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NCERT Class 11 Chemistry Chapter 6 Equilibrium
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Equilibrium
Chapter: 6
| Part – I |
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?
Ans: On increasing volume of the container, the vapour pressure will initially decrease because the same amount of vapours are now distributed over a larger space.
(b) How do rates of evaporation and condensation change initially?
Ans: On increasing the volume of the container the rate of evaporation will increase initially because now more space is available. Since the amount of the vapour per unit volume decreases on increasing the volume, therefore, the rate of condensation will decrease initially.
(c) What happens when equilibrium is restored finally and what will be the final vapour pressure?
Ans: Equilibrium will be achieved when the rates of the forward and reverse processes become equal. At this point, the vapour pressure will remain constant because it is dependent on the temperature, not on the volume of the container.
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?
2SO2(g) + O2(g) ⇌ 2SO3(g)
Ans: To find the equilibrium constant Kc for the reaction:
3. At a certain temperature and total pressure of 105 Pa, iodine vapour contains 40% by volume of I atoms.
I2 (g) ⇌ 2I (g)
Calculate Kp for the equilibrium.
Ans: The total pressure of the equilibrium mixture is 105 Pa.
Partial pressure of iodine atoms (I)
= 0.4 × 105 Pa
Partial pressure of iodine molecules (I2)
4. Write the expression for the equilibrium constant, Kc for each of the following reactions:
(i) 2NOCl (g) ⇌ 2NO (g) + Cl2 (g)
Ans:
(ii) 2Cu(NO3) 2 (s) ⇌ 2CuO (s) + 4NO2 (g) + O2 (g)
Ans:
(iii) CH3COOC2H5(aq) + H2O(I) ⇌ CH3COOH (aq) + C2H5OH (aq)
Ans:
(iv) Fe3+ (aq) + 3OH– (aq) ⇌ Fe(OH)3 (s)
Ans:
(v) I2 (s) + 5F2 ⇌ 2IF5
Ans:
5. 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 atm at 500 K.
Ans: Kp and Kc are related to each other as Kp = Kc (RT)Δng
(ii) CaCO3(s) ⇌ CaO(s) + CO2(g), Kp = 167 atm at 1073 K.
Ans:
6. 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: For the reverse reaction
7. Explain why pure liquids and solids can be ignored while writing the equilibrium constant expression?
Ans: Pure liquids or solids are excluded from the equilibrium constant expression because their molar concentrations remain constant, regardless of the quantity present.
8. Reaction between N2 and V2 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 the equilibrium mixture.
Ans: Let x moles of N2(g) participate in the reaction. According to the balanced chemical equation, x/2 moles of O2(g) will react to form x moles of N2O(g).
The molar concentration per litre of different species before the reaction and at the equilibrium point is:
The equilibrium constant value of (2.0 ×10 – 37) is extremely small. This means that only small amounts of reactants have reacted. Therefore, x is extremely small and can be omitted as far as the reactants are concerned.
Applying Law of chemical equilibrium
x2 = 43.352 × 10– 40
Or x = 6.6 × 10– 20
As x is extremely small, it can be neglected. Thus, in the equilibrium mixture
Molar conc. of N2 = 0.0482 mol L-1
Molar conc. of O2 = 0.0933 mol L-1
Molar conc. of N2O = 0.1 × x
= 0.1 × 6.6 × 10 – 22 mol L-1
= 6.6 × 10 – 21 mol L-1
9. 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: 2NO(g) +Br2(g) ⇌ 2NOBr(g)
According to the equation, 2 moles of N0(g) react with 1 mole of Br2(g) to form 2 moles NOBr(g).
The number of moles of NOBr(g) formed at equilibrium = 0.0518 mol
The number of moles of NO(g) that reacted is 0.0518 mol.
The number of moles of NO(g) remaining at equilibrium is:
0.087mol−0.0518mol = 0.0352mol
The number of moles of Br2(g) that reacted is:
1/2 × 0.0518mol = 0.0259mol
The number of moles of Br2(g) taking part in reaction = 1/2 × 0.0518 = 0.0259mol
The number of moles of Br2(g) left at equilibrium
= 0.0437 – 0.0259 = 0.0178mol
The initial and equilibrium molar concentrations of the different species can be represented as follows:
2NO(g) + Br2(g) ⇌ 2NOBr(g)
| Initial moles | 0.087 | 0.0437 | 0 |
| Moles at eqm. Point | 0.0352 | 0.0178 | 0.0518 |
10. 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:
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?
2HI(g) ⇌ H2(g) + I2(g)
Ans:
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 given as follows:
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: N2 (g) + 3H2 (g) ⇌ 2NH3(g)
Concentration quotient
Given that the equilibrium constant Kc for the reaction is 1.7 × 10−2 and Qc is not equal to Kc, this indicates that the system is not at equilibrium.
Since Qc > Kc the net reaction will shift in the reverse direction. This shift will reduce the concentration of NH3 and increase the concentrations of N2 and H2, moving the system towards equilibrium.
13. The equilibrium constant expression for a gas reaction is,
Write the balanced chemical equation corresponding to this expression.
