## NCERT Solutions for Class 11 Chemistry Chapter 7 – Free PDF Download

***According to the CBSE Syllabus 2023-24, this chapter has been renumbered as Chapter 6.**

**NCERT Solutions for Class 11 Chemistry Chapter 7 Equilibrium **is provided on this page for the CBSE Class 11 students to study and prepare well for the annual exam. Detailed, step-by-step solutions for each and every intext and exercise question listed in Chapter 7 of the NCERT Class 11 Chemistry textbook can be accessed here. Expert tutors have created these NCERT Solutions for Class 11 Chemistry according to the latest CBSE Syllabus 2023-24 and its guidelines.

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## NCERT Solutions for Class 11 Chemistry Chapter 7 Equilibrium

## NCERT Solutions for Class 11 Chemistry Chapter 7 Equilibrium

“Equilibrium” is the seventh chapter in the NCERT Class 11 Chemistry Textbook. Several important concepts, such as equilibrium constants, buffer solutions and the common-ion effect, are explained in this chapter. Along with structured questions, the **NCERT Solutions for Class 11 Chemistry** Chapter 7 provided on this page come with detailed explanations to help students learn and understand the key concepts related to chemical equilibrium in a seamless manner. The important subtopics associated with Chapter 7 – Equilibrium, are listed below.

### Subtopics included in NCERT Class 11 Chemistry Chapter 7 – Equilibrium

- Solid-liquid Equilibrium
- Liquid-vapour Equilibrium
- Solid-vapour Equilibrium
- Equilibrium Involving Dissolution of Solid or Gases in Liquids
- General Characteristics of Equilibria Involving Physical Processes

- Equilibrium in Chemical Processes – Dynamic Equilibrium
- Law of Chemical Equilibrium and Equilibrium Constant
- Homogeneous Equilibria
- Equilibrium Constant in Gaseous Systems

- Heterogeneous Equilibria
- Applications of Equilibrium Constants
- Predicting the Extent of a Reaction
- Predicting the Direction of the Reaction
- Calculating Equilibrium Concentrations

- Relationship between Equilibrium Constant K, Reaction Quotient Q and Gibbs Energy G
- Factors Affecting Equilibria
- Effect of Concentration Change
- Effect of Pressure Change
- Effect of Inert Gas Addition
- Effect of Temperature Change
- Effect of a Catalyst

- Ionic Equilibrium in Solution
- Acids, Bases and Salts
- Arrhenius Concept of Acids and Bases
- The Bronsted-lowry Acids and Bases
- Lewis Acids and Bases

- Ionisation of Acids and Bases
- The Ionisation Constant of Water and Its Ionic Product
- The pH Scale
- Ionisation Constants of Weak Acids
- Ionisation of Weak Bases
- The Relation Between K
_{a}and K_{b } - Di- and Polybasic Acids and Di- and Polyacidic Bases
- Factors Affecting Acid Strength
- Common Ion Effect in the Ionisation of Acids and Bases
- Hydrolysis of Salts and the Ph of Their Solutions

- Buffer Solutions
- Solubility Equilibria of Sparingly Soluble Salts
- Solubility Product Constant
- Common Ion Effect on Solubility of Ionic Salts

## NCERT Solutions for Class 11 Chemistry Chapter 7

**Q.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?**

** ****Ans.**

(a) On increasing the volume of the container, the vapour pressure will initially decrease because the same amount of vapours is now distributed over a large space.

(b) On increasing the volume of the container, the rates of evaporation will increase initially because now more space is available. Since the amount of vapours per unit volume decrease on increasing the volume, the rate of condensation will decrease initially.

(c) Finally, equilibrium will be restored when the rates of the forward and backward processes become equal. However, the vapour pressure will remain unchanged because it depends upon the temperature and not upon the volume of the container.

**Q.2.What is Kc for the following equilibrium when the equilibrium concentration of each substance is: [SO _{2}]= 0.60M, [O_{2}] = 0.82M and [SO_{3}] = 1.90M? **

**\(\begin{array}{l}2SO_{2}(g)+O_{2}(g) ⇌ 2SO_{3}(g)\end{array} \)**

**Ans.**

As per the question,

Hence, K for the equilibrium is 12.229 M^{–1}.

**Q.3. At a certain temperature and total pressure of 10 ^{5 }Pa, iodine vapour contains 40% by volume of I atoms**

**I _{2}(g) ⇌ 2I (g) **

**Calculate K**

_{p}for the equilibrium**Ans.**

The partial pressure of Iodine atoms (I)

The partial pressure of I_{2} molecules,

Now, for the given reaction,

**Q.4.Write the expression for the equilibrium constant, K _{c,} for each of the following reactions:**

**(i)**

**\(\begin{array}{l}2NOCl(g)\leftrightarrow 2NO(g)+Cl_{2}(g)\end{array} \)**

**(ii)**

**\(\begin{array}{l}2Cu(NO_{3})_{2}(s)\leftrightarrow 2CuO(s)+4NO_{2}(g)+O_{2}(g)\end{array} \)**

**(iii)**

**\(\begin{array}{l}CH_{3}COOC_{2}H_{5}(aq)+H_{2}O(1)\leftrightarrow CH_{3}COOH(aq)+C_{2}H_{5}OH(aq)\end{array} \)**

**(iv)**

**\(\begin{array}{l}Fe^{3+}(aq)+3OH^{-}(aq)\leftrightarrow Fe(OH)_{3}(s)\end{array} \)**

**(v)**

**\(\begin{array}{l}I_{2}(s)+5F_{2}\leftrightarrow 2IF_{5}\end{array} \)**

**Ans.**

(i)

(ii)

(iii)

(iv)

(v)

**Q.5.Find out the value of Kc for each of the following equilibria from the value of Kp: **

**(i)**

**\(\begin{array}{l}2NOCl(g) ⇌ 2NO(g)+Cl_{2}(g);\: \: \: K_{p}=1.8\times 10^{-2}\: \: at\: 500K\end{array} \)**

**(ii)**

**\(\begin{array}{l}CaCO_{3}(s) ⇌ CaO(s)+CO_{2}(g);\: \: \: K_{p}=167\: \: at\: 1073K\end{array} \)**

**Ans.**

The relation between K_{p} and K_{c} is given as follows:

**(i)** Given,

R = 0.0831 barLmol^{–1}K^{–1}

T = 500 K

K_{p}=

Now,

K_{p} = K_{c} (RT) ^{∆n}

**(ii)** Here,

∆n =2 – 1 = 1

R = 0.0831 barLmol^{–1}K^{–1}

T = 1073 K

K_{p}= 167

Now,

K_{p} = K_{c }(RT) ^{∆n}

**Q.6. For the following equilibrium, **

**\(\begin{array}{l}K_{c}=6.3\times 10^{14}\: \: \: at\: 1000K\\ \\ NO(g)+O_{3(g)} ⇌ NO_{2(g)}+O_{2(g)}\end{array} \)**

**Both the forward and reverse reactions in the equilibrium are elementary bimolecular reactions. What is K _{c} for the reverse reaction?**

**Ans.**

For the reverse reaction,

**Q.7. Explain why solids and pure liquids can be ignored while writing the equilibrium constant expression.**

**Ans.**

Solids and pure liquids can be ignored while writing the equilibrium constant expression because the molar concentration of a pure solid or liquid is independent of the amount present.

