6.15 Exercise Problems and Solutions

Q1. A liquid is in equilibrium with its vapour in a sealed container at a fixed temperature. The volume of the container is suddenly increased.
a) What is the initial effect of the change on vapour pressure?
b) How do rates of evaporation and condensation change initially?
c) What happens when equilibrium is restored finally and what will be the final vapour pressure?

Answer: (a) If the volume of the container is suddenly increased, then the vapour pressure would decrease initially. This is because the amount of vapour remains the same, but the volume increases suddenly. As a result, the same amount of vapour is distributed in a larger volume.
(b) Since the temperature is constant, the rate of evaporation also remains constant. When the volume of the container is increased, the density of the vapour phase decreases. As a result, the rate of collisions of the vapour particles also decreases. Hence, the rate of condensation decreases initially.
(c) When equilibrium is restored finally, the rate of evaporation becomes equal to the rate of condensation. In this case, only the volume changes while the temperature remains constant. The vapour pressure depends on temperature and not on volume. Hence, the final vapour pressure will be equal to the original vapour pressure of the system.

Q2. What is \(K_{\mathrm{c}}\) for the following equilibrium when the equilibrium concentration of each substance is: \(\left[\mathrm{SO}_2\right]=0.60 \mathrm{M},\left[\mathrm{O}_2\right]=0.82 \mathrm{M}\) and \(\left[\mathrm{SO}_3\right]=1.90 \mathrm{M}\)?
\(
2 \mathrm{SO}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{SO}_3(\mathrm{~g})
\)

Answer: The equilibrium constant \(\left(K_c\right)\) for the given reaction is:
\(
K_c=\frac{\left[\mathrm{SO}_3\right]^2}{\left[\mathrm{SO}_2\right]^2\left[\mathrm{O}_2\right]}=\frac{(1.9 \mathrm{M}) \times(1.9 \mathrm{M})}{(0.6 \mathrm{M}) \times(0.6 \mathrm{M}) \times(0.82 \mathrm{M})}
\)
\(
=12.229 \mathrm{~M}^{-1}=12.229 \mathrm{~L} \mathrm{~mol}^{-1}
\)

Q3. At a certain temperature and total pressure of \(10^5 \mathrm{~Pa}\), iodine vapour contains \(40 \%\) by volume of I atoms
\(
\mathrm{I}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{I}(\mathrm{g})
\)
Calculate \(K_p\) for the equilibrium.

Answer: According to available data:
Total pressure of equilibrium mixture \(=10^5 \mathrm{~Pa}\)
Partial pressure of iodine atoms \((I)=\frac{40}{100} \times\left(10^5 \mathrm{~Pa}\right)=0.4 \times 10^5 \mathrm{~Pa}\)
Partial pressure of iodine molecules \(\left(I_2\right)=\frac{60}{100} \times\left(10^5 \mathrm{~Pa}\right)=0.6 \times 10^5 \mathrm{~Pa}\)
\(I_2(g) \hspace { 2em} \rightleftharpoons \hspace {2em} 2 I(g)\)
\(\left(0.6 \times 10^5 Pa\right) \quad \left(0.4 \times 10^5 Pa\right)\)
\(
K_p=\frac{p_{I^2}}{{p_{I_2}}}=\frac{\left(0.4 \times 10^5 \mathrm{~Pa}\right)^2}{\left(0.6 \times 10^5 \mathrm{~Pa}\right)}=2.67 \times 10^4 \mathrm{~Pa}
\)

Q4. Write the expression for the equilibrium constant, \(K_c\) for each of the following reactions:
\(
\text { (i) } \quad 2 \mathrm{NOCl}(\mathrm{g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g})+\mathrm{Cl}_2 \text { (g) }
\)
\(
\text { (ii) } \quad 2 \mathrm{Cu}\left(\mathrm{NO}_3\right)_2 \text { (s) } \rightleftharpoons 2 \mathrm{CuO} \text { (s) }+4 \mathrm{NO}_2 \text { (g) }+\mathrm{O}_2 \text { (g) }
\)
\(
\text { (iii) } \mathrm{CH}_3 \mathrm{COOC}_2 \mathrm{H}_5(\mathrm{aq})+\mathrm{H}_2 \mathrm{O}(\mathrm{l}) \rightleftharpoons \mathrm{CH}_3 \mathrm{COOH}(\mathrm{aq})+\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}(\mathrm{aq})
\)
\(
\text { (iv) } \mathrm{Fe}^{3+}(\mathrm{aq})+3 \mathrm{OH}^{-}(\mathrm{aq}) \rightleftharpoons \mathrm{Fe}(\mathrm{OH})_3(\mathrm{~s})
\)
\(
\text { (v) } \quad \mathrm{I}_2(\mathrm{~s})+5 \mathrm{~F}_2 \rightleftharpoons 2 \mathrm{IF}_5
\)

Answer:

(i) \(K_c=\frac{[\mathrm{NO}(g)]^2\left[\mathrm{Cl}_2(g)\right]}{[\mathrm{NOCl}(g)]^2}\)
(ii)
\(
K_c=\frac{[\mathrm{CuO}(g)]^2\left[\mathrm{NO}_2(g)\right]^4\left[\mathrm{O}_2(g)\right]}{\left[\mathrm{Cu}\left(\mathrm{NO}_3\right)_2(s)\right]^2}=\left[\mathrm{NO}_2(g)\right]^4\left[\mathrm{O}_2(g)\right]
\)
(iii)
\(
\begin{aligned}
K_c & =\frac{\left[\mathrm{CH}_3 \mathrm{COOH}(a q)\right]\left[\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}(a q)\right]}{\left[\mathrm{CH}_3 \mathrm{COOC}_2 \mathrm{H}_5(a q)\right]\left[\mathrm{H}_2 \mathrm{O}(l)\right]} \\
& =\frac{\left[\mathrm{CH}_3 \mathrm{COOH}(a q)\right]\left[\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}(a q)\right]}{\left[\mathrm{CH}_3 \mathrm{COOC}_2 \mathrm{H}_5(a q)\right]}
\end{aligned}
\)
(iv)
\(
K_c=\frac{\left[\mathrm{Fe}(\mathrm{OH})_3(s)\right]}{\left[\mathrm{Fe}^{3+}(a q)\right]\left[\mathrm{OH}^{-}(a q)\right]^3}=\frac{1}{\left[\mathrm{Fe}^{3+}(a q)\right]\left[\mathrm{OH}^{-}(a q)\right]^3}
\)
(v)
\(
K_c=\frac{\left[\mathrm{IF}_5(l)\right]^2}{\left[\mathrm{l}_2(s)\right]\left[\mathrm{F}_2(g)\right]^5}=\frac{\left[\mathrm{IF}_5(l)\right]^2}{\left[\mathrm{~F}_2(g)\right]^5}
\)

Q5. Find out the value of \(K_c\) for each of the following equilibria from the value of \(K_p\) :
\(
\text { (i) } \quad 2 \mathrm{NOCl}(\mathrm{g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g})+\mathrm{Cl}_2(\mathrm{~g}) ; K_p=1.8 \times 10^{-2} \text { at } 500 \mathrm{~K}
\)
\(
\text { (ii) } \quad \mathrm{CaCO}_3(\mathrm{~s}) \rightleftharpoons \mathrm{CaO}(\mathrm{s})+\mathrm{CO}_2(\mathrm{~g}) ; K_p=167 \text { at } 1073 \mathrm{~K}
\)

Answer: \(K_p\) and \(K_c\) are related to each other as \(K_p=K_c(\mathrm{RT})^{\Delta n g}\) The value of \(K_c\) can be calculated as follows:
\(
\text { (i) } \quad 2 \mathrm{NOCl}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Cl}_2
\)
\(
K_p=1.8 \times 10^{-2} \mathrm{~atm},
\)
\(
\Delta^{n g}=3-2=1 ; R=0.0821 \text { litre atm } K^{-1} \mathrm{~mol}^{-1} ; T=500 \mathrm{~K}
\)
\(
K_c=\frac{K_p}{(\mathrm{R} T)^{\Delta n g}}=\frac{\left(1.8 \times 10^{-2} \mathrm{~atm}\right)}{\left(0.0821 \mathrm{~L} \mathrm{~atm} \mathrm{~K}^{-1} \mathrm{~mol}^{-1} \times 500 \mathrm{~K}\right)^1}
\)
\(
=4.33 \times 10^{-4} \text { (approximately) }
\)
\(
\text { (ii) } \quad \mathrm{CaCO}_3(\mathrm{~s}) \rightleftharpoons \mathrm{CaO}(\mathrm{s})+\mathrm{CO}_2(\mathrm{~g})
\)
\(
K_p=167 \mathrm{~atm}, \Delta^{n g}=2-1=1
\)
\(
R=0.0821 \text { liter atm } K^{-1} \mathrm{~mol}^{-1} ; T=1073 \mathrm{~K}
\)
\(
\begin{aligned}
K_c & =\frac{K_p}{(R T)^{\Delta n g}}=\frac{(167 \mathrm{~atm})}{\left(0.0821 \mathrm{~L} \mathrm{~atm} \mathrm{~K}^{-1} \mathrm{~mol}^{-1} \times 1073 \mathrm{~K}\right)^1} \\
& =1.9 \mathrm{~mol} \mathrm{~L}^{-1}
\end{aligned}
\)

Q6. For the following equilibrium, \(K_c=6.3 \times 10^{14}\) at \(1000 \mathrm{~K}\)
\(
\mathrm{NO}(\mathrm{g})+\mathrm{O}_3(\mathrm{~g}) \rightleftharpoons \mathrm{NO}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g})
\)
Both the forward and reverse reactions in the equilibrium are elementary bimolecular reactions. What is \(K_c\), for the reverse reaction?

Answer: It is given that \(K_c\) for the forward reaction is \(6.3 \times 10^{14}\) Then, \(K_c\) for the reverse reaction will be,
\(
\begin{aligned}
& K_c^{\prime}=\frac{1}{K_c} \\
& =\frac{1}{6.3 \times 10^{14}} \\
& =1.59 \times 0^{-15}
\end{aligned}
\)

Q7. Explain why pure liquids and solids can be ignored while writing the equilibrium constant expression.

Answer: This is because the molar concentration of a pure solid or liquid is independent of the amount present.
\(
\text { Molar concentration }=\frac{\text { No. of moles }}{\text { volume }} \times \frac{\text { Mass }}{\text { volume }} \times \text { Density }
\)
Since the density of pure liquid or solid is fixed and molar mass is also fixed. Therefore molar concentrations are constant.