Ans: The balanced chemical equation for the reaction is:
4NO(g) + 6H2O(g) ⇌ 4NH3(g) + SO2(g)
14. One mole of H2O and one mole of CO are taken in 10 L vessel and heated to 725K. 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: Number of moles of water originally present = 1 mol
Percentage of water reacted = 40%
Number of moles of water reacted = (1 × 40)/100 = 0.4
Number of moles of water left (10.4) 0.6 mol.
According to the reaction equation, 0.4 moles of water will react with 0.4moles of carbon monoxide to produce 0.4 moles of hydrogen and 0.4 moles of carbon dioxide.
15. 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: The reverse reaction can be written as:
HI(g) ⇌ H2(g) + I2(g) K1c = 1/54.8
16. 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: Given the equilibrium reaction:
2ICl(g) ⇌ I2(g)+Cl2(g)
Suppose at equilibrium, the molar concentrations of I2(g) and
Cl2(g) are both x mol/L.
| Initial molar conc | 0.78M | 0 | 0 |
| Wqm molar conc | (0.78-2x)M | x | x |
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?
C2H6(g) ⇌ C2H4(g) + H2(g)
Ans: The equilibrium in the reaction is
C2H6(g) ⇌ C2H4(g) + H2(g)
| Initial Pressure | 4 atm | 0 | 0 |
| Eqn. Pressure | (4-p) atm | P atm | P atm |
Or p2 = 0.04 (4-p)
Or b2+ 0.04p – 0.16 = 0
Or 0.04 = b2 / (4-p)
18. 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)
Ans:
(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.
Ans:
(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:
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 × 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 initial molar concentration of PCl5 per litre = x mol.
Molar concentration of PCI, at equilibrium = 0.05 mol
Moles of PCl3, decomposed = (x – 0.05) mol
Moles of PCI, formed = (x – 0.05) mol
Moles of Cl2 formed = (x – 0.05) mol
The molar concentrations of reactants and products per litre, both before the reaction and at the point of equilibrium, are as follows:
PCI5 ⇌ PCl3 + C2
| Initial moles/litres | x | 0 | 0 |
| Moles/litre at eqm. point | 0.05 | (x-0.05) | (x-0.05) |
Equilibrium constant, Kc
= 8.3 × 10-3
= 0.0083
Applying Law of Chemical equilibrium,
(x – 0.05)2 = 0.0083 × 0.05
= 4.15 × 10-4
(x – 0.05) = (4.15 × 10.-4)1/2
= 2.037 × 10-2 = 0.02 moles
Or x = 0.05 + 0.02 = 0.07 mol
The molar concentration of PCl₃ per litre at equilibrium = (0.07 -0.05) = 0.02 mol.
The molar concentration per litre of CI2 at eqm = 0.07- 0.05 = 0.02 mol.
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.
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 = 0.80 atm?
Ans: The reaction will shift in the backward direction to reach equilibrium. As a result, the partial pressure of CO₂ will decrease, while the partial pressure of CO will increase, allowing the system to re-establish equilibrium. Let the decrease in the partial pressure of CO₂ be denoted as p atm. Consequently, the partial pressure of CO will increase by the same amount, p atm.
pCO2 = (0.8-p) atm,
pCO(g) = (1.4 + p) atm
At equilibrium,
Kp = pCO2 / pCO
= (0.8 – p) atm / (1.4 + p) atm
= (0.8 – p) / (1.4 + p)
= 0.265 = (0.8 – p) / (1.4 + p)
0.371 + 0.265 p = 0.8 – p
Or 1.265 p = 0.8 -0.371 = 0.429
P = 0.429 /1.265 = 0.339 atm
(p co) eq = (1.4 + 0339)
= 17.39 atm
(pco) eq (0.8 + 0.339) = 0.461 atm.
21. 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:
Since the value of Qc is less than Kc (0.061), the reaction is not at equilibrium. It will proceed in the forward direction until Qc matches Kc.
22. 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 x moles of BrCl decompose to reach equilibrium. The initial molar concentrations and the molar concentrations at equilibrium for the different species can be represented as follows.
On taking the square root,
5.656 = x/2 (0.0033-x)
= x /(0.0033-x)
= 11.31
= 12.31x
= 0.037
X = 0.037 / 12.31 = 0.003
Therefor, moral concentration of BrCI at equilibrium point.
= 0.0033 – 0.003
= 0.0003 mol L-1
= 3 × 10-4 mol L-1.
23. At 1127 K and 1 atm pressure, a gaseous mixture of CO and CO2 in equilibrium with solid carbon has 90.55% CO by mass
C(s) + CO2(g) ⇌ 2CO(g)
Calculate Kc for this reaction at the above temperature.
Ans: Calculate Kc for this reaction:
Let the total mass of the gaseous mixture = 100 g
Mass of CO in the mixture = 90.55 g
Mass of CO2 in the mixture = (100 – 90.55) = 9.45g
No. of moles of CO = 90.55/(28gmo-1)
= 3.234 mol
No of moles of CO2 = 90.55/(44 g mol-1)
= pCO in the mixture = (3.234 mol) / (3.23 – 0.215) × 1 atm
= pCO2 in the mixture = (3.234 mol) / (3.449mol) × 1 atm
= 0.062 atm
Kp = p2CO/ pCO2
= (0.938 atm)2 / (0.062 atm)
= 14.19 atm.