Mole concentration=

As the density of the solid and pure liquid is fixed, the molar mass is also fixed.

**Q.8. ****The reaction between N _{2} and O_{2} takes place as follows:**

**\(\begin{array}{l}2N_{2}(g)+O_{2} ⇌ 2N_{2}O(g)\end{array} \)**

**If a solution of 0.933 mol of oxygen and 0.482 mol of nitrogen is placed in a 10 L reaction vessel and allowed to form N _{2}O at a temperature for which K_{c} =**

**\(\begin{array}{l}2.0\times 10^{-37}\end{array} \)**

**, determine the composition of the equilibrium solution.**

**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)

Initial conc. 0.482 mol 0.933 mol 0

At equilibrium(0.482-x)mol (1.933-x)mol x mol

The value of the equilibrium constant is extremely small. This means that only small amounts. Then,

Now,

** **

**Q.9. ****Nitric oxide reacts with Br2 and gives nitrosyl bromide as per the reaction given below:**

**2NO _{(g)}+Br_{2(g)} ⇌ 2NOBr_{(g)}**

**When 0.087 mol of NO and 0.0437 mol of Br _{2} are mixed in a closed container at a constant temperature, 0.0518 mol of NOBr is obtained at equilibrium. Calculate the equilibrium amount of NO and Br_{2}.**

**Ans.**

The given reaction is:

2NO_{(g)}+Br_{2(g) } ⇌ 2NOBr_{(g)}

2mol 1mol 2mol

Now, 2 mol of NOBr is formed from 2 mol of NO. Therefore, 0.0518 mol of NOBr is formed from 0.0518 mol of NO.

Again, 2 mol of NOBr is formed from 1 mol of Br.

Therefore, 0.0518 mol of NOBr is formed from

The amount of NO and Br present initially is as follows:

[NO] = 0.087 mol [Br2] = 0.0437 molTherefore, 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}] = 0.0437 – 0.0259 = 0.0178 mol.

**Q.10. At 450 K, K _{p}= **

**\(\begin{array}{l}2.0 \times 10^{10}\end{array} \)**

**/bar for the given reaction at equilibrium.**

**2SO _{2}(g)+O_{2}(g) ⇌ 2SO_{3}(g) **

**What is K**

_{c}at this temperature?**Ans.**

For the given reaction,

∆n = 2 – 3 = – 1

T = 450 K

R = 0.0831 bar L bar K^{–1} mol^{–1}

K_{p}=

We know that,

**Q.11. A sample of HI _{(g)} is placed in a flask at a pressure of 0.2 atm. At equilibrium, the partial pressure of HI_{(g)} is 0.04 atm. What is K_{p} for the given equilibrium?**

**2HI(g) ⇌ H _{2}(g)+I_{2}(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)

Initial conc. 0.2 atm 0 0

At equilibrium 0.4 atm 0.16/2 0.16/2

= 0.08atm = 0.08atm

Therefore,

Hence, the value of K_{p} for the given equilibrium is 4.0.

**Q.12. A mixture of 1.57 mol of N _{2}, 1.92 mol of H_{2} and 8.13 mol of NH_{3} is introduced into a 20 L reaction vessel at 500 K. At this temperature, the equilibrium constant, K_{c,} for the reaction is**

**N _{2}(g)+3H_{2}(g) ⇌ 2NH_{3}(g) is **

**\(\begin{array}{l}1.7\times 10^{2}\end{array} \)**

**.**

** 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 concentration of various species is

[N_{2}]=

_{2}]=

_{3}]=

Now, the reaction quotient Q_{c} is:

Since Q_{c} ≠ K_{c}, the reaction mixture is not at equilibrium.

Again, Q_{c} > K_{c}. Hence, the reaction will proceed in the reverse direction.

**Q.13. The equilibrium constant expression for a gas reaction is,**

**\(\begin{array}{l}K_{c}=\frac{[NH_{3}]^{4}[O_{2}]^{5}}{[NO]^{4}[H_{2}O]^{6}}\end{array} \)**

**.**

**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).}

**Q.14. One mole of H _{2}O and one mole of CO are taken in a 10 L vessel and heated to 725 K. At equilibrium, 60% of water (by mass) reacts with CO according to the equation, **

**H _{2}O(g) + CO(g) ⇌ H_{2}(g) + CO_{2}(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)}

Compound | H20 | CO | H_{2} |
CO_{2} |

Initial Conc. | 0.1M | 0.1M | 0 | 0 |

Equilibrium Conc. | 0.06M | 0.06M | 0.04M | 0.04M |

Therefore, the equilibrium constant for the reaction,

K_{c} = ([H_{2}][CO_{2}])/([H_{2}O][CO]) = (0.4*0.4)/(0.6*0.6) = 0.444.

**Q.15. At 700 K, the equilibrium constant for the reaction ****H _{2(g)}+I_{2(g)} ⇌ 2HI_{(g) }**

**is 54.8. If 0.5 molL**

^{–1}of HI_{(g)}is present at equilibrium at 700 K, what is the concentration of H_{2(g)}and I_{2(g),}assuming that we initially started with HI_{(g)}and allowed it to reach equilibrium at 700 K?**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)}

^{-1}will be 1/54.8.

Let the concentrations of hydrogen and iodine at equilibrium be x molL^{–1}

_{2}]=[I

_{2}]=x mol L

^{-1}

Therefore,

Hence, at equilibrium, [H_{2}]=[I_{2}]=0.068 mol L^{-1}.

**Q.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)} ⇌ I_{2(g)}+Cl_{2(g)} ; K_{c }=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 M

Now, we can write,

Hence, at equilibrium,

[H_{2}]=[I

_{2}]=0.167 M [HI]=

**Q.17. K _{p} = 0.04 atm at 899 K for the equilibrium shown below. What is the equilibrium concentration of C_{2}H_{6} when it is placed in a flask at 4.0 atm pressure and allowed to come to equilibrium?**

**C _{2}H_{6}(g) ⇌ C_{2}H_{4}(g) +H_{2}(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. 4.0 M 0 0

At equilibrium (4.0-p) p p

We can write,

Now,

Hence, at equilibrium, [C_{2}H_{6}] = 4 – p = 4 – 0.38

= 3.62 atm.

