NCERT Solutions-Equilibrium for Class 11 Chemistry

February 11, 2025December 18, 2024 by Neeraj Anand, Param Anand

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: [SO₂]= 0.60M, [O₂] = 0.82M and [SO₃] = 1.90M?

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

Ans.

As per the question,

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

(Given)

\(\begin{array}{l}K_{c}=\frac{[SO_{3}]^{2}}{[SO_{2}]^{2}[O_{2}]}\\ \\ =\frac{(1.9)^{2}M^{2}}{(0.6)^{2}(0.82)M^{3}}\\ \\ =12.229M^{-1}\end{array} \)

(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.

The partial pressure of Iodine atoms (I)

\(\begin{array}{l}p_{I}=\frac{40}{100}\times p_{total}\\ \\ =\frac{40}{100}\times 10^{5}\\ \\ =4\times 10^{4}Pa\end{array} \)

The partial pressure of I₂ molecules,

\(\begin{array}{l}p_{I}=\frac{60}{100}\times p_{total}\\ \\ =\frac{60}{100}\times 10^{5}\\ \\ =6\times 10^{4}Pa\end{array} \)

Now, for the given reaction,

\(\begin{array}{l}K_{p}=\frac{(p_{I})^{2}}{p_{I_{2}}} =\frac{(4\times 10^{4})^{2}Pa^{2}}{6\times 10^{4}Pa}\\ \\ =2.67\times 10^{4}Pa\end{array} \)

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)

\(\begin{array}{l}K_{C}=\frac{[NO_{g}]^{2}[Cl_{2_{(g)}}]}{[NOCl_{(g)}]^{2}}\end{array} \)

(ii)

\(\begin{array}{l}K_{C}=\frac{[CuO_{(s)}]^{2}[NO_{2_{(g)}}]^{4}[O_{2_{(g)}}]}{[Cu(NO_{3})_{2(g)}]^{2}}\\ \\ =[NO_{2(g)}]^{4}[O_{2(g)}]\end{array} \)

(iii)

\(\begin{array}{l}K_{C}=\frac{[CH_{3}COOH_{(aq)}][C_{2}H_{5}OH_{(aq)}]}{[CH_{3}COOC_{2}H_{5(aq)}][H_{2}O_{(l)}]}\\ \\ =\frac{[CH_{3}COOH_{(aq)}][C_{2}H_{5}OH_{(aq)}]}{[CH_{3}COOC_{2}H_{5(aq)}]}\end{array} \)

(iv)

\(\begin{array}{l}K_{C}=\frac{[Fe(OH)_{3(s)}]}{[Fe^{3+}_{(aq)}][OH^{-}_{(aq)}]^{3}}\\ \\ =\frac{1}{[Fe^{3+}_{(aq)}][OH^{-}_{(aq)}]^{3}}\end{array} \)

(v)

\(\begin{array}{l}K_{C}=\frac{[IF_{5}]^{2}}{[I_{2(s)}][F_{2}]^{5}}\\ \\ =\frac{[IF_{5}]^{2}}{[F_{2}]^{5}}\end{array} \)

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:

\(\begin{array}{l}K_{p}=K_{c}(RT)^{\Delta n}\end{array} \)

(i) Given,

R = 0.0831 barLmol^–1K^–1

\(\begin{array}{l}\Delta n = 3-2 =1\end{array} \)

T = 500 K

K_p=

\(\begin{array}{l}1.8\times 10^{-2}\end{array} \)

Now,

K_p = K_c (RT) ^∆n

\(\begin{array}{l}\Rightarrow 1.8\times 10^{-2}=K_{c}(0.0831\times 500)^{1}\\ \\ \Rightarrow K_{c}=\frac{1.8\times 10^{-2}}{0.0831\times 500}\\ \\ =4.33\times 10^{-4}(approximately)\end{array} \)

(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

\(\begin{array}{l}\Rightarrow 167=K_{c}{(0.0831\times 1073)^{\Delta n}}\\ \\ \Rightarrow K_{c}=\frac{167}{0.0831\times 1073}\\ \\ =1.87(approximately)\end{array} \)

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,

\(\begin{array}{l}K_{c}=\frac{1}{K_{c }}\\ \\ =\frac{1}{6.3\times 10^{14}}\\ \\ =1.59\times 10^{-15}\end{array} \)

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=

\(\begin{array}{l}\frac{Number\: of\: moles}{Volume}\\ \\ \frac{Mass/molecular\:mass}{Volume}\\ \\ =\frac{Mass}{Volume\,\times\, Molecular \:mass}\\ \\ =\frac{Density}{Molecular\: mass}\end{array} \)

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

\(\begin{array}{l}∴\end{array} \)

Molar concentration is constant.

Q.8. The reaction between N₂ and O₂ 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 N₂O 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 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

\(\begin{array}{l}[N_{2}]=\frac{0.482-x}{10}\cdot [O_{2}]=\frac{0.933-\frac{x}{2}}{10},[N_{2}O]=\frac{x}{10}\end{array} \)

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

\(\begin{array}{l}[N_{2}]=\frac{0.482}{10}=0.0482molL^{-1} \: \:and\:\: [O_{2}]=\frac{0.933}{10}=0.0933molL^{-1}\end{array} \)

Now,

\(\begin{array}{l}K_{c}=\frac{[N_{2}O_{(g)}]^{2}}{[N_{2(g)}][O_{2(g)}]}\\ \\ \Rightarrow 2.0\times 10^{-37}=\frac{(\frac{x}{10})^{2}}{(0.0482)^{2}(0.0933)}\\ \\ \Rightarrow \frac{x^{2}}{100}=2.0 \times 10^{-37}\times (0.0482)^{2}\times (0.0933)\\ \\ \Rightarrow x^{2}=43.35\times 10^{-40}\\ \\ \Rightarrow x=6.6\times 10^{-20}\end{array} \)

\(\begin{array}{l}[N_{2}O]=\frac{x}{10}=\frac{6.6\times 10^{-20}}{10}\\ \\ =6.6\times 10^{-21}\end{array} \)

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 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

\(\begin{array}{l}\frac{0.0518}{2}\end{array} \)

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=

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

/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=

\(\begin{array}{l}2.0 \times 10^{10}bar^{-1}\end{array} \)

We know that,

\(\begin{array}{l}K_{p}=K_{c}(RT)\Delta n\\ \\ \Rightarrow 2.0\times 10^{10}bar^{-1}=K_{c}(0.0831L\, bar\, K^{-1}mol^{-1}\times 450K)^{-1}\\ \\ \Rightarrow K_{c}=\frac{2.0\times 10^{10}bar^{-1}}{(0.0831L\, bar\, K^{-1}mol^{-1}\times 450K)^{-1}}\\ \\ =(2.0\times 10^{10}bar^{-1})(0.0831L\, bar\, K^{-1}mol^{-1}\times 450K)\\ \\ =74.79\times 10^{10}L\, mol^{-1}\\ \\ =7.48\times 10^{11}L\, mol^{-1}\\ \\ =7.48\times 10^{11}\, M^{-1}\end{array} \)

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,

\(\begin{array}{l}K_{p=}\frac{p_{H_{2}}\times p_{I_{2}}}{p^{2}_{HI}}\\ \\ =\frac{0.08\times 0.08}{(0.04)^{2}}\\ \\ =\frac{0.0064}{0.0016}\\ \\ =4.0\end{array} \)

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 is

N₂(g)+3H₂(g) ⇌ 2NH₃(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:

N₂(g)+3H₂(g) ⇌ 2NH₃(g)

The given concentration of various species is

[N₂]=

\(\begin{array}{l}\frac{1.57}{20}mol\, L^{-1}\end{array} \)

[H₂]=

\(\begin{array}{l}\frac{1.92}{20}mol\, L^{-1}\end{array} \)

[NH₃]=

\(\begin{array}{l}\frac{8.31}{20}mol\, L^{-1}\end{array} \)

Now, the reaction quotient Q_c is:

\(\begin{array}{l}Q=\frac{[NH_{3}]^{2}}{[N_{2}][H_{2}]^{3}}\\ \\ =\frac{(\frac{(8.13)}{20})^{2}}{(\frac{1.57}{20})(\frac{1.92}{20})^{3}}\\ \\ =2.4\times 10^{3}\end{array} \)

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)+6H₂o_(g) ⇌ 4NH_3(g)+5O_2(g).