Q8. Reaction between \(\mathrm{N}_2\) and \(\mathrm{O}_{2-}\) takes place as follows:
\(
\hspace { 7em} 2 \mathrm{~N}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{~N}_2 \mathrm{O}(\mathrm{g})
\)
If a mixture of \(0.482 \mathrm{~mol} \mathrm{~N}_2\) and \(0.933 \mathrm{~mol}\) of \(\mathrm{O}_2\) is placed in a \(10 \mathrm{~L}\) reaction vessel and allowed to form \(\mathrm{N}_2 \mathrm{O}\) at a temperature for which \(K_{\mathrm{c}}=2.0 \times 10^{-37}\), determine the composition of equilibrium mixture.

Answer: Let \(x\) moles of \(\mathrm{N}_2(\mathrm{~g})\) take part in the reaction. According to the equation, \({x} / 2\) moles of \(\mathrm{O}_2(\mathrm{~g})\) will react to form \(x\) moles of \(\mathrm{N}_2 \mathrm{O}(\mathrm{g})\). The molar concentration per litre of different species before the reaction and at the equilibrium point is:
\(
\hspace { 7em} 2 \mathrm{~N}_2(g)+\mathrm{O}_2(g) \rightleftharpoons 2 \mathrm{~N}_2 \mathrm{O}(g)
\)
\(
\text { Initial mole/litre: } \quad \frac{0.482}{10} \quad \frac{0.933}{10} \quad \quad \text { Zero }
\)
\(
\text { Mole/litre at eq. : } \frac{0.482-x}{10} \quad \frac{0.933-\frac{x}{2}}{10} \quad \frac{x}{10}
\)
The value of equilibrium constant \(\left(2.0 \times 10^{-37}\right)\) is extremely small. This means that only small amounts of reactants have reacted. Therefore, is extremely small and can be omitted as far as the reactants are concerned.
Applying Law of Chemical Equilibrium \(K_c=\frac{\left[\mathrm{N}_2 \mathrm{O}(g)\right]^2}{\left[\mathrm{~N}_2(g)\right]^2\left[\mathrm{O}_2(g)\right]}\)
\(
\begin{aligned}
2.0 \times 10^{-37} & =\frac{\left(\frac{x}{10}\right)^2}{\left(\frac{0.482}{10}\right)^2 \times\left(\frac{0.933}{10}\right)}=\frac{0.01 x^2}{2.1676 \times 10^{-4}} \\
x^2 & =43.352 \times 10^{-40} \text { or } x=6.6 \times 10^{-20}
\end{aligned}
\)
As \(x\) is extremely small, it can be neglected. Thus, in the equilibrium mixture
Molar conc. of \(\mathrm{N}_2=0.0482 \mathrm{~mol} \mathrm{~L}^{-1}\)
Molar conc. of \(\mathrm{O}_2=0.0933 \mathrm{~mol} \mathrm{~L}^{-1}\)
\(
\begin{aligned}
& \text { Molar conc. of } \mathrm{N}_2 \mathrm{O}=0.1 \times x=0.1 \times 6.6 \times 10^{-20} \mathrm{~mol} \mathrm{~L}^{-1} \\
& =6.6 \times 10^{-21} \mathrm{~mol} \mathrm{~L}^{-1} \\
&
\end{aligned}
\)

Q9. Nitric oxide reacts with \(\mathrm{Br}_2\) and gives nitrosyl bromide as per reaction given below:
\(
2 \mathrm{NO}(\mathrm{g})+\mathrm{Br}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NOBr}(\mathrm{g})
\)
When 0.087 moles of \(\mathrm{NO}\) and \(0.0437 \mathrm{~mole}\) of \(\mathrm{Br}_2\) are mixed in a closed container at a constant temperature, \(0.0518 \mathrm{~mol}\) of \(\mathrm{NOBr}\) is obtained at equilibrium. Calculate equilibrium amount of \(\mathrm{NO}\) and \(\mathrm{Br}_2\).

Answer: The given reaction is:
\(
2 \mathrm{NO}(\mathrm{g})+\mathrm{Br}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NOBr}(\mathrm{g})
\)
According to the equation, 2 moles of \(\mathrm{NO}(\mathrm{g})\) react with 1 mole of \(\mathrm{Br}_2(\mathrm{~g})\) to form 2 moles of \(\mathrm{NOBr}(\mathrm{g})\). The composition of the equilibrium mixture can be calculated as follows:
No. of moles of \(\mathrm{NOB}_r(\mathrm{~g})\) formed at equilibrium \(=0.0518 \mathrm{~mol}\) (given)
No. of moles of \(\mathrm{NO}(\mathrm{g})\) taking part in reaction \(=0.087 \mathrm{~mol}\)
No. of moles of \(\mathrm{NO}(\mathrm{g})\) left at equilibrium \(=0.087-0.0518=0.0352 \mathrm{~mol}\)
No. of moles of \(\mathrm{Br}_2(\mathrm{~g})\) taking part in reaction \(=1 / 2 \times 0.0518=0.0259 \mathrm{~mol}\)
No. of moles of \(\mathrm{Br}_2(\mathrm{~g})\) left at equilibrium \(=0.0437-0.0259=0.0178 \mathrm{~mol}\)

Q10. At \(450 \mathrm{~K}, K_p=2.0 \times 10^{10} / \mathrm{bar}\) for the given reaction at equilibrium.
\(
2 \mathrm{SO}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{SO}_3(\mathrm{~g})
\)
What is \(K_c\) at this temperature?

Answer:

\(
K_p=K_c(R T)^{\Delta n g} \text { or } K_c=\frac{K_p}{(\mathrm{R} T)^{\Delta n g}}=K_p(R T)^{-\Delta n g}
\)
\(
K_p=2.0 \times 10^{10} \text { bar }^{-1} ; \mathrm{R}=0.083 \mathrm{~L} \mathrm{~bar} \mathrm{~K}^{-1} \mathrm{~mol}^{-1} ; \mathrm{T}=450 \mathrm{~K} ; \Delta^{n g}=2-3=-1
\)
\(
\begin{aligned}
K_c & =\left(2.0 \times 10^{10} \mathrm{~bar}^{-1}\right) \times\left[\left(0.083 \mathrm{~L} \mathrm{~bar} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right) \times(450 \mathrm{~K})\right]^{-(-1)} \\
& =7.47 \times 10^{11} \mathrm{~mol}^{-1} \mathrm{~L}=7.47 \times 10^{11} \mathbf{~M}^{-1}
\end{aligned}
\)

Q11. A sample of \(\mathrm{HI}(\mathrm{g})\) is placed in the flask at a pressure of \(0.2 \mathrm{~atm}\). At equilibrlum the partial pressure of \(\mathrm{HI}(\mathrm{g})\) is \(0.04 \mathrm{~atm}\). What is \(K_p\) for the given equilibrium?
\(
2 \mathrm{HI}(\mathrm{g}) \rightleftharpoons \mathrm{H}_2(\mathrm{~g})+\mathrm{I}_2(\mathrm{~g})
\)

Answer: 

\(
\begin{aligned}
\mathrm{pHI} & =0.04 \mathrm{~atm}, \mathrm{pH}_2=0.08 \mathrm{~atm} ; \mathrm{pI}_2=0.08 \mathrm{~atm} \\
K_p & =\frac{\mathrm{pH}_2 \times \mathrm{pI}_2}{\mathrm{p}_{\mathrm{HI}}^2}=\frac{(0.08 \mathrm{~atm}) \times(0.08 \mathrm{~atm})}{(0.04 \mathrm{~atm}) \times(0.04 \mathrm{~atm})}=4.0
\end{aligned}
\)

Q12. A mixture of \(1.57 \mathrm{~mol}\) of \(\mathrm{N}_2, 1.92 \mathrm{~mol}\) of \(\mathrm{H}_2\) and \(8.13 \mathrm{~mol} {\text {of }} \mathrm{NH}_3\) is introduced into a \(20 \mathrm{~L}\) reaction vessel at \(500 \mathrm{~K}\). At this temperature, the equilibrium constant,\(K_{\mathrm{c}}\) for the reaction \(\mathrm{N}_2(\mathrm{~g})+3 \mathrm{H}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_3(\mathrm{~g})\) is \(1.7 \times 10^2\). Is the reaction mixture at equilibrium? If not, what is the direction of the net reaction?

Answer: The reaction is: \(\mathrm{N}_2(\mathrm{~g})+3 \mathrm{H}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_3(\mathrm{~g})\)
Concentration quotient \(\left(Q_c\right)=\frac{\left[\mathrm{NH}_3\right]^2}{\left[\mathrm{~N}_2\right]\left[\mathrm{H}_2\right]^3}=\frac{\left(8.13 / 20 \mathrm{~mol} \mathrm{~L}^{-1}\right)^2}{\left(1.57 / 20 \mathrm{~mol} \mathrm{~L}^{-1}\right) \times\left(1.92 / 20 \mathrm{~mol} \mathrm{~L}^{-1}\right)^3}\)
\(
=2.38 \times 10^3
\)
The equilibrium constant \(\left(K_c\right)\) for the reaction \(=1.7 \times 10^{-2}\) As \(Q_c \neq K_c\); this means that the reaction is not in a state of equilibrium.

Q13. The equilibrium constant expression for a gas reaction is,
\(
K_c=\frac{\left[\mathrm{NH}_3\right]^4\left[\mathrm{O}_2\right]^5}{[\mathrm{NO}]^4\left[\mathrm{H}_2 \mathrm{O}\right]^6}
\)
Write the balanced chemical equation corresponding to this expression.

Answer: A balanced chemical equation for the reaction is 
\(
4 \mathrm{NO}(g)+6 \mathrm{H}_2 \mathrm{O}(g) \rightleftharpoons 4 \mathrm{NH}_3(g)+5 \mathrm{O}_2(g)
\)

Q14. One mole of \(\mathrm{H}_2 \mathrm{O}\) and one mole of \(\mathrm{CO}\) are taken in \(10 \mathrm{~L}\) vessel and heated to \(725 \mathrm{~K}\). At equilibrium \(40 \%\) of water (by mass) reacts with \(\mathrm{CO}\) according to the equation,
\(
\mathrm{H}_2 \mathrm{O}(\mathrm{g})+\mathrm{CO}(\mathrm{g}) \rightleftharpoons \mathrm{H}_2(\mathrm{~g})+\mathrm{CO}_2(\mathrm{~g})
\)
Calculate the equilibrium constant for the reaction.