**Q.18. Ethyl acetate is formed by the reaction between ethanol and acetic acid, and the equilibrium is represented as: **

**CH _{3}COOH_{(l) }+ C_{2}H_{5}OH_{(l)} ⇌ CH_{3}COOC_{2}H_{5(l) }+ H_{2}O_{(l)}**

**(i) Write the concentration ratio (reaction quotient) Q _{c} for this reaction (note: water is not in excess and is not a solvent in this reaction).**

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

**(iii) Starting with 0.5 mol of ethanol and 1.0 mol of acetic acid and maintaining it at 293 K, 0.214 mol of ethyl acetate is found after some time. 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:

CH_{3}COOH_{(l)} + C_{2}H_{5}OH_{(l) } ⇌ CH_{3}COOC_{2}H_{5(l)} + H_{2}O_{(l)}

Initial conc. 1/V M 0.18/V M 0 0

At equilibrium (1-0.171)/V (0.18 – 0.171)/V 0.171/V M 0.171/V M

= 0.829/V M = 0.009/V M

Therefore, the equilibrium constant for the given reaction is:

(iii) Let the volume of the reaction mixture be V.

CH_{3}COOH(l)+C_{2}H_{5}OH(l)** ⇌ **CH_{3}COOC_{2}H_{5}(l)+H_{2}O(l)

Initial conc. 1.0/V M 0.5/V M 0 0

At equilibrium (10-0.214)/V (0.5-0.214)/V 0.214/V M 0.214/V M

= 0.786/V M = 0.286/V M

Therefore, the reaction quotient is,

Since Q_{c}<K_{c}, equilibrium has not been reached.

**Q.19.** **A sample of pure PCl _{5} was introduced into an evacuated vessel at 473 K. After equilibrium was attained, the concentration of PCl_{5} was found to be **

**\(\begin{array}{l}0.5\times 10^{-1}\end{array} \)**

**mol L\(\begin{array}{l}8.3\times 10^{-3}\end{array} \), what are the concentrations of PCl**

^{–1}. If the value of K_{c}is_{3}and Cl_{2}at equilibrium?**PCl _{5(g)} ⇌ PCl_{3(g)} + Cl_{2(g)}**

**Ans.**

Consider the concentration of both PCl_{3} and Cl_{2} at equilibrium to be x molL^{–1}. The given reaction is:

PCl_{5(g)}** ⇌ **PCl_{3(g)} + Cl_{2(g)}

At equilibrium

^{-1}x mol L

^{-1}

It is given that the value of the equilibrium constant, K_{c,} is

Now, we can write the expression for equilibrium as:

Therefore, at equilibrium,

[PCl_{3}]=[Cl

_{2}]=0.02mol L

^{-1}.

**Q.20. One of the reactions that take place in producing steel from iron ore is the reduction of iron (II) oxide by carbon monoxide to give iron metal and CO _{2}.**

** FeO _{(s)} + CO _{(g)} ⇌ Fe _{(s)} + CO_{2} _{(g)}; K_{p}= 0.265 at 1050 K.**

**What is the equilibrium partial pressures of CO and CO _{2} at 1050 K if the initial partial pressures are: p_{CO} = 1.4 atm and p_{CO}**

_{2}**= 0.80 atm?**

**Ans.**

For the given reaction,

FeO_{(g)} + CO_{(g)} ⇌ Fe_{(s)}+CO_{2(g)}

Initially, pCO = 1.4 atm and pCO_{2}= 0.80 atm

Q_{p}=

Since Q_{p} > K_{p, }the reaction will proceed in the backward direction.

Therefore, we can say that the pressure of CO will increase while the pressure of CO_{2 }will decrease.

Now, let the increase in pressure of CO = decrease in pressure of CO_{2} be p. Then, we can write,

Therefore, the equilibrium partial of CO_{2},p_{CO}=0.80-0.339 = 0.461 atm.

And, equilibrium partial pressure of CO,p_{CO}=1.4+0.339 = 1.739 atm.

** **

**Q.21. The equilibrium constant, K _{c,} for the reaction**

**N _{2}(g)+3H_{2}(g) ⇌ 2NH_{3 } at 500 K is 0.061 **

**At a specific time, from the analysis, we can conclude that the composition of the reaction mixture is 3.0 mol L ^{–1} N_{2}, 2.0 mol L^{–1} H_{2 }and 0.5 mol L^{–1} NH_{3}. Find out whether the reaction is at equilibrium or not. Find in which direction the reaction proceeds to reach equilibrium.**

**Ans.**

N_{2}(g) + 3H_{2}(g) **⇌** 2NH_{3}

At a particular time: 3.0 mol L^{-1} 2.0mol L^{-1} 0.5 mol L^{-1}

So,

Q_{c}=

It is given that K_{c}=0.061

The reaction proceeds in the forward direction to reach at equilibrium.

**Q.22.Bromine monochloride, BrCl decomposes into bromine and chlorine and reaches the equilibrium:**

**2BrCl _{(g)} ⇌ Br_{2(g)} + Cl_{2(g)}**

**For which K _{c}= 32 at 500 K.**

**If initially pure BrCl is present at a concentration of **

**\(\begin{array}{l}3.3 \times 10^{-3}\end{array} \)**

**molL**

^{-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.

At equilibrium

Now, we can write,

So

**Q.23.At 1127 K and 1 atm pressure, a gaseous mixture of CO and CO _{2} in equilibrium with solid carbon has 90.55% CO by mass**

** ****C(s) + CO _{2} _{(g)} ⇌ 2CO _{(g)}**

**Calculate K _{c} for this reaction at the above temperature.**

**Ans.**

Let us assume that the solution is 100g in total.