Q.14. One mole of H₂O 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,

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, 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)

[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,

\(\begin{array}{l}\frac{[H_{2}][I_{2}]}{[HI]^{2}}=K^{‘}_{c}\\ \\ \Rightarrow \frac{x\times x}{(0.5)^{2}}=\frac{1}{54.8}\\ \\ \Rightarrow x^{2}=\frac{0.25}{54.8}\\ \\ \Rightarrow x=0.06754\\ \\ x=0.068molL^{-1}(approximately)\end{array} \)

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,

\(\begin{array}{l}\frac{[I_{2}][Cl_{2}]}{[IC]^{2}}=K_{c}\\ \\ \Rightarrow \frac{x\times x}{(0.78-2x)^{2}}=0.14\\ \\ \Rightarrow \frac{x^{2}}{(0.78-2x)^{2}}=0.14\\ \\ \Rightarrow \frac{x}{0.78-2x}=0.374\\ \\ \Rightarrow x=0.292-0.748x\\ \\ \Rightarrow 1.748x=0.292\\ \\ \Rightarrow x=0.167\end{array} \)

Hence, at equilibrium,

[H₂]=[I₂]=0.167 M
[HI]=

\(\begin{array}{l}(0.78-2\times 0.167)M\\ \\ =0.446M\end{array} \)

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,

\(\begin{array}{l}\frac{p_{c_{2}H_{4}}\times p_{H_{2}}}{p_{c_{2}H_{6}}}=K_{P}\\ \\ \Rightarrow \frac{p\times p}{40-p}=0.04\\ \\ \Rightarrow p^{2}=0.16-0.04p\\ \\ \Rightarrow p^{2}+0.04p-0.16=0\end{array} \)

Now,

\(\begin{array}{l}p=\frac{-0.04\pm \sqrt{(0.04)^{2}-4\times 1\times (-0.16)}}{2\times 1}\\ \\ =\frac{-0.04\pm 0.80}{2}\\ \\ =\frac{0.76}{2}\: \: \: \: \: \: \: \: \: (Taking \:positive\: value)\\ \\ =0.38\end{array} \)

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,

\(\begin{array}{l}Q_{c}=\frac{[CH_{3}COOC_{2}H_{5}][H_{2}O]}{[CH_{3}COOH][C_{2}H_{5}OH]}\end{array} \)

(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, the equilibrium constant for the given reaction is:

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

(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,

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

Since Q_cc, 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

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

mol L^–1. If the value of K_c is

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

, what are the concentrations of PCl₃ and Cl₂ at equilibrium?

PCl_5(g) ⇌ PCl_3(g) + Cl_2(g)

Ans.

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

PCl_5(g) ⇌ PCl_3(g) + Cl_2(g)

At equilibrium

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

x mol L^-1 x mol L^-1

It is given that the value of the equilibrium constant, K_c, is

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

Now, we can write the expression for equilibrium as:

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

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) + CO₂ _(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_CO₂= 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=

\(\begin{array}{l}\frac{p_{CO_{2}}}{p_{CO}}\\ \\ =\frac{0.80}{1.4}\\ \\ =0.571\end{array} \)

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 CO₂will decrease.

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

\(\begin{array}{l}K_{p}=\frac{p_{CO_{2}}}{p_{CO}}\\ \\ \Rightarrow 0.265=\frac{0.80-p}{1.4+p}\\ \\ \Rightarrow 0.371+0.265p=0.80-p\\ \\ \Rightarrow 1.265p=0.429\\ \\ \Rightarrow p=0.339 atm\end{array} \)

Therefore, the 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=

\(\begin{array}{l}\frac{[NH_{3}]^{2}}{[N_{2}][H_{2}]^{3}}\\ \\ =\frac{(0.5)^{2}}{(3.0)(2.0)^{3}}\\ \\ =0.0104\end{array} \)

It is given that K_c=0.061

\(\begin{array}{l}∵\end{array} \)

\(\begin{array}{l}Qc \neq K_{c}\end{array} \)

, the reaction is not at equilibrium.

\(\begin{array}{l}∵\end{array} \)

\(\begin{array}{l}Qc < K_{c}\end{array} \)

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) ⇌ Br₂(g) + Cl₂(g)

Initial conc.

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

0 0

At equilibrium

\(\begin{array}{l}3.3\times 10^{-3}-2x\end{array} \)

x x

Now, we can write,

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

\(\begin{array}{l},\end{array} \)

at equilibrium

\(\begin{array}{l}[BrCl]=3.3\times 10^{-3}-(2\times 1.5\times 10^{-3})\\ \\ =3.3\times 10^{-3}-3.0\times 10^{-3}\\ \\ =0.3\times 10^{-3}\\ \\ =3.0\times 10^{-4}molL^{-1}\end{array} \)

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) + CO₂ _(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 CO₂ = (100 – 90.55) = 9.45 g

Now, the number of moles of CO,

\(\begin{array}{l}n_{CO}=\frac{90.5}{28}=3.234\: mol\end{array} \)

The number of moles of CO₂,

\(\begin{array}{l}n_{CO_{2}}=\frac{9.45}{44}=0.215\: mol\end{array} \)

The partial pressure of CO,

P_CO=

\(\begin{array}{l}\frac{n_{CO}}{n_{CO}+n_{CO_{2}}}\times p_{total}=\frac{3.234}{3.234+0.215}\times 1=0.938\, atm\end{array} \)

Partial pressure of CO₂,

\(\begin{array}{l}P_{CO_{2}}=\frac{n_{CO_{2}}}{n_{CO}+n_{CO_{2}}}\times p_{total}=\frac{0.215}{3.234+0.215}\times 1=0.062, atm\end{array} \)

Therefore, K_p=

\(\begin{array}{l}\frac{[CO]^{2}}{[CO_{2}]}\\ \\ =\frac{(0.938)^{2}}{0.062}\\ \\ =14.19\end{array} \)

For the given reaction,

∆n = 2 – 1 = 1

We know that,

K_p= K_c(RT)

\(\begin{array}{l}\Delta n\end{array} \)

\(\begin{array}{l}\Rightarrow 14.19 = K_{c}(0.082\times 1127)^{1}\\ \\ \Rightarrow K_{c}=\frac{14.19}{0.082\times 1127}\\ \\ =0.154(approximately)\end{array} \)

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

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

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

\(\begin{array}{l}K_{c}=\frac{-35.0\times 10^{-3}}{-2.303\times 8.314\times 298}\\ \\ =6.134\\ \\ ∴ K_{c}=antilog(6.134)\\ \\ =1.36\times 10^{6}\end{array} \)

Therefore, the equilibrium constant for the given reaction K_c is

\(\begin{array}{l}1.36\times 10^{6}\end{array} \)

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) 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.

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) 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 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.

