11 Chapter Test Continued Chemistry Answers

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

NCERT Solutions for Class 11 Chemistry Chapter 7 Equilibriumis provided on this page for the perusal of CBSE Class 11 Chemistry students. Detailed, step-by-step solutions for each and every intext and exercise question listed in Chapter 7 of the NCERT Class 11 Chemistry textbook categorized under the can be accessed here. Expert tutors have created these NCERT Solutions for Class 11 Chemistry according to the latest update on the CBSE Syllabus 2022-23 and its guidelines.

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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 the 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

1. 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
2. Equilibrium In Chemical Processes – Dynamic Equilibrium
3. Law Of Chemical Equilibrium And Equilibrium Constant
4. Homogeneous Equilibria
   • Equilibrium Constant In Gaseous Systems
5. Heterogeneous Equilibria
6. Applications Of Equilibrium Constants
   • Predicting The Extent Of A Reaction
   • Predicting The Direction Of The Reaction
   • Calculating Equilibrium Concentrations
7. Relationship Between Equilibrium Constant K, Reaction Quotient Q And Gibbs Energy G
8. Factors Affecting Equilibria
   • Effect Of Concentration Change
   • Effect Of Pressure Change
   • Effect Of Inert Gas Addition
   • Effect Of Temperature Change
   • Effect Of A Catalyst
9. Ionic Equilibrium In Solution
10. Acids, Bases And Salts
    • Arrhenius Concept Of Acids And Bases
    • The Bronsted-lowry Acids And Bases
    • Lewis Acids And Bases
11. Ionization Of Acids And Bases
    • The Ionization Constant Of Water And Its Ionic Product
    • The pH Scale
    • Ionization Constants Of Weak Acids
    • Ionization 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 Ionization Of Acids And Bases
    • Hydrolysis Of Salts And The Ph Of Their Solutions
12. Buffer Solutions
13. 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 are 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 the vapours per unit volume decrease on increasing the volume, therefore, 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: [SO₂]= 0.60M, [O₂] = 0.82M and [SO₃] = 1.90M ?

[latex]2SO_{2}(g)+O_{2}(g) ⇌ 2SO_{3}(g)[/latex]

Ans.

As per the question,

[latex]2SO_{2}(g)+O_{2}(g) ⇌ 2SO_{3}(g)[/latex] (Given) [latex]K_{c}=\frac{[SO_{3}]^{2}}{[SO_{2}]^{2}[O_{2}]}\\ \\ =\frac{(1.9)^{2}M^{2}}{(0.6)^{2}(0.82)M^{3}}\\ \\ =12.229M^{-1}[/latex](approximately)

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

Q.3. At a certain temperature and total pressure of 10⁵Pa, iodine vapour contains 40% by volume of I atoms

I₂(g) ⇌ 2I (g) Calculate K_p for the equilibrium

Ans.

Partial pressure of Iodine atoms (I)

[latex]p_{I}=\frac{40}{100}\times p_{total}\\ \\ =\frac{40}{100}\times 10^{5}\\ \\ =4\times 10^{4}Pa[/latex]

Partial pressure of I₂ molecules,

[latex]p_{I}=\frac{60}{100}\times p_{total}\\ \\ =\frac{60}{100}\times 10^{5}\\ \\ =6\times 10^{4}Pa[/latex]

Now, for the given reaction,

[latex]K_{p}=\frac{(p_{I})^{2}}{p_{I_{2}}} =\frac{(4\times 10^{4})^{2}Pa^{2}}{6\times 10^{4}Pa}\\ \\ =2.67\times 10^{4}Pa[/latex]

Q.4.Write the expression for the equilibrium constant, K_c for each of the following reactions:

(i)[latex]2NOCl(g)\leftrightarrow 2NO(g)+Cl_{2}(g)[/latex]

(ii)[latex]2Cu(NO_{3})_{2}(s)\leftrightarrow 2CuO(s)+4NO_{2}(g)+O_{2}(g)[/latex]

(iii)[latex]CH_{3}COOC_{2}H_{5}(aq)+H_{2}O(1)\leftrightarrow CH_{3}COOH(aq)+C_{2}H_{5}OH(aq)[/latex]

(iv)[latex]Fe^{3+}(aq)+3OH^{-}(aq)\leftrightarrow Fe(OH)_{3}(s)[/latex]

(v)[latex]I_{2}(s)+5F_{2}\leftrightarrow 2IF_{5}[/latex]

Ans.

(i) [latex]K_{C}=\frac{[NO_{g}]^{2}[Cl_{2_{(g)}}]}{[NOCl_{(g)}]^{2}}[/latex]

(ii)[latex]K_{C}=\frac{[CuO_{(s)}]^{2}[NO_{2_{(g)}}]^{4}[O_{2_{(g)}}]}{[Cu(NO_{3})_{2(g)}]^{2}}\\ \\ =[NO_{2(g)}]^{4}[O_{2(g)}][/latex]

(iii)[latex]K_{C}=\frac{[CH_{3}COOH_{(aq)}][C_{2}H_{5}OH_{(aq)}]}{[CH_{3}COOC_{2}H_{5(aq)}][H_{2}O_{(l)}]}\\ \\ =\frac{[CH_{3}COOH_{(aq)}][C_{2}H_{5}OH_{(aq)}]}{[CH_{3}COOC_{2}H_{5(aq)}]}[/latex]

(iv)[latex]K_{C}=\frac{[Fe(OH)_{3(s)}]}{[Fe^{3+}_{(aq)}][OH^{-}_{(aq)}]^{3}}\\ \\ =\frac{1}{[Fe^{3+}_{(aq)}][OH^{-}_{(aq)}]^{3}}[/latex]

(v)[latex]K_{C}=\frac{[IF_{5}]^{2}}{[I_{2(s)}][F_{2}]^{5}}\\ \\ =\frac{[IF_{5}]^{2}}{[F_{2}]^{5}}[/latex]

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

(i)[latex]2NOCl(g) ⇌ 2NO(g)+Cl_{2}(g);\: \: \: K_{p}=1.8\times 10^{-2}\: \: at\: 500K[/latex]

(ii)[latex]CaCO_{3}(s) ⇌ CaO(s)+CO_{2}(g);\: \: \: K_{p}=167\: \: at\: 1073K[/latex]

Ans.

The relation between K_p and K_c is given as:

[latex]K_{p}=K_{c}(RT)^{\Delta n}[/latex]

(i) Given,

R = 0.0831 barLmol^–1K^–1

[latex]\Delta n = 3-2 =1[/latex]

T = 500 K

K_p=[latex]1.8\times 10^{-2}[/latex]

Now,

K_p = K_c (RT) ^∆n

[latex]\Rightarrow 1.8\times 10^{-2}=K_{c}(0.0831\times 500)^{1}\\ \\ \Rightarrow K_{c}=\frac{1.8\times 10^{-2}}{0.0831\times 500}\\ \\ =4.33\times 10^{-4}(approximately)[/latex]

(ii) Here,

∆n =2 – 1 = 1

R = 0.0831 barLmol^–1K^–1

T = 1073 K

K_p= 167

Now,

K_p = K_c (RT) ^∆n

[latex]\Rightarrow 167=K_{c}{(0.0831\times 1073)^{\Delta n}}\\ \\ \Rightarrow K_{c}=\frac{167}{0.0831\times 1073}\\ \\ =1.87(approximately)[/latex]

Q.6. For the following equilibrium, [latex]K_{c}=6.3\times 10^{14}\: \: \: at\: 1000K\\ \\ NO(g)+O_{3(g)} ⇌ NO_{2(g)}+O_{2(g)}[/latex]

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, [latex]K_{c}=\frac{1}{K_{c }}\\ \\ =\frac{1}{6.3\times 10^{14}}\\ \\ =1.59\times 10^{-15}[/latex]

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

Ans.

This is because the molar concentration of a pure solid or liquid is independent of the amount present.

Mole concentration= [latex]\frac{Number\: of\: moles}{Volume}\\ \\ \frac{Mass/molecular\:mass}{Volume}\\ \\ =\frac{Mass}{Volume\,\times\, Molecular \:mass}\\ \\ =\frac{Density}{Molecular\: mass}[/latex]

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

[latex]∴[/latex]Molar concentration are constant.