Given, the mass of CO = 90.55 g

Now, the mass of CO_{2} = (100 – 90.55) = 9.45 g

Now, the number of moles of CO,

The number of moles of CO_{2},

The partial pressure of CO,

P_{CO}=

Partial pressure of CO_{2},

Therefore, K_{p}=

For the given reaction,

∆n = 2 – 1 = 1

We know that,

K_{p }= K_{c}(RT) ^{}

**Q.24. Calculate a) ∆G ^{0} and b) the equilibrium constant for the formation of NO_{2} from NO and O_{2} at 298K**

**\(\begin{array}{l}NO(g)+\frac{1}{2}O_{2}(g) ⇌ NO_{2}(g)\end{array} \)**

** Where;**

**∆ _{f}G° (NO_{2}) = 52.0 kJ/mol**

** ∆ _{f}G° (NO) = 87.0 kJ/mol**

** ∆ _{f}G° (O_{2}) = 0 kJ/mol**

**Ans.**

(a) For the given reaction, we have

∆G° = ∆G°(Products) – ∆G°(Reactants)

∆G° = 52.0 – (87.0 + 0)

= -35.0 KJ mol^{-1}

(b) We know that,

∆G° = RT log K_{c}

∆G° = 2.303 RT log K_{c}

Therefore, the equilibrium constant for the given reaction K_{c} is

**Q.25. Does the number of moles of reaction products increase, decrease or remain the same when each of the following equilibria is subjected to a decrease in pressure by increasing the volume? **

**(a) PCl _{5}(g) ⇌ PCl_{3} +Cl_{2} (g)**

**(b) CaO(s) + CO _{2} (g) ⇌ CaCO_{3}(s)**

**(c) 3Fe (s) + 4H _{2}O (g) ⇌ Fe_{3}O_{4} (s) +4H_{2} (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.

**Q.26. Which of the following reactions will get affected by increasing the pressure? Also, mention whether the change will cause the reaction to go in a forward or backward direction.**

**(I) COCl _{2} (g) ⇌ CO (g) +Cl_{2} (g)**

**(II) CH _{4} (g) +2S_{2} (g) ⇌ CS_{2} (g) + 2H_{2}S (g)**

**(III) CO _{2} (g) +C (s) ⇌ 2CO (g)**

**(IV) 2H _{2 }(g) +CO (g) ⇌ CH_{3}OH(g)**

**(V) CaCO _{3 }(s) ⇌ CaO (s) + CO_{2} (g)**

**(VI) 4NH _{3} (g) + 5O_{2} (g) ⇌ 4NO (g) + 6H_{2}O (g)**

**Ans.**

When pressure is increased:

The reactions given in (i), (iii), (iv), (v), and (vi) will get affected.

Since the number of moles of gaseous reactants is more than that of gaseous products, the reaction given in (iv) will proceed in the forward direction

Since the number of moles of gaseous reactants is less than that of gaseous products, the reactions given in (i), (iii), (v), and (vi) will shift in the backward direction.

_{ }

**Q.27. The equilibrium constant for the following reaction is **

**\(\begin{array}{l}1.6 \times 10^{5}\end{array} \)**

**at 1024 K.**

**H _{2} (g) + Br_{2} (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.**

Given, K_{p} for the reaction, i.e., H_{2} (g) + Br_{2} (g)** ⇒ **2HBr (g) is

Therefore, for the reaction 2HBr (g)** ⇒ **H_{2} (g) + Br_{2} (g), the equilibrium constant will be,

K^{’}_{p}=

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 10-2p p p

Now, we can write,

Therefore, at equilibrium,

[H_{2}]=[Br

_{2}]=

**Q.28. Dihydrogen gas is obtained from natural gas by partial oxidation with steam as per the following endothermic reaction:**

**\(\begin{array}{l}CH_{4}(g)+H_{2}O(g) ⇌ CO(g)+3H_{2}(g)\end{array} \)**

**(a) Write as an expression for K _{p} for the above reaction**

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

**(i) Increasing the pressure **

**(ii) Increasing the temperature **

**(iii) Using a catalyst**

**Ans.**

(a) For the given reaction,

K_{p}=

(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.

**Q.29. Describe the effect of: **

**a) Addition of H _{2} **

**b) Addition of CH _{3}OH**

**c) Removal of CO**

**d) Removal of CH _{3}OH on the equilibrium of the reaction:**

**2H _{2} (g)+CO (g) ⇌ CH_{3}OH (g)**

**Ans.**

(a) According to Le Chatelier’s principle, on the addition of H2, the equilibrium of the given reaction will shift in the forward direction.

(b) On addition of CH3OH, the equilibrium will shift in the backward direction.

(c) On removing CO, the equilibrium will shift in the backward direction.

(d) On removing CH3OH, the equilibrium will shift in the forward direction.

**Q.30.**** At 473 K, equilibrium constant K _{c} for decomposition of phosphorus pentachloride, PCl_{5} is **

**\(\begin{array}{l} 8.3 \times 10^{-3} \end{array} \)**

**. If decomposition is depicted as,**

**\(\begin{array}{l}PCl_{5}(g) ⇌ PCl_{3}(g)+Cl_{2}(g)\end{array} \)**

**∆ _{r}H° = 124.0 kJmol^{–1} **

**a) Write an expression for K _{c} for the reaction. **

**b) What is the value of K _{c} for the reverse reaction at the same temperature? **

**c) What would be the effect on K _{c} if **

**(i) more PCl _{5} is added?**

**(ii) pressure is increased? **

**(iii) The temperature is increased?**

**Ans.**

(a)

(b) Value of K_{c} for the reverse reaction at the same temperature is:

(c) (i) K_{c} would remain the same because, in this case, the temperature remains the same.

(ii) K_{c} is constant at a constant temperature. Thus, in this case, K_{c} would not change.

(iii) In an endothermic reaction, the value of K_{c} increases with an increase in temperature. Since the given reaction is an endothermic reaction, the value of K_{c} will increase if the temperature is increased.

**Q.31.**** Dihydrogen gas used in Haber’s process is produced by reacting methane from natural gas with high-temperature steam. The first stage of the two-stage reaction involves the formation of CO and H _{2}. In the second stage, CO formed in the first stage is reacted with more steam in the water gas shift reaction,**

**\(\begin{array}{l}CO**

_{(g)}+H_{2}O_{(g)}⇌ CO_{2}_{(g)}+H_{2}_{(g)}\end{array} \)**If a reaction vessel at 400°C is charged with an equimolar mixture of CO and steam such that P _{co}=P_{H2O} = 4.0 bar, what will be the partial pressure of H2 at equilibrium? K_{p}= 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** ⇌ **CO_{2}(g) + H_{2}(g)

Initial conc. 4.0 bar 4.0 bar 0 0

At equilibrium 4.0-p 4.0-p p p

Given K_{p} = 10.1

So, the partial pressure of H_{2 }is 3.04 bar at equilibrium.