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

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

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

K^’_p=

\(\begin{array}{l}\frac{1}{K_{p}}\\ \\ =\frac{1}{1.6\times 10^{5}}\\ \\ =6.25\times 10^{-6}\end{array} \)

Now, let p be the pressure of both H₂ and Br₂at equilibrium.

2HBr (g) ⇌ H₂ (g) + Br₂ (g)

Initial conc. 10 0 0

At equilibrium 10-2p p p

Now, we can write,

\(\begin{array}{l}\frac{p_{HBr}\times p_{2}}{p^{2}_{HBr}}=K^{‘}_{p}\\ \\ \frac{p\times p}{(10-2p)^{2}}=6.25\times 10^{-6}\\ \\ \frac{p}{10-2p}=2.5\times 10^{-3}\\ \\ p=2.5\times 10^{-2}-(5.0\times 10^{-3})p\\ \\ p+(5.0\times 10^{-3})p=2.5\times 10^{-2}\\ \\ (1005\times 10^{-3})=2.5\times 10^{-2}\\ \\ p=2.49\times 10^{-2}bar=2.5\times 10^{-2}bar (approximately)\end{array} \)

Therefore, at equilibrium,

[H₂]=[Br₂]=

\(\begin{array}{l}2.49\times 10^{-2}bar\end{array} \)

[HBr]=

\(\begin{array}{l}10-2\times (2.49\times 10^{-2})bar\\ \\ =9.95 bar=10 bar(approximately)\end{array} \)

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=

\(\begin{array}{l}\frac{p_{CO}\times p_{H_{2}}^{3}}{p_{CH_{4}}\times p_{H_{2}O}}\end{array} \)

(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.

(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

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

. If decomposition is depicted as,

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

∆_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)

\(\begin{array}{l}K_{c}=\frac{[PCl_{3}(g)][Cl_{2}(g)]}{[PCl_{3}(g)]}\end{array} \)

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

\(\begin{array}{l}K_{c}^{‘} = \frac{1}{K_{c}}\\ \\ =\frac{1}{8.3 \times 10^{-3}} = 1.2048 \times 10^{2}\\ \\ =120.48\end{array} \)

(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 H₂. 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) + 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

\(\begin{array}{l}\frac{P_{CO_{2}}\times P_{H_{2}}}{P_{CO}\times P_{H_{2}O}} = K_{P}\\ \\ \Rightarrow \frac{p\times p}{(4.0-p)(4.0-p)}=10.1\\ \\ \Rightarrow \frac{p}{4.0-p} =3.178\\ \\ \Rightarrow p =12.712-3.178p\\ \\ 4.178p = 12.712\\ \\ p =\frac{12.712}{4.178}\\ \\ p=3.04\end{array} \)

So, the partial pressure of H₂is 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 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

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

at 25°C. If the equilibrium concentration of O₂ in the air at 25°C is

\(\begin{array}{l} 1.6 \times 10^{-2} \end{array} \)

, what is the concentration of O₃?

Ans.

Given,

3O₂ (g) ⇌ 2O₃ (g)

Then, K_c =

\(\begin{array}{l} \frac{ [O_{3}(g)]^{2}}{[O_{2}(g)]^{3}}\end{array} \)

Given that K_c =

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

and [O₂(g)] =

\(\begin{array}{l}1.6 \times 10^{-2}\end{array} \)

Then,

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

So, the concentration of O₃ is

\(\begin{array}{l}2.86 \times 10^{-28}M\end{array} \)

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

\(\begin{array}{l}CH_{4}\end{array} \)

at equilibrium be y.

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

At equilibrium,

For CO –

\(\begin{array}{l}\frac{0.3}{1} \; = \; 0.3M\end{array} \)

For

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

–

\(\begin{array}{l}\frac{0.1}{1} \; = \; 0.1M\end{array} \)

For

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

–

\(\begin{array}{l}\frac{0.02}{1} \; = \; 0.02M\end{array} \)

\(\begin{array}{l}K_{c}\end{array} \)

= 3.90

Therefore,

\(\begin{array}{l}\frac{[CH_{4(g)}][H_{2}O_{(g)}]}{[CO_{(g)}][H_{2(g)}]^{3}} \; = \; K_{c}\end{array} \)

\(\begin{array}{l}\frac{y \times 0.02}{0.3 \times (0.1)^{3}} \; = \; 3.9\end{array} \)

\(\begin{array}{l}y \; = \; \frac{3.9 \times 0.3 \times (0.1)^{3}}{0.02}\end{array} \)

\(\begin{array}{l}y \; = \; \frac{0.00117}{0.02}\end{array} \)

\(\begin{array}{l}y \; = \; 0.0585M\end{array} \)

\(\begin{array}{l}y \; = \; 5.85 \times 10^{-2}M\end{array} \)

Therefore, the concentration of

\(\begin{array}{l}CH_{4}\end{array} \)

at equilibrium is

\(\begin{array}{l}5.85 \times 10^{-2}M\end{array} \)

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:

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

–

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

(Base)

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

– HCN (Acid)

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

–

\(\begin{array}{l}ClO_{4}^{-}\end{array} \)

(Base)

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

– HF (Acid)

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

–

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

(Acid)/

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

(Base)

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

–

\(\begin{array}{l}HCO_{3}^{-}\end{array} \)

(Acid)

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

–

\(\begin{array}{l}HS^{-}\end{array} \)

(Acid).

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.

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

– Lewis base

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

– Lewis acid

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

– Lewis acid

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

– Lewis acid

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 –

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

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

–

\(\begin{array}{l}HSO_{4}^{-}\end{array} \)

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

–

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

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

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

, NH₃ and

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

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 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)

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

(b)

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

(c)

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

(d) BCl₃

Ans.

(a)

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

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

(b)

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

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

(c)

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

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.

\(\begin{array}{l}pH=-log[H^{+}]\\ \\ =-log(3.8\times 10^{-3})\\ \\ =-log\: 3.8\, -\, log\: 10^{-3}\\ \\ =-log\, 3.8+3\\ \\ =−0.5798+3\\ \\ =2.423\end{array} \)

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

Ans.

\(\begin{array}{l}pH = -log [H^{+}]\\ \\ \Rightarrow log[H^{+}]=-pH\\ \\ \Rightarrow [H^{+}]=antilog(-pH)\\ \\ =antilog(-3.76)\\ \\ =0.000178\\ \\ =1.78\times 10^{-4}\end{array} \)

\(\begin{array}{l} ∴\end{array} \)

\(\begin{array}{l}1.78\times 10^{-4}\end{array} \)

is the concentration of white vinegar sample.

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^–,

\(\begin{array}{l}K_{b} = \frac{K_{w}}{K_{a}}=\frac{10^{-14}}{(6.8\times 10^{-4})}=1.47\times 10^{-11}\end{array} \)

For HCOO^–,

\(\begin{array}{l}K_{b}=\frac{10^{-14}}{(1.8\times 10^{-4})}=5.6\times 10^{-11}\end{array} \)

For CN^– =

\(\begin{array}{l}K_{b}=\frac{10^{-14}}{(4.8\times 10^{-9})}=2.08\times 10^{-6}\end{array} \)

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.

C₆H₅OH ⇌ C₆H₅O^– + H⁺

Initial 0.05M 0 0

After dissociation 0.05-x x x

∴

\(\begin{array}{l}K_{}\alpha = \frac{x^{2}}{0.05-x}\end{array} \)

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.