Q.8.

The reaction between N₂ and O₂ takes place as follows:

[latex]2N_{2}(g)+O_{2} ⇌ 2N_{2}O(g)[/latex]

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 N₂O at a temperature for which K_c =[latex]2.0\times 10^{-37}[/latex], determine the composition of the equilibrium solution.

Ans.

Let the concentration of N₂O at equilibrium be x.

The given reaction is:

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

Initial conc.     0.482 mol                     0.933 mol                                  0

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

[latex][N_{2}]=\frac{0.482-x}{10}\cdot [O_{2}]=\frac{0.933-\frac{x}{2}}{10},[N_{2}O]=\frac{x}{10}[/latex]

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

[latex][N_{2}]=\frac{0.482}{10}=0.0482molL^{-1} \: \:and\:\: [O_{2}]=\frac{0.933}{10}=0.0933molL^{-1}[/latex]

Now,

[latex]K_{c}=\frac{[N_{2}O_{(g)}]^{2}}{[N_{2(g)}][O_{2(g)}]}\\ \\ \Rightarrow 2.0\times 10^{-37}=\frac{(\frac{x}{10})^{2}}{(0.0482)^{2}(0.0933)}\\ \\ \Rightarrow \frac{x^{2}}{100}=2.0 \times 10^{-37}\times (0.0482)^{2}\times (0.0933)\\ \\ \Rightarrow x^{2}=43.35\times 10^{-40}\\ \\ \Rightarrow x=6.6\times 10^{-20}[/latex] [latex][N_{2}O]=\frac{x}{10}=\frac{6.6\times 10^{-20}}{10}\\ \\ =6.6\times 10^{-21}[/latex]

Q.9.

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

2NO_(g)+Br_2(g) ⇌ 2NOBr_(g)

When 0.087 mol of NO and 0.0437 mol of Br₂ 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 Br₂.

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 [latex]\frac{0.0518}{2}[/latex] mol or Br, or 0.0259 mol of NO.

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

[NO] = 0.087 mol [Br2] = 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:

[Br₂] = 0.0437 – 0.0259 = 0.0178 mol

Q.10. At 450 K, K_p= [latex]2.0 \times 10^{10}[/latex]/bar for the given reaction at equilibrium.

2SO₂(g)+O₂(g) ⇌ 2SO₃(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=[latex]2.0 \times 10^{10}bar^{-1}[/latex]

We know that,

[latex]K_{p}=K_{c}(RT)\Delta n\\ \\ \Rightarrow 2.0\times 10^{10}bar^{-1}=K_{c}(0.0831L\, bar\, K^{-1}mol^{-1}\times 450K)^{-1}\\ \\ \Rightarrow K_{c}=\frac{2.0\times 10^{10}bar^{-1}}{(0.0831L\, bar\, K^{-1}mol^{-1}\times 450K)^{-1}}\\ \\ =(2.0\times 10^{10}bar^{-1})(0.0831L\, bar\, K^{-1}mol^{-1}\times 450K)\\ \\ =74.79\times 10^{10}L\, mol^{-1}\\ \\ =7.48\times 10^{11}L\, mol^{-1}\\ \\ =7.48\times 10^{11}\, M^{-1}[/latex]

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) ⇌ 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)               ⇌                         H₂(g)       +      I₂(g)

Initial conc.             0.2 atm                                           0                         0

At equilibrium        0.4 atm                                         0.16/2               0.16/2

=0.08atm         =0.08atm

Therefore,

[latex]K_{p=}\frac{p_{H_{2}}\times p_{I_{2}}}{p^{2}_{HI}}\\ \\ =\frac{0.08\times 0.08}{(0.04)^{2}}\\ \\ =\frac{0.0064}{0.0016}\\ \\ =4.0[/latex]

Hence, the value of K_p for the given equilibrium is 4.0.

Q.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, K_c for the reaction

N₂(g)+3H₂(g) ⇌ 2NH₃(g) is [latex]1.7\times 10^{2}[/latex]

 Is the reaction mixture at equilibrium? If not, what is the direction of the net reaction?

Ans.

The given reaction is:

N₂(g)+3H₂(g) ⇌2NH₃(g)

The given concentration of various species is

[N₂]= [latex]\frac{1.57}{20}mol\, L^{-1}[/latex] [H₂]= [latex]\frac{1.92}{20}mol\, L^{-1}[/latex] [NH₃]= [latex]\frac{8.31}{20}mol\, L^{-1}[/latex]

Now, reaction quotient Q_c is:

[latex]Q=\frac{[NH_{3}]^{2}}{[N_{2}][H_{2}]^{3}}\\ \\ =\frac{(\frac{(8.13)}{20})^{2}}{(\frac{1.57}{20})(\frac{1.92}{20})^{3}}\\ \\ =2.4\times 10^{3}[/latex]

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,

[latex]K_{c}=\frac{[NH_{3}]^{4}[O_{2}]^{5}}{[NO]^{4}[H_{2}O]^{6}}[/latex]

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)+6H₂o_(g)  ⇌4NH_3(g)+5O_2(g)

Q.14. One mole of H₂O and one mole of CO are taken in 10 L vessel and heated to 725 K. At equilibrium, 60% 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:

H₂O_(g) +  CO_(g)⇌ H_2(g)  +  CO_2(g)

Compound	H20	CO	H2	CO2
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 = ([H₂][CO₂])/([H₂O][CO]) = (0.4*0.4)/(0.6*0.6) = 0.444

Q.15. At 700 K, 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 are 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)

[HI]=0.5 molL^-1 will be 1/54.8.

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

[H₂]=[I₂]=x mol L^-1

Therefore, [latex]\frac{[H_{2}][I_{2}]}{[HI]^{2}}=K^{'}_{c}\\ \\ \Rightarrow \frac{x\times x}{(0.5)^{2}}=\frac{1}{54.8}\\ \\ \Rightarrow x^{2}=\frac{0.25}{54.8}\\ \\ \Rightarrow x=0.06754\\ \\ x=0.068molL^{-1}(approximately)[/latex]

Hence, at equilibrium, [H₂]=[I₂]=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, [latex]\frac{[I_{2}][Cl_{2}]}{[IC]^{2}}=K_{c}\\ \\ \Rightarrow \frac{x\times x}{(0.78-2x)^{2}}=0.14\\ \\ \Rightarrow \frac{x^{2}}{(0.78-2x)^{2}}=0.14\\ \\ \Rightarrow \frac{x}{0.78-2x}=0.374\\ \\ \Rightarrow x=0.292-0.748x\\ \\ \Rightarrow 1.748x=0.292\\ \\ \Rightarrow x=0.167[/latex]

Hence, at equilibrium,

[H₂]=[I₂]=0.167 M [HI]= [latex](0.78-2\times 0.167)M\\ \\ =0.446M[/latex]

Q.17. K_p = 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,

C₂H₆(g)                   ⇌               C₂H₄(g)    +     H₂(g)

Initial conc.                 4.0 M                                           0                      0

At equilibrium          (4.0-p)                                            p                     p

We can write,

[latex]\frac{p_{c_{2}H_{4}}\times p_{H_{2}}}{p_{c_{2}H_{6}}}=K_{P}\\ \\ \Rightarrow \frac{p\times p}{40-p}=0.04\\ \\ \Rightarrow p^{2}=0.16-0.04p\\ \\ \Rightarrow p^{2}+0.04p-0.16=0[/latex]

Now,

[latex]p=\frac{-0.04\pm \sqrt{(0.04)^{2}-4\times 1\times (-0.16)}}{2\times 1}\\ \\ =\frac{-0.04\pm 0.80}{2}\\ \\ =\frac{0.76}{2}\: \: \: \: \: \: \: \: \: (Taking \:positive\: value)\\ \\ =0.38[/latex]

Hence, at equilibrium, [C₂H₆] = 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:

CH₃COOH_(l)+ C₂H₅OH_(l) ⇌ CH₃COOC₂H_5(l)+ H₂O_(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,