Calculate Kc:
Kc = Kp / (RT)Δng
Kp = 14.19 atm
R = 0.083/ L atm K-1 mol-1
T = 1127 k
Δng = 2 – 1 = 1
24. Calculate a) ∆G° and (b) the equilibrium constant for the formation of NO2 from NO and O2 at 298 K
NO(g) + ½ O2(g) ⇌ NO2(g)
where
∆f G° (NO2) = 52.0 kJ/mol
∆f G° (NO) = 87.0 kJ/mol
∆f G° (O2) = 0 kJ/mol
Ans: (a) ΔG° = Δf Gо° (NO)
= 5.20 – (87 + 0)
= – 35 Kj mol_1
= (b) ΔGо° = – 2.303 RT log Kc
= 6.134
= Kc = Antilog 6.314
= 1.36 × 106
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?
(a) PCl5(g) ⇌ PCl3 (g) + Cl2(g)
Ans: In the forward reaction, the pressure will increase, and the number of moles of products will increase.
(b) CaO (s) + CO2(g) ⇌ CaCO3(s)
Ans: In the backward reaction, the pressure will increase, and the number of moles of products will decrease.
(c) 3Fe(s) + 4H2O(g) ⇌ Fe3O4(s) + 4H2(g)
Ans: The change in pressure will have no effect on the equilibrium constant and there will be no change in the number of moles.
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?
Ans: By increasing the pressure in reactions where the number of moles of gaseous reactants and products are different (np≠nrn_p \neq n_rnp=nr), except for reactions where the number of moles on both sides are the same, the following general effects occur:
(a) The reaction will shift to the left if np > nr.
(b) The reaction will shift to the right if np < nr.
(i) COCl2(g) ⇌ CO(g) + Cl2(g)
Ans: An increase in pressure will favour the backward reaction because np = 2 (moles of products) is greater than nr = 1 (moles of reactants).
(ii) CH4(g) + 2S2(g) ⇌ CS2(g) + 2H2S(g)
Ans: An increase in pressure will not affect the equilibrium because np = nr = 3 (moles of products and reactants are equal).
(iii) CO2(g) + C(s) ⇌ 2CO(g)
Ans: Increase in pressure will favour backward reaction because np(2) > nr(1).
(iv) 2H2(g) + CO(g) ⇌ CH3OH(g)
Ans: An increase in pressure will favour the forward reaction because np = 1 (moles of products) is less than nr = 2 (moles of reactants).
(v) CaCO3(s) ⇌ CaO(s) + CO2(g)
Ans: An increase in pressure will favour the backward reaction because np = 1 (moles of products) is greater than nr = 0 (moles of reactants).
(vi) 4 NH3(g) + 5O2(g) ⇌ 4NO(g) + 6H2O(g)
Ans: An increase in pressure will favour the backward reaction because np = 10 (moles of products) is greater than nr = 9 (moles of reactants).
27. 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 1024 K.
Ans: For the equilibrium reaction H2(g) + Br2(g) ⇌ 2HBr(g) he equilibrium constant is K= 1 / 1.6 × 105
Calculation of partial pressure of gases
2HBr(g) ⇌ H2(g) + Br2(g)
| Initial pressure 10 bar | Zero | Zero |
| Eqm. pressure (10-p ) bar | P/2 bar | P/2 bar |
On taking square root,
pH2 = 0.025 bar
pBr2 = 0.025 bar
pHBr = 10-0.05 = 9.95 bar
= 10.0 bar.
28. 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 an expression for Kp for the above reaction.
Ans:
(b) How will the values of Kp and composition of equilibrium mixture be affected by:
(i) increasing the pressure.
Ans: Increasing the pressure will increase the number of moles per unit volume. To counteract this, the equilibrium will shift to the left, or in the backward direction, leading to the formation of more reactants and a decrease in Kp.
(ii) increasing the temperature.
Ans: Increasing the temperature will favour the forward reaction if it is endothermic, according to Le Chatelier’s principle. As a result, the equilibrium will shift to the right, and Kp will increase.
(iii) using a catalyst?
Ans: The addition of a catalyst will not change the equilibrium since it influences both the forward and the backward reactions to the same extent. But it will be attained more quickly.
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)
Ans: (a) The equilibrium will shift in the forward direction.
(b) The equilibrium will shift in the backward direction.
(c) The equilibrium will shift in the backward direction.
(d) Equilibrium will be shifted in the forward direction.
30. At 473 K, equilibrium constant Kc for decomposition of phosphorus pentachloride, PCl5 is 8.3 × 10-3. If decomposition is depicted as,
PCl5(g) ⇌ PCl3(g) + Cl2(g); ∆rH° = 124.0 kJ mol–1
(a) write an expression for Kc for the reaction.
Ans:
(b) what is the value of Kc for the reverse reaction at the same temperature?
Ans:
= 1/ 8.3 × 10-3
= 120.48
(c) what would be the effect on Kc if:
(i) more PCl5 is added.