\(\begin{array}{l}x = \sqrt{1\times 10^{-10}\times 0.05}\end{array} \)

\(\begin{array}{l}x = \sqrt{5\times 10^{-12}}\end{array} \)

∴ x = 2.2×10^-6M

In the 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

∴

\(\begin{array}{l}K_{\alpha }=\frac{(0.01)(y)}{0.05}=1.0\times 10^{-10}\end{array} \)

y = 5 × 10^-10

\(\begin{array}{l}\alpha = \frac{y}{c}\end{array} \)

\(\begin{array}{l}\alpha = \frac{5\times 10^{-10}}{5\times 10^{-2}}\end{array} \)

α = 10^-8.

Q.45. The first ionisation constant of H₂S 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 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

∴ The 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 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.

CH₃COOH ⇒ CH₃COO^– + H⁺

\(\begin{array}{l}K_{a}=\frac{[CH_{3}COO^{-}][H^{+}]}{[CH_{3}COOH]}=\frac{[H^{+}]^{2}}{[CH_{3}COOH]}\\ \\ \Rightarrow [H^{+}]=\sqrt{K_{a}[CH_{3}COOH]}=\sqrt{(1.74\times 10^{-5})(5\times 10^{-2})}=9.33\times 10^{-4}M\\ \\ [CH_{3}COO^{-}]=[H^{+}]=9.33\times 10^{-4}M\\ \\ pH=-log(9.33\times 1.0^{-4})=4-0.9699=4-0.97=3.03\end{array} \)

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⁺] =

\(\begin{array}{l}7.08\times 10^{-5}\, M
\end{array} \)

[A^–] = [H⁺] =

\(\begin{array}{l}7.08\times 10^{-5}\, M\end{array} \)

\(\begin{array}{l}K_{a}=\frac{[H^{+}][A^{-}]}{[HA]}=\frac{(7.08\times 10^{-5})(7.08\times 10^{-5})}{10^{-2}}=5.0\times 10^{-7}\\ \\ p_{K_{a}}=-log\, K_{a} = -log\, (5.0\times 10^{-7})=7-0.699=6.301\end{array} \)

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^–

\(\begin{array}{l}∴ [H^{+}]=[HCl]=3\times 10^{-3}M\\ \\ pH=-log(3\times 10^{-3})=2.52\end{array} \)

(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

[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) 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 =

\(\begin{array}{l}\frac{2g}{(204+16+1)g\: mol^{-1}}\times \frac{1}{2L}=4.52\times 10^{-3}M\\ \\ [OH_{-}]=[TlOH]=4.52\times 10^{-3}M\\ \\ [H^{+}]=\frac{10^{-14}}{(4.52\times 10^{-3})}=2.21\times 10^{-12}M\\ \\ ∴ pH=-log(2.21\times 10^{-12})=12-(0.3424)=11.66\end{array} \)

(II) Molar conc. Of Ca(OH)₂=

\(\begin{array}{l}\frac{0.3g}{(40+34)g\: mol^{-1}}\times \frac{1}{0.5L}=8.11\times 10^{-3}M\\ \\ [OH_{-}]=2[Ca(OH)_{2}]=2\times (8.11\times 10^{-3})M=16.22\times 10^{-3}M\\ \\ pOH=-log(16.22\times 10^{-3})=3-1.2101=1.79\\ \\ pH=14-1.79=12.21\end{array} \)

(III) Molar conc. of NaOH =

\(\begin{array}{l}\frac{0.3g}{(40+34)g\: mol^{-1}}\times \frac{1}{0.2L}=3.75\times 10^{-2}M\\ \\ [OH_{-}]=3.75\times 10^{-2}M\\ \\ pOH=-log(3.75\times 10^{-2})=2-0.0574=1.43\\ \\ pH=14-1.43=12.57\end{array} \)

(IV) M₁V₁ = M₂V₂

\(\begin{array}{l}∴ 13.6M\times \times 1mL=M_{2}\times 1000mL\\ \\ ∴ M_{2}=1.36\times 10^{-2}M\\ \\ [H^{+}]=[HCl]=1.36\times 10^{-2}M\\ \\ pH=-log(1.36\times 10^{-2})=2-0.1335\simeq 1.87\end{array} \)

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.

CH₂(Br)COOH CH₂(Br)COO^– + H⁺

Initial conc. C 0 0

Conc. at eqm. C – Cα Cα Cα

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

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 =

\(\begin{array}{l}4.27 \times 10^{-10}\end{array} \)

c = 0.001MpH =?α =?

\(\begin{array}{l}K_{b}=c\alpha ^{2}\\ \\ 4.27 \times 10^{-10}=0.001\times \alpha ^{2}\\ \\ 4270\times 10^{-10}=\alpha ^{2}\\ \\ 65.34\times 10^{-5}=\alpha =6.53\times 10^{-4}\\ \\ Then,[anion]=c\alpha =0.001\times 65.34\times10^{-5}=0.065\times10^{-5}\end{array} \)

\(\begin{array}{l}pOH=-log(0.065\times 10^{-5})\\ \\ =6.187\\ \\ pH=7.813\\ \\ Now,\\ \\ K_{a}\times K_{b}=K_{w}\\ \\ K_{a}=\frac{10^{-14}}{4.27\times 10^{-10}}\\ \\ =2.34\times 10^{-5}\end{array} \)

\(\begin{array}{l}∴ 2.34\times 10^{-5}\end{array} \)

is the ionisation constant.

Q.52. Calculate the degree of ionisation of 0.05 M acetic acid if its p_Kavalue 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)

\(\begin{array}{l}K_{a}=1.82\times 10^{-5}\\ \\ K_{a}=c\alpha ^{2}\\ \\ \alpha =\sqrt{\frac{K_{a}}{c}}\\ \\ \alpha =\sqrt{\frac{1.82\times 10^{-5}}{5\times 10^{-2}}}=1.908\times 10^{-2}\end{array} \)

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.

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.

\(\begin{array}{l}K_{a}=\frac{[CH_{3}COO^{-}][H^{+}]}{[CH_{3}COOH]}\\ \\ ∴ =\frac{(0.01)x}{0.05}\\ \\ x=\frac{1.82\times 10^{-5}\times 0.05}{0.01}\\ \\ x=1.82\times 10^{-3}\times 0.05M\end{array} \)

Now,

\(\begin{array}{l}\alpha =\frac{Amount \: of \: acid\:dissociation}{Amount\:of\:acid\:taken}\\ \\ =\frac{1.82\times 10^{-3}\times 0.05}{0.05}\\ \\ =1.82\times 10^{-3}\end{array} \)

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

\(\begin{array}{l}K_{a}=\frac{[CH_{3}COO^{-}][H^{+}]}{[CH_{3}COOH]}\\ \\∴K_{a}=\frac{(0.1)X}{0.05}\\ \\ x=\frac{1.82\times 10^{-5}\times 0.05}{0.1}\\ \\ x=1.82\times 10^{-4}\times 0.05M\end{array} \)

Now,

\(\begin{array}{l}\alpha =\frac{Amount \: of \: acid\:dissociation}{Amount\:of\:acid\:taken}\\ \\ =\frac{1.82\times 10^{-4}\times 0.05}{0.05}\\ \\ =1.82\times 10^{-4}\end{array} \)

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.