[latex]Q_{c}=\frac{[CH_{3}COOC_{2}H_{5}][H_{2}O]}{[CH_{3}COOH][C_{2}H_{5}OH]}[/latex]

(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:

CH₃COOH_(l) +  C₂H₅OH_(l) ⇌      CH₃COOC₂H_5(l)     +        H₂O_(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, equilibrium constant for the given reaction is:

[latex]K_{c}=\frac{[CH_{3}COOC_{2}H_{5}][H_{2}O]}{[CH_{3}COOH][C_{2}H_{5}OH]}\\ \\ =\frac{\frac{0.171}{V}\times \frac{0.171}{V}}{\frac{0.829}{V}\times \frac{0.009}{V}}=3.919\\ \\ =3.92(approximately)[/latex]

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

CH₃COOH(l)+C₂H₅OH(l)     ⇌CH₃COOC₂H₅(l)+H₂O(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,

[latex]K_{c}=\frac{[CH_{3}COOC_{2}H_{5}][H_{2}O]}{[CH_{3}COOH][C_{2}H_{5}OH]}\\ \\ =\frac{\frac{0.214}{V}\times \frac{0.214}{V}}{\frac{0.786}{V}\times \frac{0.286}{V}}=0.2037\\ \\ =0.204(approximately)[/latex]

Since Q_c<K_c , equilibrium has not been reached.

Q.19. A sample of pure PCl₅ was introduced into an evacuated vessel at 473 K. After equilibrium was attained, the concentration of PCl₅ was found to be [latex]0.5\times 10^{-1}[/latex]mol L^–1. If value of K_c is [latex]8.3\times 10^{-3}[/latex], what are the concentrations of PCl₃ and Cl₂ at equilibrium?

PCl_5(g) ⇌ PCl_3(g)   +   Cl_2(g)

Ans.

Consider the conc. Of both PCl₃ and Cl₂ at equilibrium be x molL^–1. The given reaction is:

PCl_5(g)      ⇌PCl_3(g)      +      Cl_2(g)

At equilibrium [latex]0.5\times 10^{-10}molL^{-1}[/latex]         x mol L^-1      x mol L^-1

It is given that the value of equilibrium constant , K_c is [latex]8.3\times 10^{-10}molL^{-3}[/latex]

Now we can write the expression for equilibrium as:

[latex]\frac{[PCl_{2}][Cl_{2}]}{[PCl_{3}]}=K_{c}\\ \\ \Rightarrow \frac{x\times x}{0.5\times 10^{-10}}=8.3\times 10^{-3}\\ \\ \Rightarrow x^{2}=4.15\times 10^{-4}\\ \\ \Rightarrow x=2.04\times 10^{-2}\\ \\ =0.0204\\ \\ =0.02(approximately)[/latex]

Therefore, at equilibrium,

[PCl₃]=[Cl₂]=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 CO₂.

 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 CO₂ 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 pCO₂= 0.80 atm

Q_p=[latex]\frac{p_{CO_{2}}}{p_{CO}}\\ \\ =\frac{0.80}{1.4}\\ \\ =0.571[/latex]

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 CO₂ will decrease.

Now, let the increase in pressure of CO = decrease in pressure of CO₂ be p. Then, we can write,

[latex]K_{p}=\frac{p_{CO_{2}}}{p_{CO}}\\ \\ \Rightarrow 0.265=\frac{0.80-p}{1.4+p}\\ \\ \Rightarrow 0.371+0.265p=0.80-p\\ \\ \Rightarrow 1.265p=0.429\\ \\ \Rightarrow p=0.339 atm[/latex]

Therefore, equilibrium partial of CO₂,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

N₂(g)+3H₂(g) ⇌ 2NH₃ 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 N₂, 2.0 mol L^–1 H₂and 0.5 mol L^–1 NH₃. Find out whether the reaction is at equilibrium or not? Find in which direction the reaction proceeds to reach equilibrium.

Ans.

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

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

So,

Q_c=[latex]\frac{[NH_{3}]^{2}}{[N_{2}][H_{2}]^{3}}\\ \\ =\frac{(0.5)^{2}}{(3.0)(2.0)^{3}}\\ \\ =0.0104[/latex]

It is given that K_c=0.061

[latex]∵[/latex]   [latex]Qc \neq K_{c}[/latex], the reaction is not at equilibrium. [latex]∵[/latex]    [latex]Qc < K_{c}[/latex],

the reaction preceeds 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 [latex]3.3 \times 10^{-3}[/latex]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)⇌                             Br₂(g)      +        Cl₂(g)

Initial conc. [latex]3.3\times 10^{-3}[/latex]                   0                         0

At equilibrium [latex]3.3\times 10^{-3}-2x[/latex]         x                          x

Now, we can write,

[latex]\frac{[Br_{2}][Cl_{2}]}{[BrCl]^{2}}=K_{c}\\ \\ \Rightarrow \frac{x\times x}{(3.3\times 10^{-3}-2x)^{2}}=32\\ \\ \Rightarrow \frac{x}{3.3\times 10^{-3}-2x}=5.66\\ \\ \Rightarrow x=18.678\times 10^{-3}-11.32x\\ \\ \Rightarrow 12.32x=18.678\times 10^{-3}\\ \\ \Rightarrow x=1.5\times 10^{-3}[/latex]

So[latex],[/latex]at equilibrium

[latex][BrCl]=3.3\times 10^{-3}-(2\times 1.5\times 10^{-3})\\ \\ =3.3\times 10^{-3}-3.0\times 10^{-3}\\ \\ =0.3\times 10^{-3}\\ \\ =3.0\times 10^{-4}molL^{-1}[/latex]

Q.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) + 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, mass of CO = 90.55 g

Now, the mass of CO₂ = (100 – 90.55) = 9.45 g

Now, number of moles of CO, [latex]n_{CO}=\frac{90.5}{28}=3.234\: mol[/latex]

Number of moles of CO₂, [latex]n_{CO_{2}}=\frac{9.45}{44}=0.215\: mol[/latex]

Partial pressure of CO,

P_CO= [latex]\frac{n_{CO}}{n_{CO}+n_{CO_{2}}}\times p_{total}=\frac{3.234}{3.234+0.215}\times 1=0.938\, atm[/latex]

Partial pressure of CO₂, [latex]P_{CO_{2}}=\frac{n_{CO_{2}}}{n_{CO}+n_{CO_{2}}}\times p_{total}=\frac{0.215}{3.234+0.215}\times 1=0.062, atm[/latex]

Therefore, K_p= [latex]\frac{[CO]^{2}}{[CO_{2}]}\\ \\ =\frac{(0.938)^{2}}{0.062}\\ \\ =14.19[/latex]

For the given reaction,

∆n = 2 – 1 = 1

We know that,

K_p= K_c(RT) ^{[latex]\Delta n[/latex]}

[latex]\Rightarrow 14.19 = K_{c}(0.082\times 1127)^{1}\\ \\ \Rightarrow K_{c}=\frac{14.19}{0.082\times 1127}\\ \\ =0.154(approximately)[/latex]

Q.24. Calculate a) ∆G⁰ and b) the equilibrium constant for the formation of NO₂ from NO and O₂ at 298K

[latex]NO(g)+\frac{1}{2}O_{2}(g) ⇌ NO_{2}(g)[/latex]

 Where;

∆_fG° (NO₂) = 52.0 kJ/mol

 ∆_fG° (NO) = 87.0 kJ/mol

 ∆_fG° (O₂) = 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

[latex]K_{c}=\frac{-35.0\times 10^{-3}}{-2.303\times 8.314\times 298}\\ \\ =6.134\\ \\ ∴ K_{c}=antilog(6.134)\\ \\ =1.36\times 10^{6}[/latex]

Therefore, the equilibrium constant for the given reaction K_c is [latex]1.36\times 10^{6}[/latex]

Q.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 +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.

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 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) 4NH₃ (g) + 5O₂ (g) ⇌ 4NO (g) + 6H₂O (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 are given in (i), (iii), (v), and (vi) will shift in the backward direction

Q.27. The equilibrium constant for the following reaction is [latex]1.6 \times 10^{5}[/latex]at 1024 K.