Ans: By adding more of PCl5, value of Kc will remain constant because there is no change in temperature.
(ii) pressure is increased.
Ans: Increasing the temperature will favour the forward reaction if it is endothermic, leading to an increase in the value of the equilibrium constant.
(iii) the temperature is increased?
Ans: When the temperature is increased, the equilibrium constant increases if the reaction is endothermic. As the temperature rises, Kf (the forward reaction constant) increases, which in turn increases Kc because Kc = Kf / Kr.
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,
CO2(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: Write the equilibrium expression for the water gas shift reaction:
CO(g) + H2O(g) → CO2(g) + H2(g)
| Initial pressure | 4.0 bar | 4.0 bar | 0 | 0 |
| Eqm pressure | (4-p) par | (4-p) bar | P bar | P bar |
or b2 / (4-p) = 0.1
or p / ( 4-p) = (0.1)½ = 0.316
or p = 0.316(4-p)
Or p = 1.264 – 0.316 p
Or 1.316 p = 1.264
32. Predict which of the following reaction will have appreciable concentration of reactants and products:
(a) Cl2 (g) ⇌ 2Cl(g); Kc = 5 × 10–39
Ans: Since the value of Kc is very small, this indicates that the concentrations of the products are much lower compared to the concentrations of the reactants.
(b) Cl2(g) + 2NO(g) ⇌ 2NOCl(g); Kc = 3.7 × 108
Ans: Since the value of Kc is quite large, this means that the molar concentrations of products is very large as compared to that of the reactants.
(c) Cl2(g) + 2NO2(g) ⇌ 2NO2Cl(g); Kc = 1.8
Ans: Since the value of Kc is 1.8, this implies that both the products and reactants have significant concentrations.
33. The value of Kc for the reaction 3O2 (g) 2O3(g) is 2.0 × 10–5 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: 3O2 (g) = 203 (g)
Or [O3]2 = 8.192 × 10-56
Or [O3] = (8.192 × 10-56)½
= 2.86 × 10-28 M.
34. 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 into 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:
35. What is meant by the conjugate acid-base pair? Find the conjugate acid/base for the following species:
HNO2, CN–, HClO4, OH–, CO32-, S2–
Ans: A conjugate acid-base pair consists of two species that differ by the presence or absence of a proton (H⁺). In this pair, the acid loses a proton to form its conjugate base, while the base gains a proton to form its conjugate acid.
Conjugate acid: HCN, H2O, HCO3– ,HS–
Conjugate base: NO2–, CIO4– , O2–
36. Which of the followings are Lewis acids?
H2O, BF3, H+, and NH4+
Ans: BF3, H+, and NH4+ ions are Lewis acids. BF₃ accepts an electron pair due to its electron-deficient nature. H⁺ accepts an electron pair as a proton. NH₄⁺ can accept an electron pair to revert to NH₃, making it a Lewis acid.
37. What will be the conjugate bases for the Brönsted acids:
HF, H2SO4 and HCO3?
Ans: Conjugate bases is F– H2SO4 – and HCO3–
38. Write the conjugate acids for the following Brönsted bases:
NH2–, NH3 and HCOO–.
Ans: NH2, NH4+ and H2CO3.
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.
Ans: Conjugate Acid: H₃O⁺, H₂CO₃, H₂SO₄ and NH₄⁺,
Conjugate Base: OH, CO₃², SO₄², NH₂.
40. Classify the following species into Lewis acids and Lewis bases and show how these act as Lewis acid/base:
(a) OH–
Ans: OH⁻ can donate an electron pair, making it a Lewis base.
(b) F–
Ans: F⁻ can also donate an electron pair, so it is a Lewis base.
(c) H+
Ans: H+ can accept electron pair. Hence it is a Lewis acid.
(d) BCl3.
Ans: BCl₃ is electron-deficient and can accept an electron pair, thus it is a Lewis acid.
41. The concentration of hydrogen ion in a sample of soft drink is 3.8 × 10–3 M. What is its pH?
Ans: pH = – log[H+]
= – log (3.8 × 10–3)
= – log 3.8 + 3
= 3 – 0.5798
= 2.4202
= 2.42
42. The pH of a sample of vinegar is 3.76. Calculate the concentration of hydrogen ion in it.
Ans:
= 1.738 × 10-4 M
= 1.74 × 10-4 M
43. 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: For HF (Hydrofluoric Acid):
Given Ka for HF = 6.8 × 10−4
For HCOOH (Formic Acid):
Given Ka for HCOOH = 1.8 × 10−4
= 5.6 × 10-11
For HCN– (Hydrocyanic Acid):
Given Ka for HCN = 4.8 × 10−9
44. 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?
or x2 / 0.05 = 1.0 × 10-10
or x2 = 5 ×10-12
x = 2.2 × 10-6 M
In the presence of 0.01 C6H5ONa, let y represent the amount of phenol that dissociates. At equilibrium, the concentrations of the species involved will be adjusted accordingly.
[C6H5ON] = 0.05 – y = 0.05
[C6H5O–] = 0.01 + y = 0.01 M
[H=] = yM
45. 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: To calculate [HS-]
Ka = x × x /(0.1)
= 9.1 × 10-8
Or x2 = 9.1 × 10-9
Or x = 9.54 × 10-5
In presence of 0.1 M HCI, let y be the amount of H2S that dissociates.