Answer: Number of moles of water originally present \(=1 \mathrm{~mol}\)
Percentage of water reacted \(=40 \%\)
Number of moles of water reacted \(=1 \times 40 / 100=0.4 \mathrm{~mol}\)
A number of moles of water left \(=(1-0.4)=0.6\) mole According to the equation, 0.4 mole of water will react with 0.4 mole of carbon monoxide to form 0.4 mole of hydrogen and 0.4 mole of carbon dioxide.
Thus, the molar conc. per litre of the reactants and products before the reaction and at the equilibrium point are as follows:
\(
\mathrm{H}_2 \mathrm{O}(g)+\mathrm{CO}(g) \rightleftharpoons \mathrm{H}_2(g)+\mathrm{CO}_2(g)
\)

Initial moles/litre
\(
\quad \frac{1}{10} \quad \quad \quad \frac{1}{10} \quad \quad \quad 0 \quad \quad \quad 0
\)
Mole/litre at the equilibrium point
\(
\frac{1-0.4}{10} \quad \quad \frac{1-0.4}{10} \quad \quad \quad \frac{0.4}{10} \quad \quad \frac{0.4}{10}
\)

\(
\quad \frac{0.6}{10} \quad \quad \quad \frac{0.6}{10} \quad \quad \quad  \frac{0.4}{10} \quad \quad \frac{0.4}{10}
\)
Applying the law of chemical equilibrium,
Equilibrium constant
\(
\begin{aligned}
\left(K_c\right) & =\frac{\left[\mathrm{H}_2(g)\right]\left[\mathrm{CO}_2(g)\right]}{\left[\mathrm{H}_2 \mathrm{O}(g)\right][\mathrm{CO}(g)]}=\frac{\left(\frac{0.4}{10} \mathrm{~mol} \mathrm{~L}^{-1}\right) \times\left(\frac{0.4}{10} \mathrm{~mol} \mathrm{~L}^{-1}\right)}{\left(\frac{0.6}{10} \mathrm{~mol} \mathrm{~L}^{-1}\right) \times\left(\frac{0.6}{10} \mathrm{~mol} \mathrm{~L}^{-1}\right)} \\
& =\frac{0.16}{0.36}=\mathbf{0 . 4 4}
\end{aligned}
\)

Q15. At \(700 \mathrm{~K}\), equilibrium constant for the reaction:
\(
\mathrm{H}_2(\mathrm{~g})+\mathrm{I}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g})
\)
1s 54.8. If \(0.5 \mathrm{~mol} \mathrm{~L}^{-1}\) of \(\mathrm{HI}(\mathrm{g})\) is present at equilibrlum at \(700 \mathrm{~K}\), what are the concentration of \(\mathrm{H}_2(\mathrm{~g})\) and \(\mathrm{I}_2(\mathrm{~g})\) assuming that we initlally started with \(\mathrm{HI}(\mathrm{g})\) and allowed it to reach equilibrium at \(700 \mathrm{~K}\)?

Answer: It is given that equilibrium constant \(K_c\) for the reaction \(\mathrm{H}_{2(\mathrm{~g})}+\mathrm{I}_{2(\mathrm{~g})} \leftrightarrow 2 \mathrm{HI}_{(\mathrm{g})}\) is 54.8.
Therefore, at equilibrium, the equilibrium constant for the reaction \(K^{\prime}{ }_c\) for the reaction
\(
2 \mathrm{HI}_{(\mathrm{g})} \leftrightarrow \mathrm{H}_{2(\mathrm{~g})}+\mathrm{I}_{2(\mathrm{~g})}
\)
\([\mathrm{HI}]=0.5 \mathrm{molL}^{-1}\) will be \(1 / 54.8\).
Let the concentrations of hydrogen and iodine at equilibrium be \(x \mathrm{~molL}^{-1}\)
\(
\left[\mathrm{H}_2\right]=\left[\mathrm{I}_2\right]=\mathrm{x} \mathrm{~mol} \mathrm{L}^{-1}
\)
Therefore, \(\frac{\left[\mathrm{H}_2\right]\left[\mathrm{I}_2\right]}{[\mathrm{HI}]^2}=\mathrm{K}_{\mathrm{c}}^{\prime}\)
\(
\begin{aligned}
& \Rightarrow \frac{\mathrm{x} \times \mathrm{x}}{(0.5)^2}=\frac{1}{54.8} \\
& \Rightarrow \mathrm{x}^2=\frac{0.25}{54.8} \\
& \Rightarrow \mathrm{x}=0.068 \mathrm{~molL}^{-1} \text { (approx.) }
\end{aligned}
\)
Hence, at equilibrium, \(\left[\mathrm{H}_2\right]=\left[\mathrm{I}_2\right]=0.068 \mathrm{~mol} \mathrm{~L}^{-1}\).

Q16. What is the equilibrium concentration of each of the substances in the equilibrium when the initial concentration of \(\mathrm{ICl}\) was \(0.78 \mathrm{M}\)?
\(
2 \mathrm{ICl}(\mathrm{g}) \rightleftharpoons \mathrm{I}_2(\mathrm{~g})+\mathrm{Cl}_2(\mathrm{~g}) ; \quad K_c=0.14
\)

Answer: Suppose at equilibrium, the molar concentration of both \(\mathrm{I}_2(\mathrm{~g})\) and \(\mathrm{Cl}_2(\mathrm{~g})\) is \(x \mathrm{~mol} \mathrm{~L}^{-1}\).
\(
\hspace {10 em} 2 \mathrm{ICl}(g) \rightleftharpoons \mathrm{I}_2(g)+\mathrm{Cl}_2(g)
\)
\(
\begin{array}{lcll}
\text { Initial molar conc. } & 0.78 \mathrm{M} & 0 & & & 0 \\
\text { Equil. molar conc. } & (0.78-2 x) \mathrm{M} & x & & & x
\end{array}
\)
\(
K_c=\frac{\left[I_2(g)\right]\left[\mathrm{Cl}_2(g)\right]}{\left[\mathrm{ICl}(g)^2\right]}=\frac{(x) \times(x)}{(0.78-2 x)^2}
\)
\(
\begin{aligned}
\frac{x}{(0.78-2 x)} & =(0.14)^{1 / 2}=0.374 \text { or } x=0.374(0.78-2 x) \\
x & =0.292-0.748 x \text { or } 1.748 x=0.292 ; x=\frac{0.292}{1.748}=0.167
\end{aligned}
\)
\(
[\mathrm{ICl}]=(0.78-2 \times 0.167)=(0.78-0.334)=\mathbf{0 . 4 4 6} \mathbf{M}
\)
\(
\left[\mathrm{I}_2\right]=0.167 \mathrm{M} ;\left[\mathrm{Cl}_2\right]=\mathbf{0 . 1 6 7} \mathrm{M}
\)

Q17. \(K_p=0.04 \mathrm{~atm}\) at \(899 \mathrm{~K}\) for the equillbrium shown below. What is the equilibrium concentration of \(\mathrm{C}_2 \mathrm{H}_6\) when it is placed in a flask at \(4.0 \mathrm{~atm}\) pressure and allowed to come to equilibrium?
\(
\mathrm{C}_2 \mathrm{H}_6(\mathrm{~g}) \rightleftharpoons \mathrm{C}_2 \mathrm{H}_4(\mathrm{~g})+\mathrm{H}_2(\mathrm{~g})
\)

Answer: The equilibrium in the reaction is:
\(
\hspace { 7em} \mathrm{C}_2 \mathrm{H}_6(\mathrm{~g}) \rightleftharpoons \mathrm{C}_2 \mathrm{H}_4(\mathrm{~g})+\mathrm{H}_2(\mathrm{~g})
\)
\(
\begin{array}{lccc}
\text { Initial pressure: } & {4} \mathrm{~atm} & 0 & 0 \\
\text { Eqn. pressure: } & (4-p) \mathrm{atm} & p \mathrm{~atm} & p \mathrm{~atm}
\end{array}
\)
\(
K_p=\frac{p_{C_2 H_4} \times p_{H_2}}{p_{C_2 H_6}} \quad \text { or } \quad 0.04=\frac{p^2}{(4-p)}
\)
\(
\begin{aligned}
p^2 & =0.04(4-p) \text { or } \mathrm{p}^2+0.04 p-0.16=0 \\
p & =\frac{(-0.04) \pm \sqrt{0.0016-4(-0.16)}}{2} \\
& =\frac{(-0.04) \pm 0.8}{2}=\frac{0.76}{2}=0.38
\end{aligned}
\)
Equilibrium pressure or concentration of \(\mathrm{C}_2 \mathrm{H}_6=(4-0.38)=3.62 \mathrm{~atm}\).

Q18. Ethyl acetate is formed by the reaction between ethanol and acetic acid and the equilibrium is represented as:
\(
\mathrm{CH}_3 \mathrm{COOH}(\mathrm{l})+\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}(\mathrm{l}) \rightleftharpoons \mathrm{CH}_3 \mathrm{COOC}_2 \mathrm{H}_5(\mathrm{l})+\mathrm{H}_2 \mathrm{O}(\mathrm{l})
\)
(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 \(293 \mathrm{~K}\), if one starts with \(1.00 \mathrm{~mol}\) of acetic acid and \(0.18 \mathrm{~mol}\) of ethanol, there is 0.171 \mathrm{~mol}[/latex] of ethyl acetate in the final equilibrium mixture. Calculate the equilibrium constant.
(iii) Starting with \(0.5 \mathrm{~mol}\) of ethanol and \(1.0 \mathrm{~mol}\) of acetic acid and maintaining it at \(293 \mathrm{~K}, 0.214 \mathrm{~mol}\) of ethyl acetate is found after some time. Has equilibrium been reached?

Answer: (i) The concentration ratio (Concentration quotient) \(Q_c\) for the reaction is:
\(
Q_c=\frac{\left[\mathrm{CH}_3 \mathrm{COOC}_2 \mathrm{H}_5(l)\right]\left[\mathrm{H}_2 \mathrm{O}(l)\right]}{\left[\mathrm{CH}_3 \mathrm{COOH}(l)\right]\left[\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}(l)\right]}
\)
(ii)
\(
\hspace { 9 em} \mathrm{CH}_3 \mathrm{COOH}(l)+\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}(l) \Longrightarrow \mathrm{CH}_3 \mathrm{COOC}_2 \mathrm{H}_5(l)+\mathrm{H}_2 \mathrm{O}(l)
\)
\(
\begin{array}{lcccc}
\text { Initial molar conc. } & 1.0 \mathrm{~mol} & 0.18 \mathrm{~mol} & & 0 & & & 0 \\
\text { Molar conc. at } & (1-0.17 \mathrm{l}) & (0.18-0.17 \mathrm{l}) & & 0.17 \mathrm{~mol} & & & 0.17 \mathrm{~mol} \\
\text { equilibrium point } & =0.829 \mathrm{~mol} & =0.009 \mathrm{~mol} & &
\end{array}
\)
Applying the Law of Chemical equilibrium,
\(
\begin{aligned}
K_c & =\frac{\left[\mathrm{CH}_3 \mathrm{COOC}_2 \mathrm{H}_5\right](l)\left[\mathrm{H}_2 \mathrm{O}(l)\right]}{\left[\mathrm{CH}_3 \mathrm{COOH}(l)\right]\left[\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}(l)\right]} \\
& =\frac{(0.171 \mathrm{~mol}) \times(0.171 \mathrm{~mol})}{(0.829 \mathrm{~mol})(0.009 \mathrm{~mol})}=\mathbf{3 . 9 2}
\end{aligned}
\)
(iii)
\(
\hspace { 9em} \mathrm{CH}_3 \mathrm{COOH}(l)+\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}(l) \rightleftharpoons \mathrm{CH}_3 \mathrm{COOC}_2 \mathrm{H}_5(l)+\mathrm{H}_2 \mathrm{O}(l)
\)
\(
\begin{array}{lcccc}
\text { Initial molar conc. } & 1.0 \mathrm{~mol} & 0.5 \mathrm{~mol} & & & 0.214 \mathrm{~mol} & & 0.214 \mathrm{~mol} \\
\text { Molar conc. at } & 1.0-0.214 & 0.5-0.214 & & \\
\text { equilibrium } & =0.786 \mathrm{~mol} & =0.286 \mathrm{~mol} &
\end{array}
\)
\(
\begin{aligned}
Q_c & =\frac{\left[\mathrm{CH}_3 \mathrm{COOC}_2 \mathrm{H}_5(l)\right]\left[\mathrm{H}_2 \mathrm{O}(l)\right]}{\left[\mathrm{CH}_3 \mathrm{COOH}(l)\right]\left[\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}(l)\right]} \\
& =\frac{(0.214 \mathrm{~mol}) \times(0.214 \mathrm{~mol})}{(0.786 \mathrm{~mol})(0.286 \mathrm{~mol})}=\mathbf{0 . 2 0 4}
\end{aligned}
\)
Since \(Q_C\) is less than \(K_c\) this means that the equilibrium has not been reached. The reactants are still taking part in the reaction to form the products.