**Q.32.**** Predict which of the following reaction will have an appreciable concentration of reactants and products:**

**(a) **

**\(\begin{array}{l}Cl_{2}(g)\leftrightarrow 2Cl(g);K_{c}=5 \times 10^{-39}\end{array} \)**

**(b) **

**\(\begin{array}{l}Cl_{2}(g)+2NO(g)\leftrightarrow 2NOCl(g);K_{c}=3.7 \times 10^{8}\end{array} \)**

**(c) **

**\(\begin{array}{l}Cl_{2}(g)+2NO_{2}(g)\leftrightarrow 2NO_{2}Cl(g);K_{c}=1.8\end{array} \)**

**Ans.**

If the value of K_{c} lies between 10^{–3} and 10^{3}, a reaction has an appreciable concentration of reactants and products. Thus, the reaction given in (c) will have an appreciable concentration of reactants and products.

**Q.33.**** The value of Kc for the reaction 3O _{2} (g) ⇌ 2O_{3} (g) is **

**\(\begin{array}{l}2.0\times 10^{-50} \end{array} \)**

**at 25°C. If the equilibrium concentration of O\(\begin{array}{l} 1.6 \times 10^{-2} \end{array} \), what is the concentration of O**

_{2}in the air at 25°C is_{3}?**Ans.**

Given,

3O_{2} (g)** ⇌ **2O_{3} (g)

Then, K_{c} =

Given that K_{c} =

_{2}(g)] =

Then,

So, the concentration of O_{3} is

**Q.34. The reaction, **

**\(\begin{array}{l}CO_{(g)} \; + \; 3H_{2(g)} \; ⇌ \; CH_{4(g)} \; + \; H_{2}O_{(g)}\end{array} \)**

**at 1300 K is at equilibrium in a 1 L container. It has 0.30 mol of CO, 0.10 mol of\(\begin{array}{l}H_{2}\end{array} \) and 0.02 mol of \(\begin{array}{l}H_{2}O\end{array} \) and an unknown amount of \(\begin{array}{l}CH_{4}\end{array} \) in the flask. Determine the concentration of \(\begin{array}{l}CH_{4}\end{array} \) in the mixture. **

**The equilibrium constant,\(\begin{array}{l}K_{c}\end{array} \), for the reaction at the given temperature, is 3.90.**

**Ans.**

Let the concentration of

At equilibrium,

For CO –

For

For

Therefore,

Therefore, the concentration of

**Q.35. What is conjugate acid-base pair? Find the conjugate acid/base of the given species:**

**\(\begin{array}{l}HNO_{2}\end{array} \)**

**,**

**\(\begin{array}{l}CN^{-}\end{array} \), **

**\(\begin{array}{l}HClO_{4}\end{array} \), **

**\(\begin{array}{l}F^{-}\end{array} \), **

**\(\begin{array}{l}OH^{-}\end{array} \), **

**\(\begin{array}{l}CO_{3}^{2-}\end{array} \) and **

**\(\begin{array}{l}S^{-}\end{array} \)**

**Ans.**

A conjugate acid-base pair is a pair that has a difference of only one proton.

The conjugate acid-base pair of the following are as follows:

**Q.36. Which of the followings are Lewis acids? **

**\(\begin{array}{l}H_{2}O\end{array} \)**

**,**

**\(\begin{array}{l}BF_{3}\end{array} \), **

**\(\begin{array}{l}H^{+}\end{array} \) and **

**\(\begin{array}{l}NH_{4}^{+}\end{array} \)**

**Ans.**

Lewis acids are acids which can accept a pair of electrons.

** **

**Q.37. What will be the conjugate bases for the Brönsted acids: ****HF, **

**\(\begin{array}{l}H_{2}SO_{4}\end{array} \)**

**and**

**\(\begin{array}{l}HCO_{3}\end{array} \)?**

**Ans.**

The following shows the conjugate bases for the Bronsted acids:

HF –

**Q.38. Write the conjugate acids for the following Brönsted bases: **

**\(\begin{array}{l}NH_{2}^{-}\end{array} \)**

**, NH\(\begin{array}{l}HCOO^{-}\end{array} \)**

_{3}and**Ans.**

The conjugate acids for the given Bronsted bases are:

NH_{2}^{– }→ NH_{3}

NH_{3} → NH_{4}^{+}

HCOO^{–} → HCOOH

**Q.39. The species: H _{2}O, HCO_{3}^{–}, HSO_{4}^{–} and NH_{3} can act both as Brönsted acids and bases. For each case, give the corresponding conjugate acid and base.**

**Ans.**

The conjugate acids and conjugate bases for the given species are listed below:

Species Conjugate Acid Conjugate Bases

H_{2}O H_{3}O^{+} OH^{–}

HCO_{3}^{–} H_{2}CO_{3} CO_{3}^{2-}

HSO_{4}^{–} H_{2}SO_{4} SO_{4}^{2-}

NH_{3} NH_{4}^{+} NH_{2}^{–}

** **

**Q.40. Classify the following species into Lewis acids and Lewis bases and show that these species act as Lewis base/acid:**

**(a) **

**\(\begin{array}{l}OH^{-}\end{array} \)**

**(b) **

**\(\begin{array}{l}F^{-}\end{array} \)**

**(c) **

**\(\begin{array}{l}H^{+}\end{array} \)**

**(d) **BCl_{3}

**Ans.**

(a)

It is a Lewis base as it has a tendency to lose a pair of electrons.

(b)

It is a Lewis base as it has a tendency to lose its lone pair of electrons.

(c)

It is a Lewis acid as it has a tendency to accept a pair of electrons.

(d) BCl_{3}:

It is a Lewis acid as it has a tendency to accept a pair of electrons.

**Q.41.The concentration of hydrogen ion in a sample of soft drink is 3.8 × 10 ^{–3} M. what is its pH?**

**Ans.**

**Q.42. The pH of a sample of vinegar is 3.76. Calculate the concentration of hydrogen ion in it. **

**Ans. **

**Q.43. The ionisation 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 ionisation constants of the corresponding conjugate base.**

**Ans.**

For F^{–},

For HCOO^{–},

For CN^{–} =

**Q.44. The ionisation constant of phenol is 1.0 × 10 ^{–10}. What is the concentration of phenolate ion in a 0.05 M solution of phenol? What will be its degree of ionisation if the solution is also 0.01 M in sodium phenolate?**

**Ans.**

C_{6}H_{5}OH **⇌** C_{6}H_{5}O^{–} + H^{+}

Initial 0.05M 0 0

After dissociation 0.05-x x x

∴

Because the value of the ionisation constant is very small, the value of x is very small. Accordingly, we may ignore x in the denominator.

∴ x = 2.2×10^{-6}M

In the presence of 0.01 sodium phenolate(C_{6}H_{5}Na), suppose y is the amount of phenol dissociated, then at equilibrium,

_{6}H

_{5}OH] = 0.05 -y [C

_{6}H

_{5}OH] ≈ 0.05 [C

_{6}H

_{5}O

^{–}] = 0.01 + y ≈ 0.01M, [H

^{+}]=y M

∴

y = 5 × 10^{-10}

α = 10^{-8}.