Then at equilibrium.
To calculate [S2-]
For the overall reaction,
In the absence of 0.1 MHCI, [H+] = 2 [S2-]
In presence of 0.1 M HCI, suppose [S2-] = y, then
[H2S] = 0.1 – y = 0.1 M
[H+] = 0.1 + y = 0.1 M
Ka = (0.1)2 × y /(0.2)
= 1.09 × 10-20
Or y = 1.09 × 10-19 M
46. The ionization constant of acetic acid is 1.74 × 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.
Ans: Let’s calculate the degree of dissociation, the concentration of acetate ions (CH3COO−), and the pH of a 0.05 M acetic acid solution given that the ionization constant (Ka) of acetic acid is 1.74 × 10−5.
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.
Ans:
48. Assuming complete dissociation, calculate the pH of the following solutions:
(a) 0.003 M HCl
Ans: HCI = aq ⟶ H+ + CI-
∴ [H+] = [HCI] = 3 × 10-3 M
pH = – log (3 × 10-3) = 2.52
(b) 0.005 M NaOH
Ans: NaOH = aq ⟶ Na+ OH-
Therefore, [OH-] = 5 × 10 -3 M
[H+] = 10-14 (5 × 10-3)
= 2 × 10-12 M
pH = – log (2 × 10-3 ) = 11.70
(c) 0.002 M HBr
Ans: HBr + aq H+ + Br–
Therefore, [OH-] = 5 × 10-3
pH = – log (2 × 10-3) = 2.70
(d) 0.002 M KOH
Ans: KOH + aq = K+ + OH–
Therefore, [OH–] = 2 × 10-3M
[H+] = 10-14 / (2 ×10-3)
= 5 × 10-12
pH = – log (5 × 10-12)
= 11.30.
49. Calculate the pH of the following solutions:
(a) 2 g of TlOH dissolved in water to give 2 litre of solution.
Ans: Molar conc. Of TIOH
= 2.21 × 10-12 M
pH = -log (2.21 × 10-12)
= 12 – (0.3424) 11.66
(b) 0.3 g of Ca(OH)2 dissolved in water to give 500 mL of solution.
Ans: Molar Conc. of Ca( OH)2
= 8.11 × 10-3 M
Ca (OH)2 → Ca2+ 2OH–
∴ [OH-] = 2 [Ca (OH)2] = 2 × (8.11 × 10-3)
M = 16.22 × 10-3 M
= pOH = – log (16.22 × 10-3)
= 3 – 1.2101 = 1.79
pH = 14 – 1.79 = 12.21
(c) 0.3 g of NaOH dissolved in water to give 200 mL of solution.
Ans: Molar concentration of NaOH
= 3.75 × 10-2 M
[OH-] = 3.75 × 10-2 M
pOH- = -log (3.75 × 10-2)
= 2 – 0.0574 = 1.43
pH = 14 – 1.43 = 12.57
(d) 1mL of 13.6 M HCl is diluted with water to give 1 litre of solution.
Ans: m1v1 = m2v2
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.
Ans: he acid dissociation constant (Ka) is given by:
= 0.1 × (0.132)2
= 1.74 × 10-3
= pKa = – log (1.74 × 10-3)
= 3 – 0.2405 = 2.76
[H+] = ca = 0.1 × 0.132
= 1.32 × 10-2 M
pH = – log (1.32 × 10-2)
= 3- 0.1206
= 1.88.
51. The pH of 0.005 M codeine (C18H21NO3) solution is 9.95. Calculate its ionization constant and pKb.
Ans: Cod + H2O ⇌ Cod H+ + OH–
pH = 9.95
Therefore, pOH = 14-9.95 = 4.05
I.e. – log [OH-] = 4.05
= 1.588 × 10-6
= pKb = – log (1.588× 10-6)
= 6-0.1987 = 5.8
52. What is the pH of 0.001M aniline solution? The ionization constant of aniline is 4.27 × 10-10. Calculate the degree of ionization of aniline in the solution. Also calculate the ionization constant of the conjugate acid of aniline.
Ans: (i) C6H5NH2 + H2O ⇌ C6H5NH3 + + OH–
= 6.534 × 10-7M
pOH = – log (6.534 × 10-7)
= (7 – log 6.534)
= (7 – 0.8152) = 6.18
pH = 14 – 6.18 = 7.82
(ii) C6H5NH2 + H2O ⇌ C6H5NH3+ + OH–
| Initial cone | c | ||
| Eqm. cone | (c-ca) | ca | ca |
= 6. 53 × 10-4
(iii) pKa + pKb = 14
(for conjugate acid – base pair)
pKb = – log Kb
= – log (4.27 × 10-10)
= (10 – log (- log 4.27)
= 10 – 0.62
= 9.38
= pKa = 14- pKb
= 14- 9.38 = 4.62
pKa = – log La = 4.62
= 2.399 × 10-5 = 2.4 × 10-5.