\(\begin{array}{l}K_{b}=5.4\times 10^{-4}\\ \\ c=0.02M\\ \\ Then,\alpha =\sqrt{\frac{K_{b}}{c}}\\ \\ =\sqrt{\frac{5.4\times 10^{-4}}{0.02}}=0.1643\end{array} \)

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,

(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

\(\begin{array}{l}\Rightarrow K_{b}=\frac{[(CH_{3})_{2}NH_{2}^{+}][OH^{-}]}{[(CH_{3})_{2}NH]}\\ \\ 5.4\times 10^{-4}=\frac{x\times 0.1}{0.02}\\ \\ x=0.0054\end{array} \)

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

[H⁺] = anitlog(–6.8)=

\(\begin{array}{l}1.5\times 10^{-7} M\end{array} \)

(II) pH of black coffee = 5.0

Since pH = –log [H⁺]

5.0 = –log [H⁺]

log[H⁺] = –5.0

[H⁺] = anitlog(–5.0)=

\(\begin{array}{l} 10^{-5} M\end{array} \)

(III) pH of tomato= 4.2

Since, pH = –log [H⁺]

4.2 = –log [H⁺]

log[H⁺] = –4.2

[H⁺] = anitlog(–4.2)=

\(\begin{array}{l}6.31\times 10^{-5} M\end{array} \)

(IV) pH of lemon juice = 2.2

Since, pH = –log [H⁺]

2.2 = –log [H⁺]

log[H⁺] = –2.2

[H⁺] = anitlog(–2.2)=

\(\begin{array}{l}6.31\times 10^{-3} M\end{array} \)

(V) pH of egg white = 7.8

Since, pH = –log [H⁺]

7.8 = –log [H⁺]

log[H⁺] = –7.8

[H⁺] = anitlog(–7.8)=

\(\begin{array}{l}1.58\times 10^{-8} M\end{array} \)

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.

[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=

⇒

\(\begin{array}{l}[H^{+}]=3.2\times 10^{-14}\end{array} \)

∴ 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 + H₂O ↔ H₃O⁺ + A^–

(0.05-0.0α) ≈ 0.05 0.05α 0.05α

\(\begin{array}{l}K_{\alpha }=\frac{[H_{3}O^{+}][A^{-}]}{[HA]}\end{array} \)

\(\begin{array}{l}\frac{(0.05\alpha )(0.05\alpha )}{0.05}\end{array} \)

= 0.05α²

\(\begin{array}{l}\alpha =\sqrt{\frac{K_{\alpha }}{0.05}}\end{array} \)

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 ionisation.

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 0.1 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

[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 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 HNO₂ (weak acid).

\(\begin{array}{l}NO_{2}^{-} + H_{2}O \leftrightarrow HNO_{2} + OH^{-}\end{array} \)

\(\begin{array}{l}K_{h} = \frac{[HNO_{2}][OH^{-}]}{[NO_{2}^{-}]}\end{array} \)

\(\begin{array}{l}\Rightarrow \;\frac{K_{w}}{K_{a}} = \frac{10^{-14}}{4.5\times 10^{-4}} = 0.22\times 10^{-10}\end{array} \)

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

\(\begin{array}{l}[NO_{2}^{-}] = 0.04 – y; 0.04\end{array} \)

\(\begin{array}{l}[HNO_{2}] = y\end{array} \)

\(\begin{array}{l}[OH^{-}] = y\end{array} \)

\(\begin{array}{l}K_{h} = \frac{y^{2}}{0.04} = 0.22\times 10^{-10}\end{array} \)

\(\begin{array}{l}y^{2} = 0.0088\times 10^{-10}\end{array} \)

\(\begin{array}{l}y = 0.093\times 10^{-5}\end{array} \)

\(\begin{array}{l} ∴ [OH^{-}]= 0.093\times 10^{-5}M\end{array} \)

\(\begin{array}{l}[H_{3}O^{+}] = \frac{10^{-14}}{0.093\times 10^{-5}} = 10.75\times 10^{-9}M\end{array} \)

Thus,

\(\begin{array}{l}pH = -\log (10.75\times 10^{-9})\end{array} \)

= 7.96.

Thus, the degree of hydrolysis is

\(\begin{array}{l} = \frac{y}{0.04} = \frac{0.093\times 10^{-5}}{0.04}\end{array} \)

\(\begin{array}{l} = 2.325\times 10^{-5}\end{array} \)

Q.61. A 0.02M solution of pyridinium hydrochloride (C₅H₆ClN) has a pH of 3.44. Calculate the ionisation constant of C₅H₅N (pyridine).

Ans.

pH = 3.44As

we know,

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

\(\begin{array}{l} ∴ [H^{+}] = 3.63\times 10^{-4}\end{array} \)

Now,

\(\begin{array}{l}K_{h} = \frac{3.63\times 10^{-4}}{0.02}\end{array} \)

;

(Given that concentration = 0.02M)

\(\begin{array}{l}\Rightarrow K_{h} = 6.6\times 10^{-6}\end{array} \)

As we know,

\(\begin{array}{l}K_{h} = \frac{K_{w}}{K_{a}}\end{array} \)

\(\begin{array}{l}K_{a} = \frac{K_{w}}{K_{h}} = \frac{10^{-14}}{6.6\times 10^{-6}}\end{array} \)

\(\begin{array}{l}1.51\times 10^{-9}\end{array} \)

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

Ans.

KBr + H₂O

\(\begin{array}{l}\leftrightarrow\end{array} \)

KOH (Strong base)+ HBr (Strong acid). Thus, it is a neutral solution.

NH₄NO₃ + H₂O

\(\begin{array}{l}\leftrightarrow\end{array} \)

NH₄OH(Weak base) + HNO₂ (Strong acid). Thus, it is an acidic solution.

KF + H₂O

\(\begin{array}{l}\leftrightarrow\end{array} \)

KOH (Strong base) + HF (weak acid). Thus, it is a basic solution.

NaNO₂ + H₂O

\(\begin{array}{l}\leftrightarrow\end{array} \)

NH₄OH(Strong base) + HNO₂(Weak acid). Thus, it is a basic solution.

NaCN + H₂O

\(\begin{array}{l}\leftrightarrow\end{array} \)

HCN (Weak acid) + NaOH (Strong base). Thus, it is a basic solution.

NaCl + H₂O

\(\begin{array}{l}\leftrightarrow\end{array} \)

NaOH (Strong base) + HCL (Strong acid). Thus, it is a neutral solution.

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 (ClCH₂COOH) is

\(\begin{array}{l}1.35\times 10^{-3}\end{array} \)

\(\begin{array}{l}\Rightarrow K_{a} = c\alpha ^{2}\end{array} \)

\(\begin{array}{l}∴ \alpha = \sqrt{\frac{K_{a}}{c}}\end{array} \)

\(\begin{array}{l}\sqrt{\frac{1.35\times 10^{-3}}{0.1}}\end{array} \)

; (given concentration = 0.1M)

\(\begin{array}{l}\alpha = \sqrt{1.35\times 10^{-2}}\end{array} \)

= 0.116

\(\begin{array}{l} ∴ [H^{+}] = c\alpha = 0.1×0.116 = 0.0116\end{array} \)

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

\(\begin{array}{l}ClCH_{2}COO^{-} + H_{2}O \leftrightarrow ClCH_{2}COOH + OH^{-}\end{array} \)

\(\begin{array}{l}K_{h} = \frac{[ClCH_{2}COO][OH^{-}]}{[ClCH_{2}COO^{-}]}\end{array} \)

Now,

\(\begin{array}{l}K_{h} = \frac{K_{w}}{K_{a}}\end{array} \)

\(\begin{array}{l}K_{h} = \frac{10^{-14}}{1.35\times 10^{-3}}\end{array} \)

\(\begin{array}{l}= 0.740\times 10^{-11}\end{array} \)

Also,

\(\begin{array}{l}K_{h} = \frac{y^2}{0.1}\end{array} \)

\(\begin{array}{l}\Rightarrow \; 0.740\times 10^{-11} = \frac{y^2}{0.1}\end{array} \)