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.

Given, K_p for the reaction i.e., H₂ (g) + Br₂ (g) ⇒ 2HBr (g) is [latex]1.6 \times 10^{5}[/latex].

Therefore, for the reaction   2HBr (g) ⇒ H₂ (g) + Br₂ (g)  the equilibrium constant will be,

K^' _p= [latex]\frac{1}{K_{p}}\\ \\ =\frac{1}{1.6\times 10^{5}}\\ \\ =6.25\times 10^{-6}[/latex]

Now, let p be the pressure of both H₂ and Br₂ at equilibrium.

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

Initial conc.                   10                                                         0                         0

At equilibrium           10-2p                                                       p                         p

Now, we can write,

[latex]\frac{p_{HBr}\times p_{2}}{p^{2}_{HBr}}=K^{'}_{p}\\ \\ \frac{p\times p}{(10-2p)^{2}}=6.25\times 10^{-6}\\ \\ \frac{p}{10-2p}=2.5\times 10^{-3}\\ \\ p=2.5\times 10^{-2}-(5.0\times 10^{-3})p\\ \\ p+(5.0\times 10^{-3})p=2.5\times 10^{-2}\\ \\ (1005\times 10^{-3})=2.5\times 10^{-2}\\ \\ p=2.49\times 10^{-2}bar=2.5\times 10^{-2}bar (approximately)[/latex]

Therefore, at equilibrium,

[H₂]=[Br₂]= [latex]2.49\times 10^{-2}bar[/latex] [HBr]= [latex]10-2\times (2.49\times 10^{-2})bar\\ \\ =9.95 bar=10 bar(approximately)[/latex]

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

[latex]CH_{4}(g)+H_{2}O(g) ⇌ CO(g)+3H_{2}(g)[/latex]

(a) Write as an expression for K_p 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,

K_p=[latex]\frac{p_{CO}\times p_{H_{2}}^{3}}{p_{CH_{4}}\times p_{H_{2}O}}[/latex]

(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 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 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, PCl₅ is [latex] 8.3 \times 10^{-3} [/latex]. If decomposition is depicted as,

[latex]PCl_{5}(g) ⇌ PCl_{3}(g)+Cl_{2}(g)[/latex]

∆_rH° = 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 PCl₅ is added

(ii) pressure is increased?

(iii) The temperature is increased?

Ans.

(a)[latex]K_{c}=\frac{[PCl_{3}(g)][Cl_{2}(g)]}{[PCl_{3}(g)]}[/latex]

(b)Value of K_c for the reverse reaction at the same temperature is:

[latex]K_{c}^{'} = \frac{1}{K_{c}}\\ \\ =\frac{1}{8.3 \times 10^{-3}} = 1.2048 \times 10^{2}\\ \\ =120.48[/latex]

(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 in 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 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,

[latex]CO_(g)+H_{2}O_(g) ⇌ CO_{2}_(g)+H_{2}_(g)[/latex]

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)       +    H₂O           ⇌CO₂(g)      +     H₂(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

[latex]\frac{P_{CO_{2}}\times P_{H_{2}}}{P_{CO}\times P_{H_{2}O}} = K_{P}\\ \\ \Rightarrow \frac{p\times p}{(4.0-p)(4.0-p)}=10.1\\ \\ \Rightarrow \frac{p}{4.0-p} =3.178\\ \\ \Rightarrow p =12.712-3.178p\\ \\ 4.178p = 12.712\\ \\ p =\frac{12.712}{4.178}\\ \\ p=3.04[/latex]

So, partial pressure of H₂ is 3.04 bar at equilibrium.

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

(a)[latex]Cl_{2}(g)\leftrightarrow 2Cl(g);K_{c}=5 \times 10^{-39}[/latex]

(b)[latex]Cl_{2}(g)+2NO(g)\leftrightarrow 2NOCl(g);K_{c}=3.7 \times 10^{8}[/latex]

(c)[latex]Cl_{2}(g)+2NO_{2}(g)\leftrightarrow 2NO_{2}Cl(g);K_{c}=1.8[/latex]

Ans.

If the value of K_c lies between 10^–3 and 10³, 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 3O₂ (g) ⇌ 2O₃ (g) is [latex]2.0\times 10^{-50} [/latex] at 25°C. If the equilibrium concentration of O₂ in air at 25°C is [latex] 1.6 \times 10^{-2} [/latex] , what is the concentration of O₃?

Ans.

Given,

3O₂ (g) ⇌2O₃ (g)

Then, K_c = [latex] \frac{ [O_{3}(g)]^{2}}{[O_{2}(g)]^{3}}[/latex]

Given that K_c = [latex] 2.0 \times 10^{-50}[/latex] and [O₂(g)] = [latex]1.6 \times 10^{-2}[/latex]

Then,

[latex] 2.0 \times 10^{-50} = \frac { [O_{3}(g)]^{2} }{ [1.6 \times 10^{ -2 }]^{3}}\\ \\ \Rightarrow [O_{3} (g)]^{2} = 2.0 \times 10^{-50} \times (1.6 \times 10^{-2})^{3}\\ \\ \Rightarrow [O_{3} (g)]^{2} = 8.192 \times 10^{-56}\\ \\ \Rightarrow [O_{3} (g)]^{2} = 2.86 \times 10^{-28} M[/latex]

So, the conc. of O₃ is [latex]2.86 \times 10^{-28}M[/latex].

Q.34. The reaction, [latex]CO_{(g)} \; + \; 3H_{2(g)} \; ⇌ \; CH_{4(g)} \; + \; H_{2}O_{(g)}[/latex] at 1300K is at equilibrium in a 1L container. It has 0.30 mol of CO, 0.10 mol of [latex]H_{2}[/latex] and 0.02 mol of [latex]H_{2}O[/latex] and an unknown amount of [latex]CH_{4}[/latex] in the flask. Determine the concentration of [latex]CH_{4}[/latex] in the mixture. The equilibrium constant, [latex]K_{c}[/latex] for the reaction at given temperature is 3.90.

Ans.

Let the concentration of [latex]CH_{4}[/latex]  at equilibrium be y.

[latex]CO_{(g)} \; + \; 3H_{2(g)} \; ⇌ \; CH_{4(g)} \; + \; H_{2}O_{(g)}[/latex]

At equilibrium,

For CO – [latex]\frac{0.3}{1} \; = \; 0.3M[/latex]

For [latex]H_{2}[/latex] – [latex]\frac{0.1}{1} \; = \; 0.1M[/latex]

For [latex]H_{2}O[/latex] – [latex]\frac{0.02}{1} \; = \; 0.02M[/latex] [latex]K_{c}[/latex] = 3.90

Therefore,

[latex]\frac{[CH_{4(g)}][H_{2}O_{(g)}]}{[CO_{(g)}][H_{2(g)}]^{3}} \; = \; K_{c}[/latex] [latex]\frac{y \times 0.02}{0.3 \times (0.1)^{3}} \; = \; 3.9[/latex] [latex]y \; = \; \frac{3.9 \times 0.3 \times (0.1)^{3}}{0.02}[/latex] [latex]y \; = \; \frac{0.00117}{0.02}[/latex] [latex]y \; = \; 0.0585M[/latex] [latex]y \; = \; 5.85 \times 10^{-2}M[/latex]

Therefore, the concentration of [latex]CH_{4}[/latex] at equilibrium is [latex]5.85 \times 10^{-2}M[/latex]

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

[latex]HNO_{2}[/latex], [latex]CN^{-}[/latex], [latex]HClO_{4}[/latex], [latex]F^{-}[/latex], [latex]OH^{-}[/latex], [latex]CO_{3}^{2-}[/latex] and [latex]S^{-}[/latex]

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:

[latex]HNO_{2}[/latex] – [latex]NO_{2}^{-}[/latex] (Base) [latex]CN^{-}[/latex] – HCN (Acid) [latex]HClO_{4}[/latex] – [latex]ClO_{4}^{-}[/latex] (Base) [latex]F^{-}[/latex] – HF (Acid) [latex]OH^{-}[/latex] – [latex]H_{2}O[/latex] (Acid)/ [latex]O^{2-}[/latex] (Base) [latex]CO_{3}^{2-}[/latex] – [latex]HCO_{3}^{-}[/latex] (Acid) [latex]S^{2-}[/latex] – [latex]HS^{-}[/latex] (Acid)

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

[latex]H_{2}O[/latex], [latex]BF_{3}[/latex], [latex]H^{+}[/latex] and [latex]NH_{4}^{+}[/latex]

Ans.