53. Calculate the degree of ionisation 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?
Ans: (i) Calculation of degree of ionisation(α)
pKa = 4.74
Or – log Ka = 4.74
= 1.908 × 10-2
In the presence of HCl, additional H+ ions are introduced, shifting the equilibrium of acetic acid dissociation to the left. This reduces the degree of dissociation of acetic acid, but the Ka value remains constant.
(ii) Calculation of degree of dissociation (a) is 0.01 M solution.
CH3COOH ⇌ CH3COO– + H+
| Initial come | 0.05 M | 0.01 + x ( = 0.01) |
| Come after dissociation | 0.05 – x = 0.05 | x |
= (1.82 × 10-5 ) × 5
Degree of inization (α).
= x/c = 9.1 × 10-5 /5.6 × 10-2
= 1.82 × 10-3
= 0.0018.
(iii) Calculation of degree of dissociation (α) is 0.1 M HCI solution
CH3COOH ⇌ CH3COO– + H+
| Initial come | 0.05 M | ||
| Come after dissociation | 0.05 – x = 0.05 | x | 0.1+ x (= 0.1) |
= 1.82 × 10-4
= 0.00018
54. The ionisation constant of dimethylamine is 5.4 × 10–4. Calculate the degree of ionisation in its 0.02 M solution. What percentage of dimethylamine is ionised if the solution is also 0.1 M in NaOH?
Ans: Degree of dissociation (α)
= (Kb/c) /1/2
= (5.4 × 10-4 /2 × 10-2)1/2
= 0.164
In the presence of 0.01 M NaOH, if x is the amount of dimethylamine dissociated, then
(CH3)2 NH + H2O ⇌ (CH3)2 NH+ OH– + OH–
| Initial Conc.0.02 M | x | 0.1 + x = 0.1 |
| Eqm. conc.0.02 – x = 0.02 |
55. Calculate the hydrogen ion concentration in the following biological fluids whose pH are given below:
(a) Human muscle-fluid, 6.83
Ans: [H+] of human muscles fluid,
(b) Human stomach fluid, 1.2
Ans: [H+] of human stomach fluid,
(c) Human blood, 7.38
Ans: [H+] of human blood,
pH = 7.38
(d) Human saliva, 6.4.
Ans: [H+] of human saliva,
pH = 6.4
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.
Ans: (a) H+ of milk,
(b) [H+] of black coffee,
pH = 5.0
= 5 × 10-5 M
(c) [H+] of tomato juice.
(d) [H+] of lemon juice.
pH = 2.2
(e) [H+] of egg white.
pH = 7.8
= 1.58 × 10-8 M.
57. 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: Moral concentration of the solution
= 0.05 M
Since KOH is a strong electrolyte, it fully dissociates in aqueous solution:
Concentration of K+,
[K+] = 0.05 M
Concentration OH–,
[OH–] = 0.05M
Concentration Of H+,
[ H3O=] = Kw / [OH–]
= (10-14m2) / (5 × 10-2m)
= 2 × 10-13M
pH = -log [ H3O+]
= – log (2 × 10 -13)
= 13 – (0.3010)
= 12.69.
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. (Atomic mass of Sr = 87.6)
Ans: The molar mass of
Sr(OH)2 = 87.6 + 34 = 121.687.6 + 34 = g/mol.
The solubility of Sr(OH)2in moles is
Assuming complete dissociation,
Sr (OH) 2 → Sr2 + 2OH –
∴ [Sr2+-] = 0.1581M
[OH–] = 2 × 0.1581 = 0.3162M
pOH = – log(0.3162) = 0.5
∴ pH = 14 – 5
= 13.5
59. The ionization constant of propanoic acid is 1.32 × 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?
Ans: To calculate the degree of dissociation (α) for propionic acid, According to Ostwald’s Dilution Law:
Calculation of pH of solution:
To calculate the degree of dissociation (α) of propionic acid in a 0.01 M HCl solution
CH3CH2COOH ⇌ CH3CH2COO− + H+
In the presence of HCl, the ionization of CH3CH2COOH will decrease. If C is the initial concentration of the acid and x is the amount dissociated at equilibrium, the increased concentration of H+ ions from HCl shifts the equilibrium to the left, reducing x compared to the dissociation in pure water.
I
–α = x/c = 1.32 ×10-3.
60. The pH of 0.1 M solution of cyanic acid (HCNO) is 2.34. Calculate ionization constant of the acid and also its degree of ionization in the solution.
Ans: To calculate the degree of dissociation (α) in a solution:
α = [H+]/c
= (4.571 × 10-3 M)/(0.1M)
= 4.571 × 10-2
= 0.04571M.
Calculation of ionization constant of acid
[H+] = (Kα × c)½
[H+] 2 = Kα × c
Kα = [H+] 2/c
= 20.89 × 10-5
Ka = 2.089 × 10-4
= 2.089 × 10-4
61. The ionization constant of nitrous acid is 4.5 × 10–4. Calculate the pH value of 0.04 M NaNO2 sodium and also its degree of hydrolysis.