\(\begin{array}{l}\Rightarrow \; 0.0740\times 10^{-11} = y^2\end{array} \)

\(\begin{array}{l}y = 0.86\times 10^{-6}\end{array} \)

\(\begin{array}{l}[OH^{-}] = 0.86\times 10^{-6}\end{array} \)

\(\begin{array}{l}∴ [H^{+}] =\frac{K_{w}}{0.86\times 10^{-6}}\end{array} \)

\(\begin{array}{l}=\frac{10^{-14}}{0.86\times 10^{-6}}\end{array} \)

\(\begin{array}{l}[H^{+}] = 1.162\times 10^{-3}\end{array} \)

pH =

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

= 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,

\(\begin{array}{l}K_{w} = [H^{+}][OH^{-}]\end{array} \)

Assuming,

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

= y

As,

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

K_w = y².K_w at 310 K is

\(\begin{array}{l}2.7\times 10^{-14}\end{array} \)

\(\begin{array}{l}∴ 2.7\times 10^{-14} = y^{2}\end{array} \)

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

\(\begin{array}{l}-\log [1.64\times 10^{-7}]\end{array} \)

= 6.78

Thus, the pH of neutral water at 310 K 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⁺ +

\(\begin{array}{l}CrO_{4}^{-}\end{array} \)

Now, K_sp = [Ag²⁺]²

\(\begin{array}{l}[CrO_{4}^{-}]\end{array} \)

Assuming the solubility of Ag₂CrO₄ is ‘x’.

Thus, [Ag²⁺] = 2x and

\(\begin{array}{l}CrO_{4}^{-}\end{array} \)

= x

Assuming the solubility of AgBr is y.

AgBr_(s) → Ag²⁺ + Br^–

y =

\(\begin{array}{l}7.07\times 10^{-7}M\end{array} \)

The ratio of molarities to their saturated solution is:

\(\begin{array}{l}\frac{x}{y} = \frac{0.65\times10^{-4}M}{7.07\times10^{-7}M} = 91.9\end{array} \)

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(IO₃)₂ → Na⁺ +

\(\begin{array}{l}IO_{3}^{-}\end{array} \)

0.0001M 0.001 M

Cu(ClO₃)₂→ Cu²⁺ +

\(\begin{array}{l}2CIO_{3}^{-}\end{array} \)

0.001M 0.001 M

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

Now, the ionic product of the copper iodate = [Cu²⁺]

\(\begin{array}{l}[IO_{3}^{-}]^{2}\end{array} \)

= (0.001)(0.001)²=

\(\begin{array}{l}1.0\times 10^{-9}M\end{array} \)

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

[H₃O⁺] =

\(\begin{array}{l}6.46\times 10^{-4}M\end{array} \)

C₆H₅COOH + H₂O →

\(\begin{array}{l}C_{6}H_{5}COO^{-}\end{array} \)

+ H₃O

\(\begin{array}{l}K_{a}=\frac{[C_{6}H_{5}COO^{-}][H_{3}O^{+}]}{C_{6}H_{5}COOH}\end{array} \)

\(\begin{array}{l}K_{a}=\frac{[C_{6}H_{5}COOH]}{C_{6}H_{5}COO^{-}} = \frac{[H_{3}O^{+}]}{K_{a}} = \frac{6.46\times 10^{-4}}{6.46\times 10^{-5}} = 10\end{array} \)

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

Now, [Ag⁺] = y

[C₆H₅COOH] =

\(\begin{array}{l}[C_{6}H_{5}COO^{-}]\end{array} \)

= y

\(\begin{array}{l}[C_{6}H_{5}COO^{-}]\end{array} \)

= y

\(\begin{array}{l}[C_{6}H_{5}COO^{-}]\end{array} \)

= y/11

K_sp= [Ag⁺]

\(\begin{array}{l}[C_{6}H_{5}COO^{-}]\end{array} \)

\(\begin{array}{l}2.5\times 10^{-13} = y\frac{y}{11}\end{array} \)

y =

\(\begin{array}{l}1.66\times 10^{-6}\end{array} \)

mol/L

Hence, solubility of C₆H₅COOAg in buffer of pH = 3.19 is

\(\begin{array}{l}1.66\times 10^{-6}\end{array} \)

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⁺]

\(\begin{array}{l}[C_{6}H_{5}COO^{-}]\end{array} \)

K_sp = (y’)² =

\(\begin{array}{l}\sqrt{K_{sp}} = \sqrt{2.5\times 10^{-13}} = 5\times 10^{-7} mol/L\end{array} \)

\(\begin{array}{l}∴ \frac{y}{x} = \frac{1.66\times 10^{-6}}{5\times 10^{-7}} = 3.32\end{array} \)

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)

\(\begin{array}{l}\leftrightarrow\end{array} \)

\(\begin{array}{l}Fe^{2+}_{(aq)} + S^{2-}_{(aq)}\end{array} \)

Ksp = [Fe²⁺][S^2-]

\(\begin{array}{l}6.3\times 10^{-18} = (\frac{y}{2})(\frac{y}{2})\end{array} \)

\(\begin{array}{l}\frac{y^{2}}{4} = 6.3\times 10^{-18}\end{array} \)

Thus, y =

\(\begin{array}{l}5.02\times 10^{-9}\end{array} \)

Thus, if the concentration of FeSO₄ and Na₂SO₄is equal to or less than that of

\(\begin{array}{l}5.02\times 10^{-9}M\end{array} \)

, then there won’t be precipitation of FeS.

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.

\(\begin{array}{l}CaSO_{4(s)}\leftrightarrow Ca^{2+}_{(aq)} + SO_{4}^{2-}(aq)\end{array} \)

K_sp =

\(\begin{array}{l}[Ca^{2+}][SO_{4}^{2-}]\end{array} \)

Assuming the solubility of calcium sulphate is x,

K_sp = x²

\(\begin{array}{l}∴ \;9.1\times 10^{-6} = x^{2}\end{array} \)

\(\begin{array}{l}∴ \;x = 3.02\times 10^{-3} mol/L\end{array} \)

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

\(\begin{array}{l}= 3.02\times 10^{-3} \times 136\end{array} \)

= 0.41 g/L.

i.e., 1 litre of H₂O will be required to dissolve 0.41 g of calcium sulphate.

Thus, the minimum volume of H₂O required to dissolve 1 gram of CaSO₄ at 298 K is =

\(\begin{array}{l}\frac{1}{0.41} L = 2.44 L\end{array} \)

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 FeSO₄, MnCl₂, ZnCl₂ 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 =

\(\begin{array}{l}1.0\times 10^{-19}M\end{array} \)

[M²⁺] = 0.04 M

Volume = 10 mL

Volume = 5 mL

After mixing:[S^2-] = ? and [M²⁺] = ?

Total volume = (10 + 5) = 15 mL

Volume = 15 mL

[S^2-] =

\(\begin{array}{l}\frac{1.0\times 10^{-19}\times 10}{15} = 6.67\times 10^{-20}M\end{array} \)

[M²⁺] =

\(\begin{array}{l}\frac{0.04\times 5}{15} = 1.33\times 10^{-2}M\end{array} \)

Now, the ionic product = [M²⁺][S^2-]=

\(\begin{array}{l}(1.33\times 10^{-2})(6.67\times 10^{-20})\end{array} \)

\(\begin{array}{l}8.87\times 10^{-22}\end{array} \)

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 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

Why are ANAND CLASSES (A School Of Competitions) NCERT Solutions a cut above the rest?

ANAND CLASSES (A School Of Competitions) 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 structured carefully to offer maximum benefits to the students to study well and obtain good marks in the annual exam. These NCERT Solutions for Class 11 Chemistry Chapter 7 are concept-focused and can, therefore, be used for revisions.