Q19. A sample of pure \(\mathrm{PCl}_5\) was introduced into an evacuated vessel at \(473 \mathrm{~K}\). After equilibrium was attalned, concentration of \(\mathrm{PCl}_5\) was found to be \(0.5 \times 10^{-1} \mathrm{~mol} \mathrm{~L}^{-1}\). If value of \(K_{\mathrm{c}}\) is \(8.3 \times 10^{-3}\), what are the concentrations of \(\mathrm{PCl}_3\) and \(\mathrm{Cl}_2\) at equillbrium?
\(
\mathrm{PCl}_5(\mathrm{~g}) \rightleftharpoons \mathrm{PCl}_3(\mathrm{~g})+\mathrm{Cl}_2(\mathrm{~g})
\)

Answer: Let the initial molar concentration of \(\mathrm{PCl}_5\) per litre \(={x}\) mol
Molar concentration of \(\mathrm{PCl}_5\) at equilibrium \(=0.05 \mathrm{~mol}\)
\(\therefore\) Moles of \(\mathrm{PCl}_5\) decomposed \(=(x-0.05) \mathrm{mol}\)
Moles of \(\mathrm{PCl}_3\) formed \(=(x-0.05) \mathrm{mol}\)
Moles of \(\mathrm{Cl}_2\) formed \(=(x-0.05) \mathrm{mol}\)
The molar conc./litre of reactants and products before the reaction and at the equilibrium point are:
\(\hspace { 10em} \mathrm{PCl}_5(\mathrm{~g}) \rightleftharpoons \mathrm{PCl}_3(\mathrm{~g})+\mathrm{Cl}_2(\mathrm{~g})\)
\(
\begin{array}{lccc}
\text { Initial moles/litre } & x & 0 & 0 \\
\text { Moles/litre at eqm. point } & 0.05 & (x-0.05) & (x-0.05)
\end{array}
\)
Applying the Law of chemical equilibrium,
\(
K_c=\frac{\left[\mathrm{PCl}_3\right]\left[\mathrm{Cl}_2\right]}{\left[\mathrm{PCl}_5\right]} ; \quad 0.0083=\frac{(x-0.05) \times(x-0.05)}{0.05}
\)
\(
\begin{aligned}
(x-0.05)^2 & =0.0083 \times 0.05=4.15 \times 10^{-4} \\
(x-0.05) & =\left(4.15 \times 10^{-4}\right)^{1 / 2}=2.037 \times 10^{-2}=0.02 \mathrm{moles} \\
x & =0.05+0.02=0.07 \mathrm{~mol}
\end{aligned}
\)
The molar concentration per litre of \(\mathrm{PCl}_3\) at eqm. \(=0.07-0.05=\mathbf{0 . 0 2} \mathbf{~ m o l}\)
The molar concentration per litre of \(\mathrm{Cl}_2 {\text {, at eqm. }}=0.07-0.05=0.02 \mathrm{~mol}\).

Q20. One of the reaction that takes place in producing steel from iron ore is the reduction of iron(II) oxide by carbon monoxide to give iron metal and \(\mathrm{CO}_2\).
\(
\mathrm{FeO}(\mathrm{s})+\mathrm{CO}(\mathrm{g}) \rightleftharpoons \mathrm{Fe}(\mathrm{s})+\mathrm{CO}_2(\mathrm{~g}) ; \quad K_p=0.265 \mathrm{~atm} \text { at } 1050 \mathrm{~K}
\)
What are the equilibrium partial pressures of \(\mathrm{CO}\) and \(\mathrm{CO}_2\) at \(1050 \mathrm{~K}\) if the initial partial pressures are: \(p_{\mathrm{CO}}=1.4 \mathrm{~atm}\) and \(=0.80 \mathrm{~atm}\)?

Answer: \(\mathrm{FeO}(s)+\mathrm{CO}(g) \rightleftharpoons \mathrm{Fe}(s)+\mathrm{CO}_2(g)\)
\(
\text { Initial pressure: } \quad 1.4 \mathrm{~atm} \quad 0.8 \mathrm{~atm}
\)
\(
Q_p=\frac{p_{\mathrm{CO}_2}}{p_{\mathrm{CO}}}=\frac{(0.8 \mathrm{~atm})}{(1.4 \mathrm{~atm})}=0.571
\)
Since \(Q_p>K_p(0.265)\), this means that the reaction will move in the backward direction to attain equilibrium. Therefore, the partial pressure of \(\mathrm{CO}_2\) will decrease while that of \(\mathrm{CO}\) will increase so that the equilibrium may be attained again. Let \(p\) atm be the decrease in the partial pressure of \(\mathrm{CO}_2\). Therefore, the partial pressure of \(\mathrm{CO}\) will increase by the same magnitude i.e., \({p}\) atm.
\(
p_{\mathrm{CO}_2}=(0.8-p) \mathrm{atm} ; p_{\mathrm{CO}}(\mathrm{g})=(1.4+p) \mathrm{atm}
\)
\(
\text { At equilibrium, } \quad K_p=\frac{p_{\mathrm{CO}_2}}{p_{\mathrm{CO}}}=\frac{(0.8-p) \mathrm{atm}}{(1.4+p) \mathrm{atm}}=\frac{(0.8-p)}{(1.4+p)}
\)
\(
\text { or } \quad 0.265=\frac{(0.8-p)}{(1.4+p)}
\)
\(
0.371+0.265 p=0.8-p \quad \text { or } 1.265 p=0.8-0.371=0.429
\)
\(
\begin{aligned}
p & =0.429 / 1.265=0.339 \mathrm{~atm} \\
\left(p_{\mathrm{CO}}\right)_{\mathrm{eq}} & =(1.4+0.339)=1.739 \mathrm{~atm} \\
\left(p_{\mathrm{CO}_2}\right)_{\mathrm{eq}} & =(0.8-0.339)=0.461 \mathrm{~atm}
\end{aligned}
\)

Q21. Equilibrium constant, \(K_c\) for the reaction
\(
\mathrm{N}_2(\mathrm{~g})+3 \mathrm{H}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_3(\mathrm{~g}) \text { at } 500 \mathrm{~K} \text { is } 0.061
\)
At a particular time, the analysis shows that composition of the reaction mixture is \(3.0 \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~N}_2, 2.0 \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{H}_2\) and \(0.5 \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{NH}_3\). Is the reaction at equilibrium? If not in which direction does the reaction tend to proceed to reach equilibrium?

Answer: The given reaction is:
\(
\mathrm{N}_2(g)+3 \mathrm{H}_2(g) \rightleftharpoons 2 \mathrm{NH}_3(g)
\)
According to available data.
\(
\begin{aligned}
& \mathrm{N}_2=[3.0] ; \mathrm{H}_2=[2.0] ; \mathrm{NH}_3=[0.50] \\
& Q_c=\frac{\left[\mathrm{NH}_3(g)\right]^2}{\left[\mathrm{~N}_2(g)\right]\left[\mathrm{H}_2(g)\right]^3}=\frac{[0.50]^2}{[3.0][2.0]^3}=\frac{0.25}{24}=\mathbf{0 . 0 1 0 4} .
\end{aligned}
\)
Since the value of \(Q_c\) is less than that of \(K_c(0.061)\), the reaction is not in a state of equilibrium. It will proceed in the forward direction till \(Q_c\) becomes the same as \(K_c\).

Q22. Bromine monochloride, \(\mathrm{BrCl}\) decomposes into bromine and chlorine and reaches the equilibrium:
\(
2 \mathrm{BrCl}(\mathrm{g}) \rightleftharpoons \mathrm{Br}_2(\mathrm{~g})+\mathrm{Cl}_2(\mathrm{~g})
\)
for which \(K_{\mathrm{c}}=32\) at \(500 \mathrm{~K}\). If initially pure \(\mathrm{BrCl}\) is present at a concentration of \(3.3 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1}\), what is its molar concentration in the mixture at equilibrium?

Answer: Let \(x\) moles of \(\mathrm{B}_{\mathrm{r}} \mathrm{Cl}\) decompose in order to attain the equilibrium. The initial molar concentration and the molar concentration at the equilibrium point of different species may be represented as follows:
\(
\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad 2 \mathrm{BrCl}(g) \rightleftharpoons \mathrm{Br}_2(g)+\mathrm{Cl}_2(g)
\)
\(
\begin{array}{lccc}
\text { Initial moles/litre } & 0.0033  & 0 &  & 0 \\
\text { Moles/litre at eqm. point } & 0.0033-x & x / 2 & x / 2
\end{array}
\)
Applying Law of chemical equilibrium, \(K_c=\frac{\left[\mathrm{Br}_2\right]\left[\mathrm{Cl}_2\right]}{\left[\mathrm{BrCl}^2\right.}\) or \(32=\frac{(x / 2) \times(x / 2)}{(0.0033-x)^2}\) On taking the square root, \(5.656=\frac{x / 2}{(0.0033-x)}\)
\(
\frac{x}{(0.0033-x)}=11.31 \text { or } 12.31 x=0.037 ; x=\frac{0.037}{12.31}=0.003
\)
\(\therefore\) Molar concentration of \(\mathrm{BrCl}\) at equilibrium point \(=0.0033-0.003=3.0 \times 10^{-4} \mathrm{molL}^{-1}\)

Q23. At \(1127 \mathrm{~K}\) and \(1 \mathrm{~atm}\) pressure, a gaseous mixture of \(\mathrm{CO}\) and \(\mathrm{CO}_2\) in equilibrium with solid carbon has \(90.55 \% \mathrm{CO}\) by mass
\(
\mathrm{C}(\mathrm{s})+\mathrm{CO}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{CO}(\mathrm{g})
\)
Calculate \(K_{\mathrm{c}}\) for this reaction at the above temperature.

