(Before dilution) (after dilution)
13.6 x 10^{3} = M_{2} x 1L
M_{2} = 1.36 x 10^{2}
[H^{+}] = 1.36 × 10^{2}
pH =  log (1.36 × 10^{2})
= ( 0.1335 + 2)
= 1.866 = 1.87
Q50 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. Ans: Degree of ionization, α = 0.132
Concentration, c = 0.1 M
Thus, the concentration of H_{3}O^{+} = c.α
= 0.1 × 0.132
= 0.0132
pH = log [H^{+}]
= log (0.0132)
= 1.879 : 1.88
Now,
K_{a} = Cα^{2}
= 0.1 x (0.132)^{2}
K_{a} = 0.0017
pK_{α} = 2.75
Q51 The pH of 0.005M codeine (C18H21NO3) solution is 9.95. Calculate its ionization constant and pKb. Ans: c = 0.005
pH = 9.95
pOH = 4.05
pH = log (4.05)
4.05 =  log [OH^{}]
[OH^{}] = 8.91x 10^{5}
cα = 8.91x 10^{5}
α = 8.91x 10^{5 }/ 5 x 10^{3} = 1.782 x 10^{2}
Thus , K_{b} = cα^{2}
= 0.005 x (1.782)^{2} x 10^{4}
= 0.005 x 3.1755 x 10^{4}
= 0.0158 x 10^{4}
K_{b} = 1.58 x 10^{6}
PK_{b} = logK_{b}
= log (1.58 x 10^{6})
= 5.80
Q52 What is the pH of 0.001 M 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. Ans: K_{b} = 4.27 x 10^{10}
c = 0.001M
pH = ?
α = ?
K_{b} = cα^{2}
4.27 x 10^{10 } = 0.001 x α^{2}
4270 x 10^{10 }= α^{2}
α = 65.34 x 10^{4}
Then (anion) = cα = 0.001 x 65.34 x 10^{4}
= 0.65 x 10^{5}
pOH = log ( 0.65 x 10^{5})
= 6.187
pH = 7.813
Now
K_{a} x K_{b} = Kw
∴ 4.27 x 10^{10} x Ka = Kw
K_{a} = 10^{14} / 4.27 x 10^{10}
= 2.34 x 10^{5}
Thus, the ionization constant of the conjugate acid of aniline is 2.34 × 10^{5}.
Q53 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) 0.01 M (b) 0.1 M in HCl? Ans: c = 0.05 M
pK_{a }= 4.74
pK_{a }= log (K_{a})
K_{a }= 1.82 x 10^{5}
When HCl is added to the solution, the concentration of H^{+} ions will increase. Therefore, the equilibrium will shift in the backward direction i.e., dissociation of acetic acid will decrease.
Case I: When 0.01 M HCl is taken.
Let x be the amount of acetic acid dissociated after the addition of HCl.
CH_{3}COOH ↔ H^{+} + CH_{3}COO^{}
Initial Conc. 0.05M 0 0
After dissociation 0.05x 0.01+x x
As the dissociation of a very small amount of acetic acid will take place, the values i.e., 0.05  x and 0.01 + x can be taken as 0.05 and 0.01 respectively.
K_{a} = [CH_{3}COO^{}] [ H^{+}] / [CH_{3}COOH]
∴ K_{a} = (0.01) (x) / (0.05)
x = 1.82 x 10^{5} x(multiply) 0.05 / 0.01
x = 1.82 x 10^{3} x(multiply) 0.05
Now
α = Amount of acid dissociated / amount of acid taken
= 1.82 x 10^{3 }x 0.05 / 0.05
= 1.82 x 10^{3}
Case II: When 0.1 M HCl is taken.
Let the amount of acetic acid dissociated in this case be X. As we have done in the first case, the concentrations of various species involved in the reaction are:
[CH_{3}COOH] = 0.05  X : 0.05M
[CH_{3}COO^{}] = X
[ H^{+}] = 0.1+X ; 0.1M
K_{a} = [CH_{3}COO^{}] [ H^{+}] / [CH_{3}COOH]
∴ K_{a} = (0.1) (X) / (0.05)
x = 1.82 x 10^{5} x(multiply) 0.05 / 0.1
x = 1.82 x 10^{4} x(multiply) 0.05
Now
α = Amount of acid dissociated / amount of acid taken
= 1.82 x 10^{4 }x 0.05 / 0.05
= 1.82 x 10^{4}
Q54 The ionization constant of dimethylamine is 5.4 x 104. Calculate its degree of ionization in its 0.02 M solution. What percentage of dimethylamine is ionized if the solution is also 0.1 M in NaOH? Ans: Kb = 5.4x104
c = 0.02M
Now, if 0.1 M of NaOH is added to the solution, then NaOH (being a strong base) undergoes complete ionization.