Designed 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 ANAND CLASSES (A School Of Competitions) support team is always available to clear doubts and solve any 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 as follows:
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

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, a 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 annual exam preparation?

Practising NCERT Solutions for Class 11 Chemistry Chapter 7 helps the students to score well in the annual exams. These solutions are devised based on the updated CBSE Syllabus, covering all the crucial concepts of the respective topic. Solving these questions will boost their confidence to face the annual exams fearlessly. Topics covered in these solutions form the basis for high scores. It also helps the students to get familiar with answering questions of all difficulty levels. Hence, it is highly recommended that the students of Class 11 should practise these solutions to prepare well for their annual exams.

Q4.

List 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

Neeraj Anand, Param Anand

Er. Neeraj K.Anand is a freelance mentor and writer who specializes in Engineering & Science subjects. Neeraj Anand received a B.Tech degree in Electronics and Communication Engineering from N.I.T Warangal & M.Tech Post Graduation from IETE, New Delhi. He has over 30 years of teaching experience and serves as the Head of Department of ANAND CLASSES. He concentrated all his energy and experiences in academics and subsequently grew up as one of the best mentors in the country for students aspiring for success in competitive examinations. In parallel, he started a Technical Publication "ANAND TECHNICAL PUBLISHERS" in 2002 and Educational Newspaper "NATIONAL EDUCATION NEWS" in 2014 at Jalandhar. Now he is a Director of leading publication "ANAND TECHNICAL PUBLISHERS", "ANAND CLASSES" and "NATIONAL EDUCATION NEWS". He has published more than hundred books in the field of Physics, Mathematics, Computers and Information Technology. Besides this he has written many books to help students prepare for IIT-JEE and AIPMT entrance exams. He is an executive member of the IEEE (Institute of Electrical & Electronics Engineers. USA) and honorary member of many Indian scientific societies such as Institution of Electronics & Telecommunication Engineers, Aeronautical Society of India, Bioinformatics Institute of India, Institution of Engineers. He has got award from American Biographical Institute Board of International Research in the year 2005.

...

CBSE Class 11 Chemistry Syllabus

CBSE Class 11 Chemistry Syllabus is a vast which needs a clear understanding of the concepts and topics. Knowing CBSE Class 11 Chemistry syllabus helps students to understand the course structure of Chemistry.

Unit-wise CBSE Class 11 Syllabus for Chemistry

Below is a list of detailed information on each unit for Class 11 Students.

UNIT I – Some Basic Concepts of Chemistry

General Introduction: Importance and scope of Chemistry.

Nature of matter, laws of chemical combination, Dalton’s atomic theory: concept of elements,
atoms and molecules.

Atomic and molecular masses, mole concept and molar mass, percentage composition, empirical and molecular formula, chemical reactions, stoichiometry and calculations based on stoichiometry.

UNIT II – Structure of Atom

Discovery of Electron, Proton and Neutron, atomic number, isotopes and isobars. Thomson’s model and its limitations. Rutherford’s model and its limitations, Bohr’s model and its limitations, concept of shells and subshells, dual nature of matter and light, de Broglie’s relationship, Heisenberg uncertainty principle, concept of orbitals, quantum numbers, shapes of s, p and d orbitals, rules for filling electrons in orbitals – Aufbau principle, Pauli’s exclusion principle and Hund’s rule, electronic configuration of atoms, stability of half-filled and completely filled orbitals.

UNIT III – Classification of Elements and Periodicity in Properties

Significance of classification, brief history of the development of periodic table, modern periodic law and the present form of periodic table, periodic trends in properties of elements -atomic radii, ionic radii, inert gas radii, Ionization enthalpy, electron gain enthalpy, electronegativity, valency. Nomenclature of elements with atomic number greater than 100.

UNIT IV – Chemical Bonding and Molecular Structure

Valence electrons, ionic bond, covalent bond, bond parameters, Lewis structure, polar character of covalent bond, covalent character of ionic bond, valence bond theory, resonance, geometry of covalent molecules, VSEPR theory, concept of hybridization, involving s, p and d orbitals and shapes of some simple molecules, molecular orbital theory of homonuclear diatomic molecules(qualitative idea only), Hydrogen bond.

UNIT V – Chemical Thermodynamics

Concepts of System and types of systems, surroundings, work, heat, energy, extensive and intensive properties, state functions. First law of thermodynamics – internal energy and enthalpy, measurement of U and H, Hess’s law of constant heat summation, enthalpy of bond dissociation, combustion, formation, atomization, sublimation, phase transition, ionization, solution and dilution. Second law of Thermodynamics (brief introduction)
Introduction of entropy as a state function, Gibb’s energy change for spontaneous and nonspontaneous processes.
Third law of thermodynamics (brief introduction).

UNIT VI – Equilibrium

Equilibrium in physical and chemical processes, dynamic nature of equilibrium, law of mass action, equilibrium constant, factors affecting equilibrium – Le Chatelier’s principle, ionic equilibrium- ionization of acids and bases, strong and weak electrolytes, degree of ionization,
ionization of poly basic acids, acid strength, concept of pH, hydrolysis of salts (elementary idea), buffer solution, Henderson Equation, solubility product, common ion effect (with illustrative examples).

UNIT VII – Redox Reactions

Concept of oxidation and reduction, redox reactions, oxidation number, balancing redox reactions, in terms of loss and gain of electrons and change in oxidation number, applications of redox reactions.

UNIT VIII – Organic Chemistry: Some basic Principles and Techniques

General introduction, classification and IUPAC nomenclature of organic compounds. Electronic displacements in a covalent bond: inductive effect, electromeric effect, resonance and hyper conjugation. Homolytic and heterolytic fission of a covalent bond: free radicals, carbocations, carbanions, electrophiles and nucleophiles, types of organic reactions.

UNIT IX – Hydrocarbons

Classification of Hydrocarbons
Aliphatic Hydrocarbons:
Alkanes – Nomenclature, isomerism, conformation (ethane only), physical properties, chemical reactions.
Alkenes – Nomenclature, structure of double bond (ethene), geometrical isomerism, physical properties, methods of preparation, chemical reactions: addition of hydrogen, halogen, water, hydrogen halides (Markovnikov’s addition and peroxide effect), ozonolysis, oxidation, mechanism of electrophilic addition.
Alkynes – Nomenclature, structure of triple bond (ethyne), physical properties, methods of preparation, chemical reactions: acidic character of alkynes, addition reaction of – hydrogen, halogens, hydrogen halides and water.

Aromatic Hydrocarbons:

Introduction, IUPAC nomenclature, benzene: resonance, aromaticity, chemical properties: mechanism of electrophilic substitution. Nitration, sulphonation, halogenation, Friedel Craft’s alkylation and acylation, directive influence of functional group in monosubstituted benzene. Carcinogenicity and toxicity.

To know the CBSE Syllabus for all the classes from 1 to 12, visit the Syllabus page of CBSE. Meanwhile, to get the Practical Syllabus of Class 11 Chemistry, read on to find out more about the syllabus and related information in this page.

CBSE Class 11 Chemistry Practical Syllabus with Marking Scheme

In Chemistry subject, practical also plays a vital role in improving their academic scores in the subject. The overall weightage of Chemistry practical mentioned in the CBSE Class 11 Chemistry syllabus is 30 marks. So, students must try their best to score well in practicals along with theory. It will help in increasing their overall academic score.

CBSE Class 11 Chemistry Practical Syllabus

The experiments will be conducted under the supervision of subject teacher. CBSE Chemistry Practicals is for 30 marks. This contribute to the overall practical marks for the subject.