Answer: Step I: Calculation of \(K_p\) for the reaction
Let the total mass of the gaseous mixture \(=100 \mathrm{~g}\)
Mass of \(\mathrm{CO}\) in the mixture \(=90.55 \mathrm{~g}\)
Mass of \(\mathrm{CO}_2\) in the mixture \(=(100-90.55)=9.45 \mathrm{~g}\)
No. of moles of \(\mathrm{CO}=\frac{90.55 \mathrm{~g}}{\left(28 \mathrm{~g} \mathrm{~mol}^{-1}\right)}=3.234 \mathrm{~mol}\)
No. of moles of \(\mathrm{CO}_2=\frac{90 \cdot 55 \mathrm{~g}}{\left(44 \mathrm{~g} \mathrm{~mol}^{-1}\right)}=0.215 \mathrm{~mol}\)
\(
p \mathrm{CO} \text { in the mixture }=\frac{(3.234 \mathrm{~mol})}{(3.234+0.215)} \times 1 \mathrm{~atm}=\frac{(3.234 \mathrm{~mol})}{(3.449 \mathrm{~mol})} \times 1 \mathrm{~atm}=0.938 \mathrm{~atm}
\)
\(
p \mathrm{CO}_2 \text { in the mixture }=\frac{(0.215 \mathrm{~mol})}{(3.449 \mathrm{~mol})} \times 1 \mathrm{~atm}=0.062 \mathrm{~atm}
\)
\(
\hspace { 6em} \mathrm{C}(s)+\mathrm{CO}_2(g) \rightleftharpoons 2 \mathrm{CO}(g)
\)
\(
\begin{array}{lll}
\text { Eqm. pressure } & 0.062 \mathrm{~atm} & 0.938 \mathrm{~atm}
\end{array}
\)
\(
\mathrm{K}_p=\frac{p^2 \mathrm{CO}}{p \mathrm{CO}_2}=\frac{(0.938 \mathrm{~atm})^2}{(0.062 \mathrm{~atm})}=14 \cdot 19 \mathrm{~atm}
\)
Step. II: Calculation of \(K_c\) for the reaction.
\(
\mathrm{K}_c=\frac{\mathrm{K}_p}{(\mathrm{RT})^{\Delta n g}}
\)
\(
\mathrm{K}_p=14.19 \mathrm{~atm}, \mathrm{R}=0.0821 \mathrm{~L} \mathrm{~atm} \mathrm{~K}^{-1} \mathrm{~mol}^{-1} \text {, }T=1127 \mathrm{~K} ; \Delta^{n g}=2-1=1
\)
\(\mathrm{K}_c=\frac{(14.19 \mathrm{~atm})}{\left(0.0821 \mathrm{~L} \mathrm{~atm} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right) \times(1127 \mathrm{~K})^1}=\mathbf{0 . 1 5 3} \mathrm{~mol} \mathrm{L}^{-1}\).

Q24. Calculate a) \(\Delta G^{\ominus}\) and b) the equilibrium constant for the formation of \(\mathrm{NO}_2\) from \(\mathrm{NO}\) and \(\mathrm{O}_2\) at \(298 \mathrm{~K}\)
\(
\mathrm{NO}(\mathrm{g})+1 / 2 \mathrm{O}_2(\mathrm{~g}) \rightleftharpoons \mathrm{NO}_2(\mathrm{~g})
\)
where
\(
\begin{aligned}
& \Delta_{\mathrm{r}} G^{\ominus}\left(\mathrm{NO}_2\right)=52.0 \mathrm{~kJ} / \mathrm{mol} \\
& \Delta_{\mathrm{r}} G^{\ominus}(\mathrm{NO})=87.0 \mathrm{~kJ} / \mathrm{mol} \\
& \Delta_{\mathrm{r}} G^{\ominus}\left(\mathrm{O}_2\right)=0 \mathrm{~kJ} / \mathrm{mol}
\end{aligned}
\)

Answer: Step I: Calculation of \(\Delta G^{\ominus}\)
\(
\begin{aligned}
\Delta \mathrm{G}^{\ominus} & =\Delta_f \mathrm{G}^{\ominus}\left(\mathrm{NO}_2\right)-\left[\Delta_f \mathrm{G}^{\ominus}(\mathrm{NO})+\Delta_f \mathrm{G}^{\ominus}\left(1 / 2 \mathrm{O}_2\right)\right] \\
& =52 \cdot 0-(87+0)=-35 \mathrm{~kJ} \mathrm{~mol}^{-1}
\end{aligned}
\)
Step II: Calculation of \(K_c\)
\(
\Delta \mathrm{G}^{\ominus}=-2.303 \mathrm{RT} \log \mathrm{K}_c
\)
\(
\log \mathrm{K}_c=-\frac{\Delta \mathrm{G}}{2 \cdot 303 \mathrm{RT}}
\)
\(
=-\frac{\left(-35 \times 10^3 \mathrm{~J} \mathrm{~mol}^{-1}\right)^3}{2 \cdot 303 \times\left(8.314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\right) \times(298 \mathrm{~K})}=6 \cdot 134
\)
\(
\mathrm{K}_c=\text { Antilog } 6.314=\mathbf{1 . 3 6} \times \mathbf{1 0 ^ { 6 }} .
\)

Q25. 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) \(\mathrm{PCl}_5(\mathrm{~g}) \rightleftharpoons \mathrm{PCl}_3(\mathrm{~g})+\mathrm{Cl}_2(\mathrm{~g})\)
(b) \(\mathrm{CaO}(\mathrm{s})+\mathrm{CO}_2(\mathrm{~g}) \rightleftharpoons \mathrm{CaCO}_3(\mathrm{~s})\)
(c) \(3 \mathrm{Fe}(\mathrm{s})+4 \mathrm{H}_2 \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{Fe}_3 \mathrm{O}_4(\mathrm{~s})+4 \mathrm{H}_2(\mathrm{~g})\)

Answer: (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) Pressure will increase in backward reaction and a number of moles of products will decrease.
(c) The change in pressure will have no effect on the equilibrium constant and there will be no change in the number of moles.

Q26. Which of the following reactions will be affected by increasing the pressure? Also, mention whether the change will cause the reaction to go into forward or backward direction.
(i) \(\mathrm{COCl}_2(\mathrm{~g}) \rightleftharpoons \mathrm{CO}(\mathrm{g})+\mathrm{Cl}_2(\mathrm{~g})\)
(ii) \(\mathrm{CH}_4(\mathrm{~g})+2 \mathrm{~S}_2(\mathrm{~g}) \rightleftharpoons \mathrm{CS}_2(\mathrm{~g})+2 \mathrm{H}_2 \mathrm{~S}(\mathrm{~g})\)
(iii) \(\mathrm{CO}_2(\mathrm{~g})+\mathrm{C}(\mathrm{s}) \rightleftharpoons 2 \mathrm{CO}(\mathrm{g})\)
(iv) \(2 \mathrm{H}_2(\mathrm{~g})+\mathrm{CO}(\mathrm{g}) \rightleftharpoons \mathrm{CH}_3 \mathrm{OH}(\mathrm{g})\)
(v) \(\mathrm{CaCO}_3(\mathrm{~s}) \rightleftharpoons \mathrm{CaO}(\mathrm{s})+\mathrm{CO}_2(\mathrm{~g})\)
(vi) \(4 \mathrm{NH}_3(\mathrm{~g})+5 \mathrm{O}_2(\mathrm{~g}) \rightleftharpoons 4 \mathrm{NO}(\mathrm{g})+6 \mathrm{H}_2 \mathrm{O}(\mathrm{g})\)

Answer: Only those reactions will be affected by increasing the pressure in which the number of moles of the gaseous reactants and products are different \(\left(n_p \neq n_r\right.\) ) (gaseous). With the exception of reaction (i); all the remaining five reactions will get affected by increasing the pressure. In general,

• The reaction will go to the left if \(n_p>n_r\).
• The reaction will go to the right if \(n_r>n_p\). Keeping this in mind,

(i) An increase in pressure will not affect equilibrium because \(n_p=n_r=3\).
(ii) Increase in pressure will favour backward reaction because \(n_p(2)>n_r\) (1)
(iii) Increase in pressure will favour backward reaction because \(n_p(10)>n_r(9)\)
(iv) Increase in pressure will favour forward reaction because \(n_p(1) <n_r\) (2)
(v) Increase in pressure will favour backward reaction because \(n_p(2)>n_r(1)\)
(vi) An increase in pressure will favour backward reaction because \(n_p(1)>n_r(0)\).

Q27. The equilibrium constant for the following reaction is \(1.6 \times 10^5\) at \(1024 \mathrm{~K}\)
\(
\mathrm{H}_2(\mathrm{~g})+\mathrm{Br}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{HBr}(\mathrm{g})
\)
F1nd the equilibrium pressure of all gases if 10.0 bar of \(\mathrm{HBr}\) is introduced into a sealed container at \(1024 \mathrm{~K}\).

Answer: Step I: Calculation of \(\mathrm{K}_{\mathrm{p}}\).
\(
\mathrm{H}_2(\mathrm{~g})+\mathrm{Br}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{HBr}(\mathrm{g})
\)
\(
K_p=K_c(\mathrm{R} T)^{\Delta n}=K_c(\mathrm{R} T)^0 \quad\left(\because \Delta_n=2-2=\text { zero }\right)
\)
\(
K_p=K_c=1.6 \times 10^5.
\)
Step II: Calculation of partial pressure of gases
\(
\hspace { 9em} 2 \mathrm{HBr}(\mathrm{g}) \rightleftharpoons \mathrm{H}_2(\mathrm{~g})+\mathrm{Br}_2(\mathrm{~g})
\)
\(
\begin{array}{lccc}
\text { Initial pressure } & 10 \mathrm{bar} & \text { zero } & \text { zero } \\
\text { Eqm. pressure } & (10-\mathrm{P}) \text { bar } & \mathrm{P} / 2 \text { bar } & \mathrm{P} / 2 \text { bar }
\end{array}
\)
\(
K_p^{\prime}=\frac{p \mathrm{H}_2 \times p \mathrm{Br}_2}{p^2 \mathrm{HBr}} \text { or } \frac{1}{1.6 \times 10^5}=\frac{\mathrm{P} / 2 \times \mathrm{P} / 2}{(10-\mathrm{P})^2}=\frac{\mathrm{P}^2}{4(10-\mathrm{P})^2}
\)
On taking square root; \(\frac{\mathrm{P}^2}{4(10-\mathrm{P})^2}=\left(\frac{1}{1.6 \times 10^5}\right)^{1 / 2}\) or \(\frac{2(10-\mathrm{P})}{\mathrm{P}}=\left(1.6 \times 10^5\right)^{1 / 2}\) \(=4 \times 10^2\)
\(
\begin{aligned}
20-2 \mathrm{P} & =4 \times 10^2 \mathrm{P} \text { or } \mathrm{P}\left(4 \times 10^2+2\right)=20 \\
\mathrm{P} & =\frac{20}{(400+2)}=\frac{20}{402}=0.050 \mathrm{bar}
\end{aligned}
\)
\(
p \mathrm{H}_2=0.025 \text { bar; } p \mathrm{Br}_2=0.025 \text { bar: } p \mathrm{HBr}=10-0.05=9.95 \text { bar } \approx 10.0 \text { bar. }
\)

Q28. Dihydrogen gas is obtained from natural gas by partial oxidation with steam as per the following endothermic reaction:
\(
\mathrm{CH}_4(\mathrm{~g})+\mathrm{H}_2 \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{CO}(\mathrm{g})+3 \mathrm{H}_2(\mathrm{~g})
\)
(a) Write an expression for \(K_p\) for the above reaction.
(b) How will the values of \(K_p\) and the composition of the equilibrium mixture be affected by
(i) Increasing the pressure
(ii) increasing the temperature
(iii) using a catalyst?