Ans: To calculate the degree of hydrolysis (h)
Sodium nitrite (NaNO₂), we need to consider that it is a salt formed from a strong base (NaOH) and a weak acid (HNO₂). The hydrolysis of sodium nitrite can be described by the following equation:
Degree of hydrolysis
Calculation of pH of the solution
62. A 0.02 M solution of pyridinium hydrochloride has pH = 3.44. Calculate the ionization constant of pyridine.
Ans: For pyridinium hydrochloride (C6H5NH⁺Cl⁻), the pH of the solution can be calculated using the following expression:
pH = – 1 2 [log Kw -log Kb + log(c)]
= 1.51 × 10-9
63. Predict if the solutions of the following salts are neutral, acidic or basic: NaCl, KBr, NaCN, NH4NO3, NaNO2 and KF
Ans: Neutral: NaCI, KBr
Basic: NaCN, NaNO2, KF
Acidic: NH4NO3
64. The ionization constant of chloroacetic acid is 1.35 × 10–3. What will be the pH of 0.1M acid and its 0.1M sodium salt solution?
Ans: Calculation of pH of 0.1 M sodium salt solution
Ka = 1.35 × 10-3,
C = 0.1 M
= – log [H+]
= – log (1.16 × 10-2)
= 2 – log 1.16
= 2 – 0.06
= 1.94
Calculation of pH of 0.1 M sodium salt of acid
Degree of hydrolysis
When a sodium salt of a weak acid undergoes hydrolysis, it forms a basic solution
[OH–] = ch = 0.1 × 8.6 × 10– 6 = 8.6 × 10 – 7
= 1.16 × 10-8
= pH= -log[H+]
= -log [1.16 × 10-8]
= [8 – log 1.16]
= 8 -0.064
= 7.94.
65. Ionic product of water at 310 K is 2.7 × 10–14. What is the pH of neutral water at this temperature?
Ans:
66. Calculate the pH of the resultant mixtures:
(a) 10 mL of 0.2M Ca(OH)2 + 25 mL of 0.1M HCl
Ans: 10 mL of 0.2 M Ca(OH)₂:
Moles of Ca(OH)₂ = Volume × Molarity
Moles = 0.01 L × 0.2 M = 0.002 moles = 2 millimoles
25 mL of 0.1 M HCl:
Moles of HCl = Volume × Molarity
Moles = 0.025 L × 0.1 M = 0.0025 moles = 2.5 millimoles of Ca(OH)2
∴ Ca (OH) 2 left = 2 – 1.25 = 0.75 milli moles [HCl is the limiting reagent]
∴ Ca (OH) 2 left = 2 – 1.25 = 0.75 milli moles
Total volume of the solution = Volume of Ca(OH)₂ solution + Volume of HCl
solution
Total Volume = 10 mL + 25 mL = 35 mL
(b) 10 mL of 0.01M H2SO4 + 10 mL of 0.01M Ca(OH)2
Ans: 10 mL of 0.01M × H2SO4
= 10 × 0.01 millimole
= 0.1 milli mole
= 10mL of 0.01M Ca (OH)2
= 10 × 0.01 milli mole
= 0.1 milli mole
Ca(OH)2 + H2SO4 → CaSO4 + 2H2O
1 mole of Ca(OH)2 reacts with 1 mole of H2SO4.
∴ 0.1 millimole of Ca(OH)2 will completely react with 0.1 milli mole of H2SO4.
Therefore, the resulting solution will be neutral, with a pH of 7.0.
(c) 10 mL of 0.1M H2SO4 + 10 mL of 0.1M KOH
Ans: 10 mL of 0.1 MH2SO4
= 1 milli mole
10 mL of 0.1 М КОН
= 1 milli mole
2KOH + H2SO4 → K2SO4 + 2H2O
1 milli moles of KOH will react with 0.5 milli mole of H2SO4
∴ H2SO4 left = 1 – 0.5 = 0.5 milli mole Volume of reaction mixture
= 10 + 10 = 20 ml
∴ Molarity of H2SO4 in the mixture solution
= 0.5 / 20
= 2.5 × 10-12 M
[H+]= 2× (2.5 × 10-2)
= 5 × 10-2
pH = – log (5 × 10-2)
= 2 – 0.699
= 1.3
67. Determine the solubilities of silver chromate, barium chromate, ferric hydroxide, lead chloride and mercurous iodide at 298 K. from their solubility product constants given in Table 6.9. Determine also the molarities of individual ions.
Ans: (i) Silver chromate,
Ag2CrO4 ⇌ 2Ag ++ CrO42-
s 2s s
Ksp = (25)2 (5) = 4s2
∴ S3 = Ksp / 4 = (1.1 × 10-12) /4
= 0.275 × 10-12
= 0.275 × 10-13
∴ 3 log S = log (2.75 × 10-13)
= -13 × 0.4393
= -12.5607
= 13. 0 × 10-5 M
= 1.30 × 10-4 M
[CrO42] = 6.5 × 10-5 M
(ii) For Barium Chromate
Ba CrO4(s) ⇌ Ba2+ (aq) + CrO42 (aq)
Let the solubility of salt in water = S [Ba2+ (aq)] = S
And [CrO42- (aq)] = S
K sp = [Ba2+ (aq)][CO42- (aq)]
1.2 × 10-10 = (S) × (S) = S2
S = (1.2 × 10-10)1/2
= 1.10 × 10-5 M
Molarity of Ba2+ (aq) = 1.1 × 10-5 M
Molarity ofCrO42 (aq) = 1.1 × 10-5 M
(iii) For ferric hydroxide solution:
Fe(OH)3 ⇌ Fe3+ (aq) + 3OH−(aq)
Let the solubility of Fe(OH)3 in water be s.