The table below consists of evaluation scheme of practical exams.

Evaluation Scheme	Marks
Volumetric Analysis	08
Salt Analysis	08
Content Based Experiment	06
Project Work	04
Class record and viva	04
Total	30

CBSE Syllabus for Class 11 Chemistry Practical

Micro-chemical methods are available for several of the practical experiments. Wherever possible such techniques should be used.

A. Basic Laboratory Techniques
1. Cutting glass tube and glass rod
2. Bending a glass tube
3. Drawing out a glass jet
4. Boring a cork

B. Characterization and Purification of Chemical Substances
1. Determination of melting point of an organic compound.
2. Determination of boiling point of an organic compound.
3. Crystallization of impure sample of any one of the following: Alum, Copper Sulphate, Benzoic Acid.

C. Experiments based on pH

1. Any one of the following experiments:

Determination of pH of some solutions obtained from fruit juices, solution of known and varied concentrations of acids, bases and salts using pH paper or universal indicator.
Comparing the pH of solutions of strong and weak acids of same concentration.
Study the pH change in the titration of a strong base using universal indicator.

2. Study the pH change by common-ion in case of weak acids and weak bases.

D. Chemical Equilibrium
One of the following experiments:

1. Study the shift in equilibrium between ferric ions and thiocyanate ions by increasing/decreasing the concentration of either of the ions.
2. Study the shift in equilibrium between [Co(H₂O)₆] ²⁺ and chloride ions by changing the concentration of either of the ions.

E. Quantitative Estimation
i. Using a mechanical balance/electronic balance.
ii. Preparation of standard solution of Oxalic acid.
iii. Determination of strength of a given solution of Sodium hydroxide by titrating it against standard solution of Oxalic acid.
iv. Preparation of standard solution of Sodium carbonate.
v. Determination of strength of a given solution of hydrochloric acid by titrating it against standard Sodium Carbonatesolution.

F. Qualitative Analysis
1) Determination of one anion and one cation in a given salt
Cations‐ Pb²⁺, Cu²⁺, As³⁺, Al³⁺, Fe³⁺, Mn²⁺, Ni²⁺, Zn²⁺, Co²⁺, Ca²⁺, Sr²⁺, Ba²⁺, Mg²⁺, NH₄ ⁺
Anions – (CO₃)^2‐ , S^2‐, NO₂^‐ , SO₃^2‐, SO^2‐ , NO ^‐ , Cl^‐ , Br^‐, I‐, PO₄^3‐ , C₂O²‐ ,CH₃COO^‐
(Note: Insoluble salts excluded)

2) Detection of ‐ Nitrogen, Sulphur, Chlorine in organic compounds.

G) PROJECTS
Scientific investigations involving laboratory testing and collecting information from other sources.

A few suggested projects are as follows:

Checking the bacterial contamination in drinking water by testing sulphide ion
Study of the methods of purification of water.
Testing the hardness, presence of Iron, Fluoride, Chloride, etc., depending upon the regional
variation in drinking water and study of causes of presence of these ions above permissible
limit (if any).
Investigation of the foaming capacity of different washing soaps and the effect of addition of
Sodium carbonate on it.
Study the acidity of different samples of tea leaves.
Determination of the rate of evaporation of different liquids Study the effect of acids and
bases on the tensile strength of fibres.
Study of acidity of fruit and vegetable juices.

Note: Any other investigatory project, which involves about 10 periods of work, can be chosen with the approval of the teacher.

Practical Examination for Visually Impaired Students of Class 11

Below is a list of practicals for the visually impaired students.

A. List of apparatus for identification for assessment in practicals (All experiments)
Beaker, tripod stand, wire gauze, glass rod, funnel, filter paper, Bunsen burner, test tube, test tube stand,
dropper, test tube holder, ignition tube, china dish, tongs, standard flask, pipette, burette, conical flask, clamp
stand, dropper, wash bottle
• Odour detection in qualitative analysis
• Procedure/Setup of the apparatus

B. List of Experiments A. Characterization and Purification of Chemical Substances
1. Crystallization of an impure sample of any one of the following: copper sulphate, benzoic acid
B. Experiments based on pH
1. Determination of pH of some solutions obtained from fruit juices, solutions of known and varied
concentrations of acids, bases and salts using pH paper
2. Comparing the pH of solutions of strong and weak acids of same concentration.

C. Chemical Equilibrium
1. Study the shift in equilibrium between ferric ions and thiocyanate ions by increasing/decreasing
the concentration of eitherions.
2. Study the shift in equilibrium between [Co(H2O)6]2+ and chloride ions by changing the
concentration of either of the ions.

D. Quantitative estimation
1. Preparation of standard solution of oxalic acid.
2. Determination of molarity of a given solution of sodium hydroxide by titrating it against standard
solution of oxalic acid.

E. Qualitative Analysis
1. Determination of one anion and one cation in a given salt
2. Cations – NH+4
Anions – (CO3)2-, S2-, (SO3)2-, Cl-, CH3COO-
(Note: insoluble salts excluded)
3. Detection of Nitrogen in the given organic compound.
4. Detection of Halogen in the given organic compound.

Note: The above practicals may be carried out in an experiential manner rather than recording observations.

We hope students must have found this information on CBSE Syllabus useful for their studying Chemistry. Learn Maths & Science in interactive and fun loving ways with ANAND CLASSES (A School Of Competitions) App/Tablet.

Frequently Asked Questions on CBSE Class 11 Chemistry Syllabus

How many units are in the CBSE Class 11 Chemistry Syllabus?

There are 9 units in the CBSE Class 11 Chemistry Syllabus. Students can access various study materials for the chapters mentioned in this article for free at ANAND CLASSES (A School Of Competitions).

What is the total marks for practicals examination as per the CBSE Class 11 Chemistry Syllabus?

The total marks for the practicals as per the CBSE Class 11 Chemistry Syllabus is 30. It includes volumetric analysis, content-based experiment, salt analysis, class record, project work and viva.

Which chapter carries more weightage as per the CBSE Syllabus for Class 11 Chemistry?

The organic chemistry chapter carries more weightage as per the CBSE Syllabus for Class 11 Chemistry.

NCERT Solutions-Equilibrium for Class 11 Chemistry

NCERT Solutions for Class 11 Chemistry Chapter 7

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

Why are ANAND CLASSES (A School Of Competitions) NCERT Solutions a cut above the rest?

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.

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

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

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

CBSE Class 11 Chemistry Syllabus

Unit-wise CBSE Class 11 Syllabus for Chemistry

UNIT I – Some Basic Concepts of Chemistry

UNIT II – Structure of Atom

UNIT III – Classification of Elements and Periodicity in Properties

UNIT IV – Chemical Bonding and Molecular Structure

UNIT V – Chemical Thermodynamics

UNIT VI – Equilibrium

UNIT VII – Redox Reactions

UNIT VIII – Organic Chemistry: Some basic Principles and Techniques

UNIT IX – Hydrocarbons

CBSE Class 11 Chemistry Practical Syllabus with Marking Scheme

CBSE Class 11 Chemistry Practical Syllabus

CBSE Syllabus for Class 11 Chemistry Practical

Practical Examination for Visually Impaired Students of Class 11

Frequently Asked Questions on CBSE Class 11 Chemistry Syllabus

How many units are in the CBSE Class 11 Chemistry Syllabus?

What is the total marks for practicals examination as per the CBSE Class 11 Chemistry Syllabus?

Which chapter carries more weightage as per the CBSE Syllabus for Class 11 Chemistry?

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