**Q.45. The first ionisation constant of H _{2}S is 9.1 × 10^{–8}. Calculate the concentration of HS^{–} ion in its 0.1 M solution. How will this concentration be affected if the solution is 0.1 M in HCl also? If the second dissociation constant of H_{2}S is 1.2 × 10^{–13}, calculate the concentration of S^{2–} under both conditions.**

**Ans.**

**To calculate [HS ^{–}]**

To find [HS^{–}]:

**Case 1** – HCl is absent.

Now

K_{a} = ([H^{+}][HS^{–}])/[H_{2}S] = 9.1×10^{-8} (given)

∴ x^{2}/(0.1-x) = 9.1×10^{-8}

But 0.1-x is approximately equal to 0.1. Substituting this value in the equation,

x^{2}/0.1 = 9.1×10^{-8}

x^{2} = 9.1×10^{-9}

x = 9.54× 10^{-5} M

∴ The concentration of HS^{–} is 9.54×10^{-5} M.

**Case 2** – HCl is present

Now,

K_{a} = ([H^{+}][HS^{–}])/[H_{2}S] = (y× (0.1+y))/(0.1-y) = 9.1×10^{-8} (given)

But (0.1 + y) and (0.1 – y) can be approximated to 0.1.

9.1× 10^{-8} = (0.1*y)/0.1

∴ y = [HS^{–}] = 9.1× 10^{-8} M

To calculate [S^{2-}]:

**Case 1 –** HCl is absent.

The dissociation of HS^{–} is given by the equation:

HS^{–} ⇌ H^{+} + S^{2-}[HS^{–}] = 9.1×10^{-5} M

^{+}] = 9.54×10

^{-5}M

K_{a} = ([H^{+}][S^{2-}])/[HS^{–}] = 1.2×10^{-13} (given)

K_{a} = (9.1× 10^{-5} × [S^{2-}])/9.1×10^{-5}

∴ [S^{2-}] = 1.2×10^{-13} M

**Case 2** – HCl is present

^{–}] = 9.1×10

^{-8}M [H

^{+}] = 0.1 M

K_{a} = 1.2×10^{-13} M

= (0.1×[S^{2-}])/ 9.1×10^{-8}

Therefore, [S^{2-}] = 1.092×10^{-19} M.

**Q.46. The ionisation 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.**

CH_{3}COOH** ⇒ ** CH_{3}COO^{–} + H^{+}

**Q.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 ionisation constant of the acid and its pK _{a}.**

**Ans.**

HA ⇌ H^{+} + A^{–}

pH = -log[H^{+}]

log[H^{+}] = -4.15

^{+}] =

^{–}] = [H

^{+}] =

**Q.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.**

**Ans.**

(a) HCl + aq → H^{+} + Cl^{–}

(b) NaOH + aq → Na^{+} + OH^{–}

^{–}] = 5 × 10

^{-3}[H

^{+}] = 10

^{-14}/5 × 10

^{-3}[H

^{+}] = 2 × 10

^{-12}M

pH = -log (2 × 10^{-12})

pH = 11.70

(c) HBr + aq → H^{+} + Br^{–}

^{+}] = 2 × 10

^{-3}M

pH = -log (2 × 10^{-3})

pH = 2.70

(d) KOH + aq → K^{+} + OH^{–}

^{–}] = 2 × 10

^{-3}[H

^{+}] = 10

^{-14}/2 × 10

^{-3}[H

^{+}] = 5 × 10

^{-12}M

pH = -log (5 × 10^{-12})

pH = 11.30

**Q.49. Calculate the pH of the following solutions: ****(I) 2 g of TIOH dissolved in water to give 2 litres of the solution, ****(II) 0.3 g of Ca(OH) _{2} dissolved in water to give 500 mL of the solution, **

**(III) 0.3 g of NaOH dissolved in water to give 200 mL of the solution,**

**(IV) 1 mL of 13.6 M HCl is diluted with water to given 1 litre of the solution.**

**Ans.**

(I) Molar conc. Of TlOH =

(II) Molar conc. Of Ca(OH)_{2}=

(III) Molar conc. of NaOH =

(IV) M_{1}V_{1} = M_{2}V_{2}

**Q.50. The degree of ionisation of a 0.1 M bromoacetic acid solution is 0.132. Calculate the pH of the solution and the p _{Ka} of bromoacetic acid.**

**Ans.**

CH_{2}(Br)COOH CH_{2}(Br)COO^{–} + H^{+}

Initial conc. C 0 0

Conc. at eqm. C – Cα Cα_{ } Cα

**Q.51. What is the pH of 0.001 M aniline solution? The ionisation constant of aniline can be taken from Table 7.7. Calculate the degree of ionisation of aniline in the solution. Also, calculate the ionisation constant of the conjugate acid of aniline.**

**Ans.**

K_{b} =

**Q.52. Calculate the degree of ionisation of 0.05 M acetic acid if its p _{Ka }value is 4.74. How is the degree of dissociation affected when its solution also contains **

**(I) 0.01 M and**

**(II) 0.1 M in HCl?**

**Ans.**

c=0.05M p_{Ka}=4.74 p_{Ka}=-log(K_{a})

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. the dissociation of acetic acid will decrease.

Case 1: 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 0.05-x 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.

Case 2: 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: 0.05 M [CH

_{3}COO

^{–}]=X[H

^{+}]=0.1+X ; 0.1M

Now,

**Q.53. The ionisation constant of dimethylamine is **

**\(\begin{array}{l}5.4 \times × 10^{-4}\end{array} \)**

**. Calculate its 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.**

Now, if 0.1 M of NaOH is added to the solution, then NaOH (being a strong base) undergoes complete ionisation.

NaOH_{(aq)} ↔ Na^{+}_{(aq)} + OH^{–}_{(aq)}

0.1M 0.1M

And,

(CH_{3})_{2} NH + H_{2}O ↔ (CH_{3})_{2} NH_{2}^{+} + OH

(0.02-x) x x

0.02M 0.1M

Then,[ (CH_{3})_{2} NH^{+}_{2}]=x

^{–}]=x+0.1;0.1

It means that in the presence of 0.1 M NaOH, 0.54% of dimethylamine will get dissociated.** **

**Q.54. Calculate the hydrogen ion concentration in the following biological fluids whose pH values are given below: ****(I) Human saliva, 6.4 ****(II) Human stomach fluid, 1.2 ****(III) Human muscle-fluid, 6.83 and ****(IV) Human blood, 7.38.**

**Ans.**

(I) Human saliva, 6.4

pH = 6.4

6.4 = – log [H^{+}]
[H^{+}] = 3.98 x 10^{-7}

(II) Human stomach fluid, 1.2

pH =1.2

1.2 = – log [H^{+}]

∴ [H^{+}] = 0.063

(III) Human muscle fluid 6.83

pH = 6.83

pH = – log [H^{+}]

∴ 6.83 = – log [H^{+}]
[H^{+}] =1.48 x 10^{-7} M

(IV) Human blood, 7.38

pH = 7.38 = – log [H^{+}]

∴ [H^{+}] = 4.17 x 10^{-8} M.