Answer: The expression for \(K_p\) for the reaction is:
\(
K_p=\frac{(p \mathrm{CO}) \times\left(p \mathrm{H}_2\right)^3}{\left(p \mathrm{CH}_4\right) \times\left(p \mathrm{H}_2 \mathrm{O}\right)}
\)
(i) By increasing the pressure, the number of moles per unit volume will increase. In order to decrease the same, the equilibrium gets shifted to the left or in the backward direction. As a result, more of reactants will be formed and the value of \(K_p\) will decrease.
(ii) If the temperature is increased, according to Le Chatelier’s principle, the forward reaction will be favoured as it is endothermic. Therefore, the equilibrium gets shifted to the right and the value of \(K_p\) will increase.
(iii) The addition of catalyst will not change the equilibrium since it influences both the forward and the backward reactions to the same extent. But it will be attained more quickly.

Q29. Describe the effect of :
a) addition of \(\mathrm{H}_2\)
b) addition of \(\mathrm{CH}_3 \mathrm{OH}\)
c) removal of \(\mathrm{CO}\)
d) removal of \(\mathrm{CH}_3 \mathrm{OH}\)
on the equilibrium of the reaction
\(
2 \mathrm{H}_2(\mathrm{~g})+\mathrm{CO}(\mathrm{g}) \rightleftharpoons \mathrm{CH}_3 \mathrm{OH}(\mathrm{g})
\)

Answer: (i) Equilibrium will be shifted in the forward direction.
(ii) Equilibrium will be shifted in the backward direction.
(iii) Equilibrium will be shifted in the backward direction.
(iv) Equilibrium will be shifted in the forward direction.

Q30. At \(473 \mathrm{~K}\), equilibrium constant \(K_c\) for decomposition of phosphorus pentachloride, \(\mathrm{PCl}_5\) is \(8.3 \times 10^{-3}\). If decomposition is depicted as,
\(
\mathrm{PCl}_5(\mathrm{~g}) \rightleftharpoons \mathrm{PCl}_3(\mathrm{~g})+\mathrm{Cl}_2(\mathrm{~g}) \quad \Delta_{\mathrm{r}} H^{\ominus}=124.0 \mathrm{~kJ} \mathrm{~mol}{ }^{-1}
\)
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 \(K_{\mathrm{c}}\) if
(i) more \(\mathrm{PCl}_5\) is added
(ii) pressure is increased
(iii) the temperature is increased?

Answer: (a) The expression for
\(
K_c=\frac{\left[\mathrm{PCl}_3(\mathrm{~g})\right]\left[\mathrm{Cl}_2(\mathrm{~g})\right]}{\left[\mathrm{PCl}_5(\mathrm{~g})\right]}
\)
(b) \(
\text { For reverse reaction }\left(K_c^{\prime}\right)=\frac{\mathrm{PCl}_5(\mathrm{~g})}{\left[\mathrm{PCl}_3(\mathrm{~g})\right]\left[\mathrm{Cl}_2(\mathrm{~g})\right]}=\frac{1}{8.3 \times 10^{-3}}=120.48
\)
(c) (i) By adding more of \(\mathrm{PCl}_5\), the value of \(\mathrm{K}_{\mathrm{c}}\) will remain constant because there is no change in temperature.
(ii) \(\mathrm{K}_{\mathrm{c}}\) is constant at a constant temperature. Thus, in this case, \(\mathrm{K}_{\mathrm{c}}\) would not change.
(iii) By increasing the temperature, the forward reaction will be favoured since it is endothermic in nature. Therefore, the value of the equilibrium constant will increase.

Q31. 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 \(\mathrm{CO}\) and \(\mathrm{H}_2\). In the second stage, CO formed in the first stage is reacted with more steam in the water gas shift reaction,
\(
\mathrm{CO}(\mathrm{g})+\mathrm{H}_2 \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{CO}_2(\mathrm{~g})+\mathrm{H}_2(\mathrm{~g})
\)
If a reaction vessel at \(400^{\circ} \mathrm{C}\) is charged with an equimolar mixture of \(\mathrm{CO}\) and steam such that \(p_{\mathrm{co}}=p_{\mathrm{H}_2 \mathrm{O}}=4.0\) bar, what will be the partial pressure of \(\mathrm{H}_2\) at equilibrium? \(K_p=10.1\) at \(400^{\circ} \mathrm{C}\)

Answer: Let the partial pressure of hydrogen \(\left(\mathrm{H}_2\right)\) at equilibrium point \(=p\) bar
\(
\hspace { 9em} \mathrm{CO}(g)+\mathrm{H}_2 \mathrm{O}(g) \rightleftharpoons \mathrm{CO}_2(g)+\mathrm{H}_2(g)
\)
\(
\begin{array}{lcccc}
\text { Initial pressure: } & 4.0 \mathrm{~bar} & 4.0 \mathrm{~bar} & 0 & 0 \\
\text { Eqm. pressure: } & (4-p) \text { bar } & (4-p) \text { bar } & p \text { bar } & p \text { bar }
\end{array}
\)
\(
K_p=\frac{p_{\mathrm{CO}_2} \times p_{\mathrm{H}_2}}{p_{\mathrm{CO}} \times p_{\mathrm{H}_2 \mathrm{O}}} \quad \text { or } \quad 0.1=\frac{(p \text { bar }) \times(p \text { bar })}{(4-p) \text { bar } \times(4-p) \text { bar }}
\)
\(
\frac{p^2}{(4-p)^2}=0.1 \text { or } \frac{p}{(4-p)}=(0.1)^{1 / 2}=0.316
\)
\(
p=0.316(4-p) \quad \text { or } \quad p=1.264-0.316 p
\)
\(
1.316 p=1.264 \text { or } p=\frac{1.264}{1.316}=0.96 \mathrm{bar}
\)

Q32. Predict which of the following reaction will have an appreciable concentration of reactants and products:
a) \(\mathrm{Cl}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{Cl}(\mathrm{g}) \quad K_c=5 \times 10^{-39}\)
b) \(\mathrm{Cl}_2(\mathrm{~g})+2 \mathrm{NO}(\mathrm{g}) \rightleftharpoons 2 \mathrm{NOCl}(\mathrm{g}) \quad K_{\mathrm{c}}=3.7 \times 10^8\)
c) \(\mathrm{Cl}_2(\mathrm{~g})+2 \mathrm{NO}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}_2 \mathrm{Cl}(\mathrm{g}) \quad K_c=1.8\)

Answer: The following conclusions can be drawn from the values of \({K_c}\).
(a) Since the value of \(\mathrm{K}_c\) is very small, this means that the molar concentration of the products is very small as compared to that of the reactants.
(b) Since the value of \(\mathrm{K}_c\) is quite large, this means that the molar concentration of the products is very large as compared to that of the reactants.
(c) Since the value of \(\mathrm{K}_c\) is 1.8, this means that both the products and reactants have appreciable concentration.

Q33. The value of \(K_c\) for the reaction \(3 \mathrm{O}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{O}_3(\mathrm{~g})\) is \(2.0 \times 10^{-50}\) at \(25^{\circ} \mathrm{C}\). If the equilibrium concentration of \(\mathrm{O}_2\) in alr at \(25^{\circ} \mathrm{C}\) is \(1.6 \times 10^{-2}\), what is the concentration of \(\mathrm{O}_3\)?

Answer: \(
3 \mathrm{O}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{O}_3(\mathrm{~g})
\)
\(
K_c=\frac{\left[\mathrm{O}_3\right]^2}{\left[\mathrm{O}_2\right]^3} \quad \text { or } \quad\left(2.0 \times 10^{-50}\right)=\frac{\left[\mathrm{O}_3\right]^2}{\left(1.6 \times 10^{-2}\right)^3}
\)
\(
\left[\mathrm{O}_3\right]^2=\left(2.0 \times 10^{-50}\right) \times\left(1.6 \times 10^{-2}\right)^3
\)
\(
\left[\mathrm{O}_3\right]^2=8.192 \times 10^{-56} \text { or }\left[\mathrm{O}_3\right]=\left(8.192 \times 10^{-56}\right)^{1 / 2}=2.86 \times 10^{-28} \mathrm{M}
\)

Q34. The reaction, \(\mathrm{CO}(\mathrm{g})+3 \mathrm{H}_2(\mathrm{~g}) \rightleftharpoons \mathrm{CH}_4(\mathrm{~g})+\mathrm{H}_2 \mathrm{O}(\mathrm{g})\)
is at equilibrium at \(1300 \mathrm{~K}\) in a \(1 \mathrm{~L}\) flask. It also contain \(0.30 \mathrm{~mol}\) of \(\mathrm{CO}, 0.10 \mathrm{~mol}\) of \(\mathrm{H}_2\) and \(0.02 \mathrm{~mol}\) of \(\mathrm{H}_2 \mathrm{O}\) and an unknown amount of \(\mathrm{CH}_4\) in the flask. Determine the concentration of \(\mathrm{CH}_4\) in the mixture. The equilibrium constant, \(K_c\) for the reaction at the given temperature is 3.90.