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

[latex]H_{2}O[/latex] – Lewis base [latex]BF_{3}[/latex] – Lewis acid [latex]H^{+}[/latex] – Lewis acid [latex]NH_{4}^{+}[/latex] – Lewis acid

Q.37. What will be the conjugate bases for the Brönsted acids: HF, [latex]H_{2}SO_{4}[/latex] and [latex]HCO_{3}[/latex]

Ans.

The following shows the conjugate bases for the Bronsted acids:

HF – [latex]F^{-}[/latex]

[latex]H_{2}SO_{4}[/latex] – [latex]HSO_{4}^{-}[/latex] [latex]HCO_{3}[/latex] – [latex]CO_{3}^{2-}[/latex]

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

[latex]NH_{2}^{-}[/latex], NH₃ and [latex]HCOO^{-}[/latex]

Ans.

The conjugate acids for the given Bronsted bases are:

NH₂ ^–→   NH₃

NH₃    →   NH₄ ⁺

HCOO^–    →  HCOOH

Q.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.

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

Species        Conjugate acid            Conjugate bases

H₂O                  H₃O⁺                           OH^–

HCO₃ ^–              H₂CO₃                         CO₃ ^2-

HSO₄ ^–               H₂SO₄                        SO₄ ^2-

NH₃                   NH₄ ⁺                          NH₂ ^–

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

(a) [latex]OH^{-}[/latex]

(b) [latex]F^{-}[/latex]

(c) [latex]H^{+}[/latex]

(d) BCl₃

Ans.

(a) [latex]OH^{-}[/latex]

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

(b) [latex]F^{-}[/latex]

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

(c) [latex]H^{+}[/latex]

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

(d) BCl₃:

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.

[latex]pH=-log[H^{+}]\\ \\ =-log(3.8\times 10^{-3})\\ \\ =-log\: 3.8\, -\, log\: 10^{-3}\\ \\ =-log\, 3.8+3\\ \\ =−0.5798+3\\ \\ =2.423[/latex]

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

Ans.

[latex]pH = -log [H^{+}]\\ \\ \Rightarrow log[H^{+}]=-pH\\ \\ \Rightarrow [H^{+}]=antilog(-pH)\\ \\ =antilog(-3.76)\\ \\ =0.000178\\ \\ =1.78\times 10^{-4}[/latex][latex] ∴[/latex] [latex]1.78\times 10^{-4}[/latex] is the concentration of white vinegar sample.

Q.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 F^–,  [latex]K_{b} = \frac{K_{w}}{K_{a}}=\frac{10^{-14}}{(6.8\times 10^{-4})}=1.47\times 10^{-11}[/latex]

For HCOO^–, [latex]K_{b}=\frac{10^{-14}}{(1.8\times 10^{-4})}=5.6\times 10^{-11}[/latex]

For CN^– = [latex]K_{b}=\frac{10^{-14}}{(4.8\times 10^{-9})}=2.08\times 10^{-6}[/latex]

Q.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?

Ans.

C₆H₅OH⇌                          C₆H₅O^– + H⁺

Initial  0.05M                                                   0             0

After dissociation    0.05-x                               x          x

∴ [latex]K_{}\alpha = \frac{x^{2}}{0.05-x}[/latex]

Because the value of the ionization constant is very small, the value of x is very small. Accordingly, w e may ignore x in the denominator.

[latex]x = \sqrt{1\times 10^{-10}\times 0.05}[/latex] [latex]x = \sqrt{5\times 10^{-12}}[/latex]

∴ x = 2.2×10^-6M

In presence of 0.01  sodium phenolate(C₆H₅Na), suppose y is the amount of phenol dissociated, then at equilibrium

[C₆H₅OH] = 0.05 -y [C₆H₅OH] ≈ 0.05 [C₆H₅O^–] = 0.01 + y  ≈ 0.01M, [H⁺]=y M

∴ [latex]K_{\alpha }=\frac{(0.01)(y)}{0.05}=1.0\times 10^{-10}[/latex]

y = 5 × 10^-10

[latex]\alpha = \frac{y}{c}[/latex] [latex]\alpha = \frac{5\times 10^{-10}}{5\times 10^{-2}}[/latex]

α = 10^-8

Q.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 S^2– under both conditions.

Ans.

To calculate [HS^–]

To find [HS^–]:

Case 1 – HCl is absent.

Now

K_a = ([H⁺][HS^–])/[H₂S] = 9.1×10^-8 (given)

∴  x²/(0.1-x) = 9.1×10^-8

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

x²/0.1 = 9.1×10^-8

x² = 9.1×10^-9

x = 9.54× 10^-5 M

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

Case 2 – HCl is present

Now,

K_a = ([H⁺][HS^–])/[H₂S] = (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

[H⁺] = 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

[HS^–] = 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 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.

CH₃COOH ⇒ CH₃COO^– + H⁺[latex]K_{a}=\frac{[CH_{3}COO^{-}][H^{+}]}{[CH_{3}COOH]}=\frac{[H^{+}]^{2}}{[CH_{3}COOH]}\\ \\ \Rightarrow [H^{+}]=\sqrt{K_{a}[CH_{3}COOH]}=\sqrt{(1.74\times 10^{-5})(5\times 10^{-2})}=9.33\times 10^{-4}M\\ \\ [CH_{3}COO^{-}]=[H^{+}]=9.33\times 10^{-4}M\\ \\ pH=-log(9.33\times 1.0^{-4})=4-0.9699=4-0.97=3.03[/latex]

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 ionization constant of the acid and its pK_a.

Ans.

HA          ⇌               H⁺   +   A^–

pH = -log[H⁺]

log[H⁺] = -4.15

[H⁺] = [latex]7.08\times 10^{-5}\, M [/latex][A^–] = [H⁺] = [latex]7.08\times 10^{-5}\, M[/latex] [latex]K_{a}=\frac{[H^{+}][A^{-}]}{[HA]}=\frac{(7.08\times 10^{-5})(7.08\times 10^{-5})}{10^{-2}}=5.0\times 10^{-7}\\ \\ p_{K_{a}}=-log\, K_{a} = -log\, (5.0\times 10^{-7})=7-0.699=6.301[/latex]

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^–

[latex]∴ [H^{+}]=[HCl]=3\times 10^{-3}M\\ \\ pH=-log(3\times 10^{-3})=2.52[/latex]

(b) NaOH + aq → Na⁺ + OH^–

[OH^–] = 5 × 10^-3 [H⁺] = 10^-14/5 × 10^-3 [H⁺] = 2 × 10^-12M

pH = -log (2 × 10^-12)

pH = 11.70

(c) HBr + aq → H⁺ + Br^–

[H⁺] = 2 × 10^-3M

pH = -log (2 × 10^-3)

pH = 2.70

(d) KOH + aq → K⁺ + OH^–

[OH^–] = 2 × 10^-3 [H⁺] = 10^-14/2 × 10^-3 [H⁺] = 5 × 10^-12M

pH = -log (5 × 10^-12)

pH = 11.30

Q.49. Calculate the pH of the following solutions: (I)2g of TIOH dissolved in water to give 2 litres of the solution (II)0.3g of Ca(OH)₂ dissolved in water to given 500mL of the solution (III)0.3g of NaOH dissolved in water to give 200mL 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 = [latex]\frac{2g}{(204+16+1)g\: mol^{-1}}\times \frac{1}{2L}=4.52\times 10^{-3}M\\ \\ [OH_{-}]=[TlOH]=4.52\times 10^{-3}M\\ \\ [H^{+}]=\frac{10^{-14}}{(4.52\times 10^{-3})}=2.21\times 10^{-12}M\\ \\ ∴ pH=-log(2.21\times 10^{-12})=12-(0.3424)=11.66[/latex]