**Q.55. 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 the 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

^{+}] = anitlog(–6.8)=

(II) pH of black coffee = 5.0

Since pH = –log [H^{+}]

5.0 = –log [H^{+}]

log[H^{+}] = –5.0

^{+}] = anitlog(–5.0)=

(III) pH of tomato= 4.2

Since, pH = –log [H^{+}]

4.2 = –log [H^{+}]

log[H^{+}] = –4.2

^{+}] = anitlog(–4.2)=

(IV) pH of lemon juice = 2.2

Since, pH = –log [H^{+}]

2.2 = –log [H^{+}]

log[H^{+}] = –2.2

^{+}] = anitlog(–2.2)=

(V) pH of egg white = 7.8

Since, pH = –log [H^{+}]

7.8 = –log [H^{+}]

log[H^{+}] = –7.8

^{+}] = anitlog(–7.8)=

**Q.56. If 0.561 g of KOH is dissolved in water and given 200 mL of solution at 298 K, calculate the concentrations of potassium, hydrogen and hydroxyl ions. What is its pH?**

**Ans.**

_{(aq)}] = 0.561

**/**(1/5)g/L

= 2.805 g/L

= 2.805 x (1/56.11)

= 0.05M

KOH_{(aq) }→ K^{+}_{ (aq)} + OH^{–}_{(aq)}

^{–}] = 0.05M = [K

^{+}] [H

^{+}][OH

^{–}] = K

_{w}[H+] = Kw/[OH

^{–}] [H+] = 10

^{-14}/0.05 [H+] = 2 × 10

^{-13}M

pH = -log[H^{+}]

pH = -log[2 × 10^{-13}]

pH = 12.70.

**Q.57. 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}=

Sr(OH)_{2(aq)}→ Sr^{2+}_{(aq)} + 2(OH^{–})_{(aq)}

∴ [ Sr^{2+}]= 0.1581M

^{–}]= 2 × 0.1581M

= 0.3126.

Now,

K_{w}=[OH^{–}][H^{+}]

⇒

⇒

∴ pH = 13.50.

**Q.58. The ionisation constant of propanoic acid is **

**\(\begin{array}{l}1.32 \times × 10^{-5}\end{array} \)**

**. Calculate the degree of ionisation of the acid in its 0.05M solution and also its pH. What will be its degree of ionisation if the solution is 0.01 M in HCl also?**

**Ans.**

Let the degree of ionisation of propanoic acid be α.

Then, representing propionic acid as HA, we have:

HA + H_{2}O ↔ H_{3}O^{+} + A^{–}

(0.05-0.0α) ≈ 0.05 0.05α 0.05α

=

= 0.05α^{2}

α = 1.63 × 10^{-2}

Then, [ H_{3}O^{+} ]= 0.05α = 0.05 × 1.63× 10^{-2}= K_{b}. 15×10^{-4}M

∴ pH = 3.09.

In the presence of 0.1M of HCl, let α´ be the degree of ionisation.

Then, [H_{3}O^{+}] = 0.01

^{–}] = 0.05α’ [HA] = 0.05

K_{a} = 0.01 x 0.05α’ / 0.05

⇒ 1.32 x 10^{-5} = 0. 01 α’

α’ = 1.32 x 10^{-3}.

**Q.59. The pH of 0.1 M solution of cyanic acid (HCNO) is 2.34. Calculate the ionisation constant of the acid and its degree of ionisation in the solution. **

**Ans. **

c = 0.1 M

pH = 2.34

-log [H^{+}] = pH

-log [H^{+}] = 2.34

^{+}] = 4.5× 10

^{-3}

Also,

[H^{+}]=cα

4.5 × 10^{-3 }= 0.1× α

α = 0.1/(4.5 × 10^{-3})

α = 0.045

K_{a} = cα^{2}

K_{a} = 0.1 × (0.045)^{2}

K_{a} = 0.0002025

K_{a} = 2.025 x 10^{-4}.

**Q.60. The ionisation constant of nitrous acid is 4.5 × 10 ^{–4}. Calculate the pH of 0.04 M sodium nitrite solution and also its degree of hydrolysis. **

**Ans.**

Sodium nitrite is a salt of NaOH (strong base) and HNO_{2} (weak acid).

Let y mole of salt has undergone hydrolysis, then the concentration of various species present in the solution will be:

Thus,

Thus, the degree of hydrolysis is

**Q.61. A 0.02M solution of pyridinium hydrochloride (C _{5}H_{6}ClN) has a pH of 3.44. Calculate the ionisation constant of C_{5}H_{5}N (pyridine).**

**Ans.**

pH = 3.44As

we know,

Now,

(Given that concentration = 0.02M)

As we know,

=

**Q.62. Predict if the solutions of the following salts are neutral, acidic or basic: NaCl, KBr, NaCN, NH _{4}NO_{3}, NaNO_{2} and KF.**

**Ans.**

- KBr

KBr + H_{2}O

- NH
_{4}NO_{3}

NH_{4}NO_{3} + H_{2}O

_{4}OH(Weak base) + HNO

_{2}(Strong acid). Thus, it is an acidic solution.

- KF

KF + H_{2}O

- NaNO
_{2}

NaNO_{2} + H_{2}O

_{4}OH(Strong base) + HNO

_{2}(Weak acid). Thus, it is a basic solution.

- NaCN

NaCN + H_{2}O

- NaCl

NaCl + H_{2}O

**Q.63. The ionisation constant of chloroacetic acid is 1.35 × 10 ^{–3}. What will be the pH of 0.1 M acid and its 0.1 M sodium salt solution?**

**Ans.**

The K_{a} for chloroacetic acid (ClCH_{2}COOH) is

=

= 0.116

- pH = \(\begin{array}{l}-\log [H^{+}]\end{array} \)= 1.94

ClCH_{2}COONa is a salt of strong base, i.e. NaOH, and weak acid, i.e. ClCH_{2}COOH

Now,

Also,

pH =

**Q.64. Ionic product of water at 310 K is 2.7 × 10 ^{–14}. What is the pH of neutral water at this temperature? **

**Ans.**

Ionic Product,

Assuming,

As,

K_{w} = y^{2}.K_{w} at 310 K is

- y = \(\begin{array}{l}1.64\times 10^{-7}\end{array} \)
- \(\begin{array}{l}[H^{+}] = 1.64\times 10^{-7}\end{array} \)
- pH = \(\begin{array}{l}-\log [H^{+}]\end{array} \)

=

Thus, the pH of neutral water at 310 K temperature is 6.78.