Answer:

\(
\mathrm{CO}(g)+3 \mathrm{H}_2(g) \rightleftharpoons \mathrm{CH}_4(g)+\mathrm{H}_2 \mathrm{O}(g)
\)
According to available data,
\(
\mathrm{K}_c=\frac{\left[\mathrm{CH}_4\right] \times\left[\mathrm{H}_2 \mathrm{O}\right]}{[\mathrm{CO}] \times\left[\mathrm{H}_2\right]^3} \text { or } 3.90=\frac{\left[\mathrm{CH}_4\right][0 \cdot 02]}{[0 \cdot 30] \times[0 \cdot 1]^3}
\)
\(
\left[\mathrm{CH}_4\right]=\frac{(3.9) \times(0.30) \times(0.001)}{(0.02)}=\mathbf{5 . 8 5} \times \mathbf{1 0 ^ { – 2 }} \mathbf~{M}
\)

Q35. What is meant by the conjugate acid-base pair? Find the conjugate acid/base for the following species:
\(
\mathrm{HNO}_2, \mathrm{CN}^{-}, \mathrm{HClO}_4 \mathrm{~F}^{-}, \mathrm{OH}^{-}, \mathrm{CO}_3^{2-} \text {, and } \mathrm{S}^{2-}
\)

Answer: A conjugate acid-base pair is a pair that differs only by one proton.
The conjugate acid base for the given species is mentioned in the table below.
\(
\begin{array}{|c|c|}
\hline \text { Species } & \text { Conjugate acid-base } \\
\hline \mathrm{HNO}_2 & \mathrm{NO}_2^{-} \text {(base) } \\
\hline \mathrm{CN}^{-} & \mathrm{HCN} \text { (acid) } \\
\hline \mathrm{HCIO}_4 & \mathrm{CIO}_4^{-} \text {(base) } \\
\hline \mathrm{F}^{-} & \text { HF(acid) } \\
\hline \mathrm{OH}^{-} & \mathrm{H}_2 \mathrm{O}\left(\text { acid) } / \mathrm{O}^{2-}\right. \text { (base) } \\
\hline \mathrm{CO}_3^{2-} & \mathrm{HCO}_3^{-} \text {(acid) } \\
\hline \mathrm{S}^{2-} & \mathrm{HS}^{-} \text {(acid) } \\
\hline
\end{array}
\)

Q36. Which of the following are Lewis acids? \(\mathrm{H}_2 \mathrm{O}, \mathrm{BF}_3, \mathrm{H}^{+}\), and \(\mathrm{NH}_4^{+}\)

Answer: Lewis acids are those acids that can accept a pair of electrons. For example, \(\mathrm{BF}_3, \mathrm{H}^{+}\), and \(\mathrm{NH}_4^{+}\)are Lewis acids.

Q37. What will be the conjugate bases for the Brönsted acids: \(\mathrm{HF}, \mathrm{H}_2 \mathrm{SO}_4\) and \(\mathrm{HCO}_3^{-}\)?

Answer: The table below lists the conjugate bases for the given Bronsted acids.
Bronsted acid Conjugate base
\(
\begin{array}{ll}
\mathrm{HF} & & & \mathrm{F}^{-} \\
\mathrm{H}_2 \mathrm{SO}_4 & & & \mathrm{HSO}_4^{-} \\
\mathrm{HCO}_3^{-} & & & \mathrm{CO}_3^{2-}
\end{array}
\)

Q38. Write the conjugate acids for the following Brönsted bases: \(\mathrm{NH}_2^{-}, \mathrm{NH}_3\) and \(\mathrm{HCOO}^{-}\).

Answer: The table below lists the conjugate acids for the given Bronsted bases.

Bronsted base Conjugate acid
\(
\begin{array}{ll}
\mathrm{NH}_2^{-} & & & \mathrm{NH}_3 \\
\mathrm{NH}_3 & & & \mathrm{NH}_4^{+} \\
\mathrm{HCOO}^{-} & & & \mathrm{HCOOH}
\end{array}
\)

Q39. The species: \(\mathrm{H}_2 \mathrm{O}, \mathrm{HCO}_3^{-}, \mathrm{HSO}_4^{-}\)and \(\mathrm{NH}_3\) can act both as Brönsted acids and bases. For each case give the corresponding conjugate acid and base.

Answer: 

Q40. Classify the following species into Lewis acids and Lewis bases and show how these act as Lewis acid/base:
(a) \(\mathrm{OH}^{-}\)
(b) \(\mathrm{F}^{-}\)
(c) \(\mathrm{H}^{+}\)
(d) \(\mathrm{BCl}_3\)

Answer: (a) \(\mathrm{OH}^{-}\)ions can donate an electron pair and act as Lewis base.
(b) \(\mathrm{F}^{-}\)ions can donate an electron pair and act as Lewis base.
(c) \(\mathrm{H}^{+}\)ions can accept an electron pair and act as Lewis acid.
(d) \(\mathrm{BCl}_3\) can accept an electron pair since the Boron atom is electron deficient. It is a Lewis acid.

Q41. The concentration of hydrogen ion in a sample of soft drink is \(3.8 \times 10^{-3} \mathrm{M}\). what is its \(\mathrm{pH}\)?

Answer: 

\(
\begin{aligned}
& \mathrm{pH}=-\log \left[\mathrm{H}^{+}\right]=-\log \left(3.8 \times 10^{-3}\right) \\
& =(\log 3-\log 3.8)=3-0.5798=2.42
\end{aligned}
\)

Q42. The \(\mathrm{pH}\) of a sample of vinegar is 3.76. Calculate the concentration of hydrogen ion in it.

Answer: \(\mathrm{pH}=-\log \left[\mathrm{H}^{+}\right]\)or \(\log \left[\mathrm{H}^{+}\right]=-\mathrm{pH}=-3.76=4.24\)
\(\therefore\left[\mathrm{H}^{+}\right]=\)Antilog \(4.24=1.738 \times 10^{-4}=1.74 \times 10^{-4} \mathrm{M}\)

Q43. The Ionization constant of \(\mathrm{HF}, \mathrm{HCOOH}\) and \(\mathrm{HCN}\) at \(298 \mathrm{~K}\) are \(6.8 \times 10^{-4}\), \(1.8 \times 10^{-4}\) and \(4.8 \times 10^{-9}\) respectively. Calculate the 1onization constants of the corresponding conjugate base.

Answer: We know that : \(\mathrm{K}_a \times \mathrm{K}_b=\mathrm{K}_w\)
For HF: \(\mathrm{K}_a=6.8 \times 10^{-4} ; \mathrm{K}_b=\frac{\mathrm{K}_w}{\mathrm{~K}_a}=\frac{10^{-14}}{6.8 \times 10^{-4}}=1.5 \times 10^{-11}\)
For HCOOH: \(\mathrm{K}_a=1.8 \times 10^{-4} ; \mathrm{K}_b=\frac{\mathrm{K}_w}{\mathrm{~K}_a}=\frac{10^{-14}}{1.8 \times 10^{-4}}=\mathbf{5 . 6} \times 10^{-11}\)
For HCN: \(\mathrm{K}_a=4 \cdot 8 \times 10^{-9} ; \mathrm{K}_b=\frac{\mathrm{K}_w}{\mathrm{~K}_a}=\frac{10^{-14}}{4 \cdot 8 \times 10^{-9}}=2 \cdot 08 \times 10^{-6}\)

Q44. The ionization constant of phenol is \(1.0 \times 10^{-10}\). What is the concentration of phenolate ion in \(0.05 \mathrm{M}\) solution of phenol? What will be its degree of ionization if the solution is also \(0.01 \mathrm{M}\) in sodium phenolate?

Answer: Case I: Let \(\mathrm{C}\) moles of phenol be dissolved in water to form the solution and let \(\alpha\) be the degree of ionisation of phenol. The concentration of different species at the equilibrium point is.
\(
\hspace { 7em} \mathrm{PhOH}+\mathrm{H}_2 \mathrm{O} \rightleftharpoons \mathrm{PhO}^{-}+\mathrm{H}_3 \mathrm{O}^{+}
\)
\(
\begin{array}{lccc}
\text { Initial molar conc. } & \mathrm{C} & & 0 & & 0 \\
\text { Eqm. molar conc. } & \mathrm{C}(1-\alpha) & & \mathrm{C} \alpha & & \mathrm{C} \alpha
\end{array}
\)
\(
\begin{aligned}
\alpha & =\sqrt{\frac{\mathrm{K}_a}{\mathrm{C}}}=\sqrt{\frac{1.0 \times 10^{-10}}{0.05}}=\sqrt{\frac{1.0 \times 10^{-10}}{5.0 \times 10^{-2}}}=\sqrt{0.2 \times 10^{-8}} \\
& =0.447 \times 10^{-4}=4.47 \times 10^{-5}
\end{aligned}
\)
\(
\left[\mathrm{PhO}^{-}\right]=\mathrm{C} \alpha=(0.05 \mathrm{M}) \times 4.47 \times 10^{-5}=\mathbf{2 . 2 } \times 10^{-6} \mathbf{M}
\)
\(
\begin{aligned}
p \mathrm{H} & =-\log \left[\mathrm{H}_3 \mathrm{O}^{+}\right]=-\log \left(2.2 \times 10^{-6}\right) \\
& =(6-\log 2.2)=6-0 \cdot 3483=\mathbf{5 . 6 5}
\end{aligned}
\)
Case II: When phenol \((\mathrm{PhOH})\) is mixed with \(0.01 \mathrm{M}\) sodium phenate solution, the ionisation may be represented as :
\(
\mathrm{PhOH} \rightleftharpoons \mathrm{PhO}^{-}+\mathrm{H}^{+} ; \mathrm{PhONa} \longrightarrow \mathrm{PhO}^{-}+\mathrm{Na}^{+}
\)
Conc. of \(\mathrm{PhO}^{-}\)ions due to ionisation of sodium phenate (completely ionised) \(=0.01 \mathrm{M}\) Let the concentration of \(\mathrm{PhO}^{-}\)ions from \(\mathrm{PhOH}=x \mathrm{M}\)
\(
\therefore \text { Total concentration of } \mathrm{PhO}^{-} \text {ions i.e., }\left[\mathrm{PhO}^{-}\right]=0.01+x \approx 0.01 \mathrm{M}
\)
( \(x\) being very small can be neglected)
Concentration of \(\mathrm{PhOH}\) left unionised \(=0.05-x \approx 0.05 \mathrm{M}\)
\(
\text { Ionisation constant }\left(\mathrm{K}_a\right) \text { for } \mathrm{PhOH}=\frac{\left[\mathrm{PhO}^{-}\right]\left[\mathrm{H}^{+}\right]}{[\mathrm{PhOH}]}
\)
\(
1 \times 10^{-10}=\frac{(0 \cdot 01 \mathrm{M}) \times\left[\mathrm{H}^{+}\right]}{(0 \cdot 05 \mathrm{M})}
\)
\(
\left[\mathrm{H}^{+}\right]=\frac{1 \times 10^{-10} \times 0.05}{0.01}=5.0 \times 10^{-10}
\)
Degree of ionization of phenol i.e., \(\alpha=\frac{5 \cdot 0 \times 10^{-10}}{5 \cdot 0 \times 10^{-2}}=10^{-8}\)

Q45. The first ionization constant of \(\mathrm{H}_2 \mathrm{S}\) is \(9.1 \times 10^{-8}\). Calculate the concentration of \(\mathrm{HS}^{-}\) ion in its \(0.1 \mathrm{M}\) solution. How will this concentration be affected if the solution 1s \(0.1 \mathrm{M}\) in \(\mathrm{HCl}\) also ? If the second dissociation constant of \(\mathrm{H}_2 \mathrm{S}\) is \(1.2 \times 10^{-13}\), calculate the concentration of \(\mathrm{S}^{2-}\) under both conditions.