NaOH_{(aq)} ↔ Na^{+}_{(aq) } + OH^{} _{(aq) }
0.1M 0.1M
and
(CH_{3})_{2} NH + H_{2}O ↔ (CH_{3})_{2} NH^{+}_{2} + O^{}H
0.02x x x
Then (CH_{3})_{2} NH^{+}_{2 }= x
[OH^{}] = x + 0.1 ; 0.1
⇒ K_{b} = [(CH_{3})_{2} NH^{+}_{2} ] [OH^{}] / [CH_{3})_{2} NH]
5.4x104 = x x 0.1 / 0.02
x = 0.0054
It means that in the presence of 0.1 M NaOH, 0.54% of dimethylamine will get dissociated.
Q55 Calculate the hydrogen ion concentration in the following biological fluids whose pH are given below: (a) Human musclefluid, 6.83 (b) Human stomach fluid, 1.2 (c) Human blood, 7.38 (d) Human saliva, 6.4. Ans: a) Human muscle fluid 6.83:
pH = 6.83
pH =  log [H^{+}]
∴6.83 =  log [H^{+}]
[H^{+}] =1.48 x 10^{7} M
(b) Human stomach fluid, 1.2:
pH =1.2
1.2 =  log [H^{+}]
∴[H^{+}] = 0.063
(c) Human blood, 7.38:
pH = 7.38 =  log [H^{+}]
∴ [H^{+}] = 4.17 x 10^{8} M
(d) Human saliva, 6.4:
pH = 6.4
6.4 =  log [H^{+}]
[H^{+}] = 3.98 x 10^{7}
Q56 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. Ans: The hydrogen ion concentration in the given substances can be calculated by using the given relation:
pH = log [H^{+}]
(i) pH of milk = 6.8
Since, pH = log [H^{+}]
6.8 = log [H^{+}]
log [H^{+}] = 6.8
[H^{+}] = anitlog(6.8)
= 1.5x107M
(ii) pH of black coffee = 5.0
Since, pH = log [H^{+}]
5.0 = log [H^{+}]
log [H^{+}] = 5.0
[H^{+}] = anitlog(5.0)
= 10^{5}M
(iii) pH of tomato juice = 4.2
Since, pH =log [H^{+}]
4.2 =log [H^{+}]
log [H^{+}] = 4.2
[H^{+}] = anitlog(4.2)
= 6.31x10^{5}M
(iv) pH of lemon juice = 2.2
Since, pH = log [H^{+}]
2.2 = log [H^{+}]
log [H^{+}] = 2.2
[H^{+}] = anitlog(2.2)
=6.31x10^{3}M
(v) pH of egg white = 7.8
Since, pH = log [H^{+}]
7.8 = log[H^{+}]
log[H^{+}] = 7.8
[H^{+}] = anitlog(7.8)
= 1.58x10^{8}M
Q57 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? Ans: [KOH_{(aq)}] = 0.561 / (1/5)g/L
= 2.805 g/L
= 2.805 x 1/56.11 M
= 0.05M
KOH_{(aq)} → K^{+}_{(aq)} + OH^{}_{(aq) }
[OH^{}] = 0.05M = [K^{+}]
[H^{}] [H^{+}] = K_{w}
[H^{+}] K_{w} / [OH^{}]
= 10^{14} / 0.05 = 2x10^{13} M
∴ pH = 12.70
Q58 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. Ans: Solubility of Sr(OH)_{2} = 19.23 g/L
Then, concentration of Sr(OH)_{2}
= 19.23 / 121.63 M
= 0.1581 M
Sr(OH)_{2(aq) }→ Sr^{2+}_{(aq) }+ 2 (OH^{})_{(aq)}
∴Sr^{2+ }= 0.1581M
[OH^{}] = 2 x 0.1581M = 0.3126 M
Now
Kw = [OH^{}] [H^{+}]
10^{14} / 0.3126 = [H^{+}]
⇒ [H^{+}] = 3.2 x 10^{14}
∴ pH = 13.495 = 13.50
Q59 The ionization constant of propanoic acid is 1.32 x 105. 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? Ans: Let the degree of ionization of propanoic acid be α.