= 1.387 × 10-10 M
Molarity of
Fe3+ (aq) = 1.387 × 10-10 M
Molarity os = OH– (aq) = 3 × 1.387× 10-10
= 4.16 × 10-19 M
68. The solubility product constant of Ag2CrO4 and AgBr are 1.1 × 10–12 and 5.0 × 10–13 respectively. Calculate the ratio of the molarities of their saturated solutions.
Ans: Calculation of molar solubility (molarity) of Ag2CrO4 solution.
Let the solubility of the salt in water be S
[Ag+ (aq)] = 2s
And [CrO42- (aq)] = s
Calculation of moral solubility (molarity) of AgBr solution
Calculation of the ratio of the molarities of the saturated solution
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 × 10–8).
Ans: Cu(IO3)2 ⇌ Cu2+ (aq) + 2IO3–(aq).
The Cu2+ (aq) ions are supplied from a copper chlorate solution, while the IO3− (aq) ions come from a sodium iodate solution.
Since equal volumes of the solutions have been mixed. Therefore, concentrations of both Cu 2+ (aq) and IO3– (aq)) in solution after mixing will be reduced to half, i.e.
(0.002M)/2 = 0.001M
The ionic product
= [Cu2+] [IO3–]2
= (0.001) × (0.001)2
Ksp Value of
Cu(IO3)2 = 1 × 10– 9
= 7.4 × 10– 8 (given)
Since the ionic product is less than the Ksp value, copper iodate will not be precipitated.
70. The ionisation constant of benzoic acid is 6.46 × 10–5 and Ksp for silver benzoate is 2.5 × 10–13. How many times is silver benzoate more soluble in a buffer of pH 3.19 compared to its solubility in pure water?
Ans: Calculation of solubility in water:
C6H5COOAg → C6H5COO– + Ag+
Suppose solubility in water = x mol L-1
Then [C6H5COO-] = [Ag+]
= x mol L-1
x 2 =Ksp
X = (Ksp)½
= (2.5 × 10-13)½
= 5 × 10-7 mol L-1
Calculation of solubility in buffer of pH = 3.19
pH = 3.19
Or -log [H+] = 3.19
[H+] = 6.457 × 10-4
The C6H5COO− ions now react with the H+ ions to form benzoic acid, but the concentration of [H+] stays nearly constant due to the presence of the buffer solution.
Suppose the solubility in the buffer solution is y mol L−1. Since most of the benzoate ions are converted into benzoic acid molecules (which largely remain un-ionized), we have y = [Ag+].
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 × 10–18).
Ans: Let the concentrations of both FeSO, and Na₂S solutions before mixing = x mol L– 1 = xM
Since equal volumes of the solutions are mixed, the concentrations of the solutions, as well as the ions upon mixing, will be reduced by half, i.e., x/2 M. For FeS, the solubility product.
= 6.3 × 10-18 × 2
= 4 × 6.3 × 10-18
= 25.2 × 10 -18
Or x = (25.2 × 10– 18) (1/2)
= 5.02 × 10– 9 M
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 × 10–6).
Ans: Sol. Calculation of molar solubility of CaSO4
CaSO4(s) ⇌ Ca2 + (aq) + SO42-(aq)
Let the solubility of salt in water be s
[Ca2+ (aq)] = s
and [SO42- (aq)] = s
Ksp = [Ca2+ (aq)] [SO42- (aq)]
9.1 × 10– 6 = s × s
s = (9.1 × 10 – 6)(1/2)
= 3.02 ог 10– 3 M
= 3.02 × 10 – 3 mol L– 1
Calculation of amount of water required Molar mass of
CaSO4 = (40 + 32 + 64)
= 136 gmo l – 1
Mass of CaSO4 = (3.02 × 10-3 mol L-¹) × (136 g mol-1)
= 0.411 g L-1
Volume of water required to dissolve 0.411 g of CaSO4 = 1 L
Volume of water required to dissolve 1.0 g of CaSO4
= (1L × 1g) / (0.411g)
= 2.43L
73. The concentration of sulphide ion in 0.1M HCl solution saturated with hydrogen sulphide is 1.0 × 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?
Ans: Precipitation will occur in the solution when the ionic product exceeds the solubility product. If 10 mL of a solution containing S²⁻ ions is combined with 5 mL of a metal salt solution, a precipitate will form after mixing if the ionic product of the resulting solution is greater than the solubility product of the metal sulphide.
Ionic product of [M2+] and [S2]
= (1.33 × 10-2) × (6.67 × 10-20)
= 8.87 × 10-22
Since this ionic product exceeds the Ksp of both ZnS and CdS, precipitation will occur for both ZnS and CdS. Consequently, solutions of ZnCl₂ and CdCl₂ will cause precipitation of ZnS and CdS, respectively.