**Q.65. The solubility product constant of Ag _{2}CrO_{4} and AgBr are 1.1 × 10^{–12} and 5.0 × 10^{–13}, respectively. Calculate the ratio of the molarities of their saturated solutions.**

**Ans.**

Ag_{2}CrO_{4}→ 2Ag^{2+} +

Now, K_{sp} = [Ag^{2+}]^{2}

Assuming the solubility of Ag_{2}CrO_{4} is ‘x’.

Thus, [Ag^{2+}] = 2x and

- K
_{sp}= (2x)^{2}×x - \(\begin{array}{l}1.1\times 10^{-12} = 4x^{3}\end{array} \)
- x = \(\begin{array}{l}0.65\times 10^{-4}M\end{array} \)

Assuming the solubility of AgBr is y.

AgBr_{(s)} → Ag^{2+} + Br^{–}

- K
_{sp}= (y)^{2} - \(\begin{array}{l}5.0\times 10^{-13} = y^{2}\end{array} \)

y =

The ratio of molarities to their saturated solution is:

**Q.66. Equal volumes of 0.002 M solutions of sodium iodate and cupric chlorate are mixed together. Will it lead to the precipitation of copper iodate? (For cupric iodate K _{sp} = 7.4 × 10^{–8} ). **

**Ans.**

If cupric chlorate and sodium iodate having equal volumes are mixed together, then the molar concentration of cupric chlorate and sodium iodate will reduce to half. So, the molar concentration of cupric chlorate and sodium iodate in a mixture is 0.001 M.

Na(IO_{3})_{2} → Na^{+} +

0.0001M 0.001 M

Cu(ClO_{3})_{2}→ Cu^{2+} +

0.001M 0.001 M

The Solubility for Cu(IO_{3})_{2} ⇒ Cu^{2+ }(aq) + 2IO_{3}^{– }(aq)

Now, the ionic product of the copper iodate = [Cu^{2+}]

^{2}=

As the value of K_{sp} is more than the ionic product, precipitation will not occur.

**Q.67. 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.**

Here, pH = 3.19

[H_{3}O

^{+}] =

C_{6}H_{5}COOH + H_{2}O →

_{3}O

Assuming the solubility of silver benzoate (C6H5COOAg) is y mol/L,

Now, [Ag^{+}] = y

_{6}H

_{5}COOH] =

10

K_{sp}= [Ag^{+}]

y =

Hence, solubility of C_{6}H_{5}COOAg in buffer of pH = 3.19 is

For water: Assuming the solubility of silver benzoate (C_{6}H_{5}COOAg) is x mol/L,

Now, [Ag^{+}] = y’ M and [CH_{3}COO^{–}] = y’M

K_{sp} = [Ag^{+}]

K_{sp} = (y’)^{2} =

Thus, the solubility of silver benzoate in water is 3.32 times the solubility of silver benzoate in pH = 3.19.

**Q.68. 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.**

Assuming the maximum concentration of each solution is y mol/L,

On mixing the solutions, the volume of the concentration of each solution is reduced to half. After mixing, the maximum concentration of each solution is y/2 mol/L.

Thus, [FeSO_{4}] = [Na_{2}S] = y/2 MSo, [Fe^{2+}] = [FeSO_{4}] = y/2 M

FeS(s)

Ksp = [Fe^{2+}][S^{2-}]

Thus, y =

Thus, if the concentration of FeSO_{4} and Na_{2}SO_{4 }is equal to or less than that of

**Q.69. What is the minimum volume of water required to dissolve 1 g of calcium sulphate at 298 K? (For calcium sulphate, K _{sp} is 9.1 × 10^{–6}).**

**Ans.**

K_{sp} =

Assuming the solubility of calcium sulphate is x,

K_{sp} = x^{2}

Now, the molecular mass os calcium sulphate is 136g/mol. Solubility in calcium sulphate in g/mol is

i.e., 1 litre of H_{2}O will be required to dissolve 0.41 g of calcium sulphate.

Thus, the minimum volume of H_{2}O required to dissolve 1 gram of CaSO_{4} at 298 K is =

**Q.70. The concentration of sulphide ion in 0.1 M 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 FeSO_{4}, MnCl_{2}, ZnCl_{2} and CdCl_{2,} in which of these solutions precipitation will take place? **

**Ans.**

If the ionic product exceeds the K_{sp} value, then only precipitation can take place.

Before mixing: [S^{2-}] = K_{sp} =

^{2+}] = 0.04 M

Volume = 10 mL

Volume = 5 mL

After mixing:[S^{2-}] = ? and [M^{2+}] = ?

Total volume = (10 + 5) = 15 mL

Volume = 15 mL

[S^{2-}] =

^{2+}] =

^{2+}][S

^{2-}]=

Here, the ionic product of CdS and ZnS exceeds its corresponding K_{sp} value. Thus, precipitation will occur in ZnCl_{2} and CdCl_{2} solutions.

### Important Topics Covered in NCERT Solutions for Class 11 Chemistry Chapter 7 are listed below:

- Law of chemical equilibrium
- Equilibrium constant
- Homogeneous and heterogeneous equilibria
- Applications of equilibrium constants
- Relationship between different equilibrium constants
- Factors affecting equilibria
- Ionic equilibrium in solution
- Ionisation of acids and bases
- Buffer Solutions

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## Frequently Asked Questions on NCERT Solutions for Class 11 Chemistry Chapter 7

### List out the important topics covered in NCERT Solutions for Class 11 Chemistry Chapter 7.

Solid-liquid Equilibrium

Equilibrium in Chemical Processes – Dynamic Equilibrium

Law of Chemical Equilibrium and Equilibrium Constant

Homogeneous Equilibria

Heterogeneous Equilibria

Applications of Equilibrium Constants

Relationship between Equilibrium Constant K, Reaction Quotient Q and Gibbs Energy G

Factors Affecting Equilibria

Ionic Equilibrium in Solution

Acids, Bases And Salts

Ionisation of Acids and Bases

Buffer Solutions

Solubility Equilibria of Sparingly Soluble Salts

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