(II) Molar conc. Of Ca(OH)₂=[latex]\frac{0.3g}{(40+34)g\: mol^{-1}}\times \frac{1}{0.5L}=8.11\times 10^{-3}M\\ \\ [OH_{-}]=2[Ca(OH)_{2}]=2\times (8.11\times 10^{-3})M=16.22\times 10^{-3}M\\ \\ pOH=-log(16.22\times 10^{-3})=3-1.2101=1.79\\ \\ pH=14-1.79=12.21[/latex]

(III) Molar conc. of NaOH = [latex]\frac{0.3g}{(40+34)g\: mol^{-1}}\times \frac{1}{0.2L}=3.75\times 10^{-2}M\\ \\ [OH_{-}]=3.75\times 10^{-2}M\\ \\ pOH=-log(3.75\times 10^{-2})=2-0.0574=1.43\\ \\ pH=14-1.43=12.57[/latex]

(IV) M₁V₁ = M₂V₂[latex]∴ 13.6M\times \times 1mL=M_{2}\times 1000mL\\ \\ ∴ M_{2}=1.36\times 10^{-2}M\\ \\ [H^{+}]=[HCl]=1.36\times 10^{-2}M\\ \\ pH=-log(1.36\times 10^{-2})=2-0.1335\simeq 1.87[/latex]

Q.50. The degree of ionization of a 0.1M bromoacetic acid solution is 0.132. Calculate the pH of the solution and the p_Ka of bromoacetic acid.

Ans.

CH₂(Br)COOH                    CH₂(Br)COO^–  +  H⁺

Initial conc.                                  C                                     0                  0

Conc. at eqm.                         C – Cα                                   Cα        Cα

[latex]K_{a}=\frac{C\alpha \cdot C\alpha }{C(1-\alpha )}=\frac{C\alpha ^{2}}{1-\alpha }\simeq C\alpha ^{2}=0.1\times (0.132)^{2}=1.74\times 10^{-3}\\ \\ p_{K_{a}}=-log(1.74\times 10^{-3})=3-0.2405=2.76\\ \\ [H^{+}]=C\alpha =0.1\times 0.132=1.32\times 10^{-2}M\\ \\ pH=-log(1.32\times 10^{-2})=2-0.1206=1.88[/latex]

Q.51. 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 = [latex]4.27 \times 10^{-10}[/latex]c = 0.001MpH =?α =?[latex]K_{b}=c\alpha ^{2}\\ \\ 4.27 \times 10^{-10}=0.001\times \alpha ^{2}\\ \\ 4270\times 10^{-10}=\alpha ^{2}\\ \\ 65.34\times 10^{-5}=\alpha =6.53\times 10^{-4}\\ \\ Then,[anion]=c\alpha =0.001\times 65.34\times10^{-5}=0.065\times10^{-5}[/latex][latex]pOH=-log(0.065\times 10^{-5})\\ \\ =6.187\\ \\ pH=7.813\\ \\ Now,\\ \\ K_{a}\times K_{b}=K_{w}\\ \\ K_{a}=\frac{10^{-14}}{4.27\times 10^{-10}}\\ \\ =2.34\times 10^{-5}[/latex][latex]∴ 2.34\times 10^{-5}[/latex] is the ionization constant.

Q.52. Calculate the degree of ionization of 0.05M 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 (II)0.1 M in HCl?

Ans.

c=0.05M   p_Ka=4.74 p_Ka=-log(K_a)[latex]K_{a}=1.82\times 10^{-5}\\ \\ K_{a}=c\alpha ^{2}\\ \\ \alpha =\sqrt{\frac{K_{a}}{c}}\\ \\ \alpha =\sqrt{\frac{1.82\times 10^{-5}}{5\times 10^{-2}}}=1.908\times 10^{-2}[/latex]

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 1: When 0.01 M HCl is taken.Let x be the amount of acetic acid dissociated after the addition of HCl.

CH₃COOH    ↔       H⁺  +  CH₃COO^–

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.

[latex]K_{a}=\frac{[CH_{3}COO^{-}][H^{+}]}{[CH_{3}COOH]}\\ \\ ∴ =\frac{(0.01)x}{0.05}\\ \\ x=\frac{1.82\times 10^{-5}\times 0.05}{0.01}\\ \\ x=1.82\times 10^{-3}\times 0.05M[/latex]Now,[latex]\alpha =\frac{Amount \: of \: acid\:dissociation}{Amount\:of\:acid\:taken}\\ \\ =\frac{1.82\times 10^{-3}\times 0.05}{0.05}\\ \\ =1.82\times 10^{-3}[/latex]

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:

[CH₃COOH]=0.05 – X: 0.05 M [CH₃COO^–]=X[H⁺]=0.1+X ; 0.1M[latex]K_{a}=\frac{[CH_{3}COO^{-}][H^{+}]}{[CH_{3}COOH]}\\ \\∴K_{a}=\frac{(0.1)X}{0.05}\\ \\ x=\frac{1.82\times 10^{-5}\times 0.05}{0.1}\\ \\ x=1.82\times 10^{-4}\times 0.05M[/latex]

Now,[latex]\alpha =\frac{Amount \: of \: acid\:dissociation}{Amount\:of\:acid\:taken}\\ \\ =\frac{1.82\times 10^{-4}\times 0.05}{0.05}\\ \\ =1.82\times 10^{-4}[/latex].

Q.53. The ionization constant of dimethylamine is[latex]5.4 \times × 10^{-4}[/latex]. 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.

[latex]K_{b}=5.4\times 10^{-4}\\ \\ c=0.02M\\ \\ Then,\alpha =\sqrt{\frac{K_{b}}{c}}\\ \\ =\sqrt{\frac{5.4\times 10^{-4}}{0.02}}=0.1643[/latex]

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,

(CH₃)₂ NH   +  H₂O        ↔                   (CH₃)₂ NH₂ ⁺   +  OH

(0.02-x)                                                     x                    x

0.02M                                                                             0.1M

Then,[ (CH₃)₂ NH⁺ ₂]=x

[OH^– ]=x+0.1;0.1 [latex]\Rightarrow K_{b}=\frac{[(CH_{3})_{2}NH_{2}^{+}][OH^{-}]}{[(CH_{3})_{2}NH]}\\ \\ 5.4\times 10^{-4}=\frac{x\times 0.1}{0.02}\\ \\ x=0.0054[/latex]

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 are given below: (I) Human saliva, 6.4 (II) Human stomach fluid, 1.2 (III) Human muscle-fluid, 6.83 (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 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)= [latex]1.5\times 10^{-7} M[/latex]

(II) pH ofblack coffee = 5.0

Since, pH = –log [H⁺]

5.0 = –log [H⁺]

log[H⁺] = –5.0

[H⁺] = anitlog(–5.0)= [latex] 10^{-5} M[/latex]

(III) pH of tomato= 4.2

Since, pH = –log [H⁺]

4.2 = –log [H⁺]

log[H⁺] = –4.2

[H⁺] = anitlog(–4.2)= [latex]6.31\times 10^{-5} M[/latex]

(IV) pH of lemon juice= 2.2

Since, pH = –log [H⁺]

2.2 = –log [H⁺]

log[H⁺] = –2.2

[H⁺] = anitlog(–2.2)= [latex]6.31\times 10^{-3} M[/latex]

(V) pH of egg white= 7.8

Since, pH = –log [H⁺]

7.8 = –log [H⁺]

log[H⁺] = –7.8

[H⁺] = anitlog(–7.8)= [latex]1.58\times 10^{-8} M[/latex]

Q.56. 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)

= 0.05M

KOH_(aq)→ K⁺ _(aq) + OH^– _(aq)

[OH^–] = 0.05M = [K⁺] [H⁺][OH^–] = K_w [H+] = Kw/[OH^–] [H+] = 10^-14/0.05 [H+] = 2 × 10^-13M

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=[latex]\frac{19.23}{121.63}M\\ \\ =0.1581M

Sr(OH)_2(aq)→  Sr²⁺ _(aq) + 2(OH^–)_(aq)