Then, representing propionic acid as HA, we have:
HA + H_{2}O ↔ H_{3}O^{+} + A^{}
(0.050.0α) ≈ 0.05 0.05α 0.05α
K_{a } = [H_{3}O^{+}] [A^{}] / [HA]
= (0.05α)(0.05α) / 0.05
= 0.05 α^{2}
Then, [H_{3}O^{+}] = 0.05α = 0.05 x 1.63 x 10^{2} = Kb . 15 x 10^{4} M
∴ pH = 3.09
In the presence of 0.1M of HCl, let α' be the degree of ionization.
Then, [H_{3}O^{+}] = 0.01
[A^{}] = 0.05α'
[HA] = 0.05
Ka = 0.01 x 0.05α' / 0.05
⇒ 1.32 x 10^{5} = 0.1 α'
α' = 1.32 x 10^{3}
Q60 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. Ans: c = 0.1 M
pH = 2.34
log [H^{+}] = pH
log [H^{+}] = 2.34
[H^{+}] = 4.5 x 10^{3}
also
[H^{+}] = cα
4.5 x 10^{3 }= 0.1 α
α = 4.5 x 10^{3} / 0.1
α = 45 x 10^{3 }
Then,
K_{a} = cα^{2}
= 0.1 (45 x 10^{3})^{2}
= 202.5 x 10^{6}
= 2.02 x 10^{4}
Q61 The ionization constant of nitrous acid is 4.5 x 104. Calculate the pH of 0.04 M sodium nitrite solution and also its degree of hydrolysis. Ans: NaNO_{2} is the salt of a strong base (NaOH) and a weak acid (HNO_{2}).
NO^{}_{2} + H_{2}O ↔ HNO_{2} + OH^{}
K_{h} = [ HNO_{2} ] [ OH^{}] / [NO^{}_{2}]
⇒ K_{w} / k_{a} = 10^{14} / 4.5 x 10^{4 }= 0.22 x 10^{10}
Now, If x moles of the salt undergo hydrolysis, then the concentration of various species present in the solution will be:
[NO^{}_{2} ] = 0.04  x ; 0.04
[ HNO_{2} ] = x
[ OH^{}] = x
K_{h} = x^{2} / 0.04 = 0.22 x 10^{10}
x^{2} = 0.0088 x 10^{10}
x = 0.093 x 10^{5}
∴ [ OH^{}] = 0.093 x 10^{5}M
[H_{3}O^{+}] = 10^{14} / 0.093 x 10^{5 }= 10.75 x 10^{9}M
⇒ pH = log(10.75 x 10^{9})
= 7.96
Therefore, degree of hydrolysis
= x / 0.04 = (0.093 x 10^{5} ) / 0.04 = 2.325 x 10^{5}
Q62 A 0.02 M solution of pyridinium hydrochloride has pH = 3.44. Calculate the ionization constant of pyridine Ans: Given, pH = 3.44
We know that
pH = log [H^{+}]
∴ [H^{+}] = 3.63x10^{4}
Then K_{h} = ( 3.63x10^{4 })^{2} / 0.02 (∵ Concentration is 0.02M)
⇒ K_{h} = 6.6 x 10^{6}
Now , K_{h} = K_{w} / K_{a }
⇒ K_{a }= K_{w} / K_{h }
= 10^{14} / 6.6 x 10^{6}
= 1.51 x 10^{9}
Q65 Ionic product of water at 310 K is 2.7 x 1014. What is the pH of neutral water at this temperature? Ans: Ionic product,
K_{w} = [H^{+}] [ OH^{}]
Let [H^{+}] = x.
Since [H^{+}] = [ OH^{}] , K_{w} = x^{2}
⇒ K_{w} at 310K is 2.7x 10^{14}
∴ 2.7x 10^{14} = x^{2}
⇒ x = 1.64 x 10^{7}
⇒ [H^{+}] = 1.64 x 10^{7}
⇒ pH = log [H^{+}]
= log (1.64 x 10^{7})
= 6.78
Hence, the pH of neutral water is 6.78.
Q72 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 106). Ans: CaSO_{4(s)} ↔ Ca^{ 2+} _{(aq)} + SO^{2}_{4(aq)}
K_{sp } = [ Ca^{ 2+ }] [ SO^{2 }]
Let the solubility of CaSO_{4} be s.
Then, K_{sp } = s^{2}
9.1 x 10^{6 }= s^{2}
s = 3.02 x 10^{3} mol/L
Molecular mass of CaSO_{4} = 136 g/mol
Solubility of CaSO_{4} in gram/L = 3.02 × 10^{3} × 136 = 0.41 g/L
This means that we need 1L of water to dissolve 0.41g of CaSO_{4}
Therefore, to dissolve 1g of CaSO_{4} we require = 1/0.41 L = 2.44 Lof water.