∴[ Sr²⁺]= 0.1581M

[OH^–]=  2 × 0.1581M

=0.3126

Now,

K_w=[OH^–][H⁺]

⇒ [latex]\frac{10^{-14}}{0.3126}=[H^{+}][/latex]

⇒ [latex][H^{+}]=3.2\times 10^{-14}[/latex]

∴ pH = 13.50

Q.58. The ionization constant of propanoic acid is [latex]1.32 \times × 10^{-5}[/latex]. 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 + H₂O      ↔       H₃O⁺       +       A^–

(0.05-0.0α) ≈ 0.05      0.05α              0.05α

[latex]K_{\alpha }=\frac{[H_{3}O^{+}][A^{-}]}{[HA]}[/latex]

= [latex]\frac{(0.05\alpha )(0.05\alpha )}{0.05}[/latex]

= 0.05α²

[latex]\alpha =\sqrt{\frac{K_{\alpha }}{0.05}}[/latex]

α = 1.63 × 10^-2

Then,[ H₃O⁺ ]= 0.05α = 0.05 × 1.63× 10^-2= K_b. 15×10^-4M

∴ pH=3.09

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

Then, [H₃O⁺]  = 0.01

[A^–]  =  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 0.1M 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× 10^-3

Also,

[H⁺]=cα

4.5 × 10^-3= 0.1× α

α = 0.1/(4.5 × 10^-3)

α = 0.045

K_a = cα²

K_a = 0.1 × (0.045)²

K_a = 0.0002025

K_a = 2.025 x 10^-4

Q.60. The ionization 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 HNO₂ (weak acid).

[latex]NO_{2}^{-} + H_{2}O \leftrightarrow HNO_{2} + OH^{-}[/latex] [latex]K_{h} = \frac{[HNO_{2}][OH^{-}]}{[NO_{2}^{-}]}[/latex] [latex]\Rightarrow \;\frac{K_{w}}{K_{a}} = \frac{10^{-14}}{4.5\times 10^{-4}} = 0.22\times 10^{-10}[/latex]

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

[latex][NO_{2}^{-}] = 0.04 – y; 0.04[/latex] [latex][HNO_{2}] = y[/latex] [latex][OH^{-}] = y[/latex] [latex]K_{h} = \frac{y^{2}}{0.04} = 0.22\times 10^{-10}[/latex] [latex]y^{2} = 0.0088\times 10^{-10}[/latex] [latex]y = 0.093\times 10^{-5}[/latex] [latex]∴ [OH^{-}]= 0.093\times 10^{-5}M[/latex] [latex][H_{3}O^{+}] = \frac{10^{-14}}{0.093\times 10^{-5}} = 10.75\times 10^{-9}M[/latex]

Thus, [latex]pH = -\log (10.75\times 10^{-9})[/latex]= 7.96

Thus, the degree of hydrolysis is

[latex]= \frac{y}{0.04} = \frac{0.093\times 10^{-5}}{0.04}[/latex] [latex]= 2.325\times 10^{-5}[/latex]

Q.61. A 0.02M solution of pyridinium hydrochloride (C₅H₆ClN)  is having pH = 3.44. Calculate the ionization constant of C₅H₅N (pyridine).

Ans.

pH = 3.44As

we know,

[latex]pH = \log [H^{+}][/latex] [latex]∴ [H^{+}] = 3.63\times 10^{-4}[/latex]

Now, [latex]K_{h} = \frac{3.63\times 10^{-4}}{0.02}[/latex];

(Given that concentration = 0.02M)

[latex]\Rightarrow K_{h} = 6.6\times 10^{-6}[/latex]

As we know that,

[latex]K_{h} = \frac{K_{w}}{K_{a}}[/latex] [latex]K_{a} = \frac{K_{w}}{K_{h}} = \frac{10^{-14}}{6.6\times 10^{-6}}[/latex]

= [latex]1.51\times 10^{-9}[/latex]

Q.62 Predict if the solutions of the following salts are neutral, acidic or basic: NaCl, KBr, NaCN, NH₄NO₃, NaNO₂ and KF.

Ans.

• KBr

KBr + H₂O [latex]\leftrightarrow[/latex] KOH (Strong base)+ HBr (Strong acid)Thus, it is a neutral solution.

• NH₄NO₃

NH₄NO₃ + H₂O [latex]\leftrightarrow[/latex] NH₄OH(Weak base) + HNO₂ (Strong acid)Thus, it is an acidic solution.

• KF

KF + H₂O [latex]\leftrightarrow[/latex] KOH (Strong base) + HF (weak acid)Thus, it is a basic solution.

• NaNO₂

NaNO₂ + H₂O [latex]\leftrightarrow[/latex] NH₄OH(Strong base) + HNO₂(Weak acid)Thus, it is a basic solution.

• NaCN

NaCN + H₂O [latex]\leftrightarrow[/latex] HCN (Weak acid) + NaOH (Strong base)Thus, it is a basic solution.

• NaCl

NaCl + H₂O [latex]\leftrightarrow[/latex] NaOH (Strong base) + HCL (Strong acid)Thus, it is a neutral solution.

Q.63. 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.

The K_a for chloroacetic acid (ClCH₂COOH) is [latex]1.35\times 10^{-3}[/latex].

[latex]\Rightarrow K_{a} = c\alpha ^{2}[/latex] [latex]∴ \alpha = \sqrt{\frac{K_{a}}{c}}[/latex]

= [latex]\sqrt{\frac{1.35\times 10^{-3}}{0.1}}[/latex]; (given concentration = 0.1M)

[latex]\alpha = \sqrt{1.35\times 10^{-2}}[/latex]

= 0.116

[latex] ∴ [H^{+}] = c\alpha = 0.1×0.116 = 0.0116[/latex]

• pH = [latex]-\log [H^{+}][/latex] = 1.94

ClCH₂COONa is a salt of strong base i.e. NaOH, and weak acid i.e. ClCH₂COOH

[latex]ClCH_{2}COO^{-} + H_{2}O \leftrightarrow ClCH_{2}COOH + OH^{-}[/latex] [latex]K_{h} = \frac{[ClCH_{2}COO][OH^{-}]}{[ClCH_{2}COO^{-}]}[/latex]

Now, [latex]K_{h} = \frac{K_{w}}{K_{a}}[/latex]

[latex]K_{h} = \frac{10^{-14}}{1.35\times 10^{-3}}[/latex] [latex]= 0.740\times 10^{-11}[/latex]

Also, [latex]K_{h} = \frac{y^2}{0.1}[/latex] [latex]\Rightarrow \; 0.740\times 10^{-11} = \frac{y^2}{0.1}[/latex] [latex]\Rightarrow \; 0.0740\times 10^{-11} = y^2[/latex]

[latex]y = 0.86\times 10^{-6}[/latex] [latex][OH^{-}] = 0.86\times 10^{-6}[/latex] [latex]∴ [H^{+}] =\frac{K_{w}}{0.86\times 10^{-6}}[/latex] [latex]=\frac{10^{-14}}{0.86\times 10^{-6}}[/latex] [latex][H^{+}] = 1.162\times 10^{-3}[/latex]

pH = [latex]- \log [H^{+}][/latex]= 7.94

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,[latex]K_{w} = [H^{+}][OH^{-}][/latex]

Assuming, [latex][H^{+}][/latex] = y

As, [latex][H^{+}] = [OH^{-}][/latex],

K_w = y².K_w at 310 K is [latex]2.7\times 10^{-14}[/latex] [latex]∴ 2.7\times 10^{-14} = y^{2}[/latex]

• y = [latex]1.64\times 10^{-7}[/latex]
• [latex][H^{+}] = 1.64\times 10^{-7}[/latex]
• pH = [latex]-\log [H^{+}][/latex]

= [latex]-\log [1.64\times 10^{-7}][/latex]= 6.78

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

Q.65. 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.

Ag₂CrO₄→ 2Ag²⁺ + [latex]CrO_{4}^{-}[/latex]

Now, K_sp = [Ag²⁺]²[latex][CrO_{4}^{-}][/latex]

Assuming the solubility of Ag₂CrO₄ is 'x'.

Thus,[Ag²⁺] = 2x and [latex]CrO_{4}^{-}[/latex]  = x

• K_sp = (2x)²×x
• [latex]1.1\times 10^{-12} = 4x^{3}[/latex]
• x = [latex]0.65\times 10^{-4}M[/latex]

Assuming the solubility of AgBr is y.

AgBr_(s) → Ag²⁺ + Br^–

• K_sp = (y)²
• [latex]5.0\times 10^{-13} = y^{2}[/latex]

y = [latex]7.07\times 10^{-7}M[/latex]

The ratio of molarities to their saturated solution is:

[latex]\frac{x}{y} = \frac{0.65\times10^{-4}M}{7.07\times10^{-7}M} = 91.9[/latex]

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

Ans.

Cupric chlorate and sodium iodate having equal volume 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 0.001M.

Na(IO₃)₂ → Na⁺ + [latex]IO_{3}^{-}[/latex]

0.0001M                0.001M

Cu(ClO₃)₂→ Cu²⁺ + [latex]2CIO_{3}^{-}[/latex]

0.001M                0.001M

The Solubility for Cu(IO₃)₂ ⇒ Cu²⁺ (aq) + 2IO₃ ^–(aq)

Now, the ionic product of the copper iodate is:= [Cu²⁺] [latex][IO_{3}^{-}]^{2}[/latex]= (0.001)(0.001)²= [latex]1.0\times 10^{-9}M[/latex]

As the value of K_sp is more than Ionic product. Thus, precipitation will not occur.

Q.67. The ionization 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

[H₃O⁺] = [latex]6.46\times 10^{-4}M[/latex]

C₆H₅COOH + H₂O →[latex]C₆H₅COO^{-}[/latex] + H₃O

[latex]K_{a}=\frac{[C_{6}H_{5}COO^{-}][H_{3}O^{+}]}{C_{6}H_{5}COOH}[/latex] [latex]K_{a}=\frac{[C_{6}H_{5}COOH]}{C_{6}H_{5}COO^{-}} = \frac{[H_{3}O^{+}]}{K_{a}} = \frac{6.46\times 10^{-4}}{6.46\times 10^{-5}} = 10[/latex]

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

Now,  [Ag⁺] = y

[C₆H₅COOH] = [latex][C₆H₅COO^{-}][/latex] = y

10[latex][C₆H₅COO^{-}][/latex] + [latex][C₆H₅COO^{-}][/latex] = y

[latex][C₆H₅COO^{-}][/latex] = y/11

K_sp= [Ag⁺][latex][C₆H₅COO^{-}][/latex] [latex]2.5\times 10^{-13} = y\frac{y}{11}[/latex]

y = [latex]1.66\times 10^{-6}[/latex] mol/L

Hence, solubility of C₆H₅COOAg in buffer of pH = 3.19 is [latex]1.66\times 10^{-6}[/latex] mol/L.

For, water: Assuming the solubility of silver benzoate (C₆H₅COOAg) is x mol/L.

Now,  [Ag⁺] = y' M and [CH₃COO^–] = y'M

K_sp = [Ag⁺][latex][C₆H₅COO^{-}][/latex]

K_sp = (y')² = [latex]\sqrt{K_{sp}} = \sqrt{2.5\times 10^{-13}} = 5\times 10^{-7} mol/L[/latex]

[latex]∴ \frac{y}{x} = \frac{1.66\times 10^{-6}}{5\times 10^{-7}} = 3.32[/latex]

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, [FeSO₄] = [Na₂S] = y/2 MSo, [Fe²⁺] = [FeSO₄] = y/2 M

FeS(s) [latex]\leftrightarrow[/latex] [latex]Fe^{2+}_{(aq)} + S^{2-}_{(aq)}[/latex]

Ksp = [Fe²⁺][S^2-] [latex]6.3\times 10^{-18} = (\frac{y}{2})(\frac{y}{2})[/latex] [latex]\frac{y^{2}}{4} = 6.3\times 10^{-18}[/latex]

Thus, y = [latex]5.02\times 10^{-9}[/latex]

Thus, if the concentration of FeSO₄ and Na₂SO₄ are equal to or less than that of [latex]5.02\times 10^{-9}M[/latex], then there won't be precipitation of FeS.

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

Ans.

[latex]CaSO_{4(s)}\leftrightarrow Ca^{2+}_{(aq)} + SO_{4}^{2-}(aq)[/latex]

K_sp = [latex][Ca^{2+}][SO_{4}^{2-}][/latex]

Assuming the solubility of calcium sulphate is x. So,

K_sp = x²

[latex]∴ \;9.1\times 10^{-6} = x^{2}[/latex] [latex]∴ \;x = 3.02\times 10^{-3} mol/L[/latex]

Now, molecular mass os calcium sulphate is 136g/mol. Solubility in calcium sulphate in g/mol is[latex]= 3.02\times 10^{-3} \times 136[/latex]= 0.41 g/L

i.e. 1 litre H₂O will be required to dissolve 0.41g of calcium sulphate.Thus, minimum volume of H₂O required to dissolve 1 gram of CaSO₄ at 298K is= [latex]\frac{1}{0.41} L = 2.44 L[/latex]

Q.70. 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 0.04M solution of the following: FeSO₄, MnCl₂, ZnCl₂ and CdCl₂ . 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 = [latex]1.0\times 10^{-19}M

[/latex] [M²⁺] = 0.04M

Volume = 10mL

Volume = 5mL

After mixing:[S^2-] = ? and    [M²⁺] = ?

Total volume = (10 + 5) = 15mL

Volume = 15mL

[S^2-] = [latex]\frac{1.0\times 10^{-19}\times 10}{15} = 6.67\times 10^{-20}M[/latex] [M²⁺] = [latex]\frac{0.04\times 5}{15} = 1.33\times 10^{-2}M[/latex] Now, the ionic product = [M²⁺][S^2-]= [latex](1.33\times 10^{-2})(6.67\times 10^{-20})[/latex]= [latex]8.87\times 10^{-22}[/latex]

Here, the ionic product of CdS and ZnS exceeds its corresponding K_sp value. Thus, precipitation will occur in  ZnCl₂ and CdCl₂ solutions.

Important topics covered in NCERT Solutions for Class 11 Chemistry Chapter 7 are:

• 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
• Ionization of acids and bases
• Buffer Solutions

Why BYJU'S NCERT Solutions are a cut above the rest?

BYJU'S primary mission is to create top-notch educational content accessible to each and every student in the country. The NCERT Solutions that we offer have been thoughtfully structured to offer several benefits to the students that utilize them. For example, these NCERT Solutions for Class 11 Chemistry Chapter 7 are crafted to be concept focused and can, therefore, be used even for chapter revisions.

Crafted by subject matter experts to be highly student-friendly, these NCERT Solutions for Class 11 Chemistry feature simple, explanative solutions to even relatively complex questions. Also, the BYJU'S support team is always available to solve any doubts and issues faced by students.

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.

The important concepts covered in NCERT Solutions for Class 11 Chemistry Chapter 7 are
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
Ionization Of Acids And Bases
Buffer Solutions
Solubility Equilibria Of Sparingly Soluble Salts

What is the meaning of buffer solution in NCERT Solutions for Class 11 Chemistry Chapter 7?

According to NCERT Solutions for Class 11 Chemistry Chapter 7 Buffer Solution is a water solvent-based solution that consists of a mixture containing a weak acid and the conjugate base of the weak acid, or a weak base and the conjugate acid of the weak base. They resist a change in pH upon dilution or upon the addition of small amounts of acid/alkali to them.

How are the NCERT Solutions for Class 11 Chemistry Chapter 7 helpful for board exam preparation?

Practising NCERT Solutions for Class 11 Chemistry Chapter 7 help the students to top the board exams and ace a subject. These solutions are devised, based on the recently updated CBSE Syllabus, covering all the crucial topics of the respective subjects. Hence, solving these questions will make the students more confident to face the board exams. Topics given in these solutions form the basis for top scores. It also helps the students to get familiar with answering questions of all difficulty levels. These solutions are highly recommended to the students for referencing and to practise for their board exams.

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