Answer: Case I: Calculation of \(\left[\mathrm{HS}^{-}\right]\)in \(0.1 \mathrm{M} \mathrm{~H}_2 \mathrm{~S}\) solution
Let the degree of dissociation of \(\mathrm{H}_2 \mathrm{S}=\alpha\)
\(
\hspace { 9em} \mathrm{H}_2 \mathrm{S} \rightleftharpoons \mathrm{H}^{+}+\mathrm{HS}^{-}
\)
\(
\begin{array}{lccc}
\text { Initial molar conc. } & 0.1 \mathrm{M} & 0 & & 0 \\
\text { Eqm. molar conc. } & 0.1-\alpha & \alpha & & \alpha
\end{array}
\)
According to the Ostwald dilution formula,
\(
\mathrm{K}_{a_1}=\frac{\left[\mathrm{H}^{+}\right]\left[\mathrm{HS}^{-}\right]}{\left[\mathrm{H}_2 \mathrm{~S}\right]}=\frac{[\alpha][\alpha]}{[0 \cdot 1-\alpha]}=\frac{[\alpha][\alpha]}{0 \cdot 1}
\)
\(
9.1 \times 10^{-8}=\frac{[\alpha][\alpha]}{0 \cdot 1} \text { or }[\alpha]^2=9.1 \times 10^{-9}=91 \times 10^{-10}
\)
\(
\alpha=\left(91 \times 10^{-10}\right)^{1 / 2}=9.54 \times 10^{-5} \text { or }\left[\mathrm{HS}^{-}\right]=\mathbf{9 . 5 4} \times 10^{-5}
\)
Case II:
\(
\text { Calculation of }\left[\mathrm{HS}^{-}\right] \text {in } 0.1 \mathrm{M} \mathrm{~HCl} \text { solution }
\)
When \(\mathrm{H}_2 \mathrm{~S}\) solution is mixed with \(0 \cdot 1 \mathrm{~M} \mathrm{~HCl}\) solution, the dissociation may be represented as
\(
\mathrm{H}_2 \mathrm{S} \rightleftharpoons \mathrm{H}^{+}+\mathrm{HS}^{-} ; \mathrm{HCl} \longrightarrow \mathrm{H}^{+}+\mathrm{Cl}^{-}
\)
\(\left[\mathrm{H}^{+}\right]\)due to dissociation of \(\mathrm{HCl}\) (strong acid) \(=0.1 \mathrm{M}\)
Let \(\left[\mathrm{H}^{+}\right]\)due to dissociation of \(\mathrm{H}_2 \mathrm{S}\) (weak acid) \(=x \mathrm{M}\)
Total concentration of \(\mathrm{H}^{+}\)ions i.e., \(\left[\mathrm{H}^{+}\right]=0.1+x \approx 0.1 \mathrm{M}\)
( \(x\) being very small can be neglected)
\(\left[\mathrm{HS}^{-}\right]\)in solution \(=x \mathrm{M}\)
Concentration of \(\mathrm{H}_2 \mathrm{~S}\) left undissociated i.e., \(\left[\mathrm{H}_2 \mathrm{~S}\right]=0.1-x \approx 0.1 \mathrm{M}\)
\(
\begin{aligned}
\mathrm{K}_{a_1} & =\frac{\left[\mathrm{H}^{+}\right]\left[\mathrm{HS}^{-}\right]}{\left[\mathrm{H}_2 \mathrm{S}\right]} \text { or } 9 \cdot 1 \times 10^{-8}=\frac{(0 \cdot 1) \times(x)}{(0 \cdot 1)} \\
x & =9 \cdot 1 \times 10^{-8} \text { or }\left[\mathrm{HS}^{-}\right]=\mathbf{9 . 1} \times \mathbf{1 0}^{-8}
\end{aligned}
\)
Case-III: Calculation of \(\left[\mathrm{S}^{2-}\right]\) in the absence of \(\mathrm{HCl}\)
\(
\begin{aligned}
& \mathrm{H}_2 \mathrm{~S} \rightleftharpoons \mathrm{H}^{+}+\mathrm{HS}^{-} ; \mathrm{K}_{a_1}=9 \cdot 1 \times 10^{-8} \\
& \mathrm{HS}^{-} \rightleftharpoons \mathrm{H}^{+}+\mathrm{S}^{2-} ; \mathrm{K}_{a_2}=1.2 \times 10^{-13}
\end{aligned}
\)
In order to calculate \(\mathrm{K}_a\) for the overall reaction, add the two equations.
\(
\mathrm{K}_a=\mathrm{K}_{a_1} \times \mathrm{K}_{a_2}=\left(9.1 \times 10^{-8}\right) \times\left(1.2 \times 10^{-13}\right)=1.092 \times 10^{-20}
\)
\(
\mathrm{H}_2 \mathrm{S} \rightleftharpoons 2 \mathrm{H}^{+}+\mathrm{S}^{2-}
\)
\(
0 \cdot 1-y \approx 0 \cdot 1 \quad 2 y \quad y
\)
\(
\mathrm{K}_a=\frac{\left[\mathrm{H}^{+}\right]\left[\mathrm{S}^{2-}\right]}{\left[\mathrm{H}_2 \mathrm{~S}\right]} \text { or } 1.092 \times 10^{-20}=\frac{4 y^2 \times y}{0 \cdot 1}
\)
\(
4 y^3=1.092 \times 10^{-21} \text { or } y=\left(\frac{1.092 \times 10^{-21}}{4}\right)^{\frac{1}{3}}=\left(273 \times 10^{-24}\right)^{1 / 3}
\)
\(
y=6.49 \times 10^{-8} \mathrm{~M} \approx \mathbf{6} \cdot \mathbf{5} \times 10^{-8} \mathbf{~M}
\)
Case-IV:
\(
\text { Calculation of }\left[\mathrm{S}^{2-}\right] \text { in the presence of } 0 \cdot 1 \mathrm{M} \mathrm{~HCl}
\)
Let \(\left[\mathrm{S}^{2-}\right]\) due to dissociation of \(\mathrm{H}_2 \mathrm{~S}=\mathrm{Z} \mathrm{M}\)
Dissociation of \(\mathrm{H}_2 \mathrm{S}\) may be represented as :
\(
\mathrm{H}_2 \mathrm{~S} \rightleftharpoons 2 \mathrm{H}^{+}+\mathrm{S}^{2-}
\)
\(
0 \cdot 1-Z \approx 0.1 \quad 2 Z \quad Z
\)
Total concentration of \(\mathrm{H}^{+}\)ions i.e., \(\left[\mathrm{H}^{+}\right]=0.1+2 \mathrm{Z} \approx 0.1 \mathrm{M}\) \(\left[\mathrm{S}^{2-}\right]\) in solution \(=\mathrm{Z}\)
\(
\mathrm{K}_a=\frac{\left[\mathrm{H}^{+}\right]^2\left[\mathrm{~S}^{2-}\right]}{\left[\mathrm{H}_2 \mathrm{S}\right]} \text { or } 1.092 \times 10^{-20}=\frac{(0.1)^2 \times \mathrm{Z}}{0.1}
\)
\(
Z=\frac{1.092 \times 10^{-20} \times 0.1}{(0.1)^2}=\mathbf{1 . 0 9 2} \times 10^{-19} \mathbf{~M}
\)

Q46. The 1onization constant of acetic acid is \(1.74 \times 10^{-5}\). Calculate the degree of dissociation of acetic acid in its \(0.05 \mathrm{~M}\) solution. Calculate the concentration of acetate ion in the solution and its \(\mathrm{pH}\).

Answer: \(\mathrm{CH}_3 \mathrm{COOH} \rightleftharpoons \mathrm{CH}_3 \mathrm{COO}^{-}+\mathrm{H}^{+}\)
\(
K_a=\frac{\left[\mathrm{CH}_3 \mathrm{COO}^{-}\right]\left[\mathrm{H}^{+}\right]}{\left[\mathrm{CH}_3 \mathrm{COOH}\right]}=\frac{\left[\mathrm{H}^{+}\right]^2}{\left[\mathrm{CH}_3 \mathrm{COOH}\right]}
\)
\(
\left[\mathrm{H}^{+}\right]=\sqrt{K_a\left[\mathrm{CH}_3 \mathrm{COOH}\right]}=\sqrt{\left(1.74 \times 10^{-5}\right)\left(5 \times 10^{-2}\right)}=9.33 \times 10^{-4} \mathrm{M}
\)
\(
\left[\mathrm{CH}_3 \mathrm{COO}^{-}\right]=\left[\mathrm{H}^{+}\right]=\mathbf{9 . 3 3} \times 10^{-4} \mathbf{~ M}
\)
\(
\mathrm{pH}=-\log \left(9.33 \times 10^{-4}\right)=4-0.9699=4-0.97=3.03
\)

Q47. It has been found that the \(\mathrm{pH}\) of a \(0.01 \mathrm{M}\) solution of an organic acid is 4.15. Calculate the concentration of the anion, the ionization constant of the acid and its \(\mathrm{p} K_{\mathrm{a}}\).

Answer: \(p \mathrm{H}=-\log \left[\mathrm{H}^{+}\right] \text {or } \log \left[\mathrm{H}^{+}\right]=-p \mathrm{H}=-4 \cdot 15=5.85\)
\(
\left[\mathrm{H}^{+}\right]=\text {Antilog }(5.85)=7.08 \times 10^{-5} \mathrm{~M}
\)
\(
\left[\mathrm{A}^{-}\right]=\left[\mathrm{H}^{+}\right]=7.08 \times 10^{-5} \mathrm{~M}
\)
\(
\left[\mathrm{K}_a\right]=\frac{\left[\mathrm{H}^{+}\right]\left[\mathrm{A}^{-}\right]}{[\mathrm{HA}]}=\frac{\left(7 \cdot 08 \times 10^{-5}\right) \times\left(7 \cdot 08 \times 10^{-5}\right)}{(0 \cdot 01)}=\mathbf{5 \cdot 0} \times \mathbf{1 0}^{-\mathbf{7}}
\)
\(
p \mathrm{~K}_a=-\log \mathrm{K}_a=-\log \left(5 \cdot 0 \times 10^{-7}\right)=(7-\log 5)=(7-0 \cdot 699)=\mathbf{6.30}
\)

Q48. Assuming complete dissociation, calculate the \(\mathrm{pH}\) of the following solutions:
(a) \(0.003 \mathrm{~M} \mathrm{~HCl}\) (b) \(0.005 \mathrm{~M} \mathrm{~NaOH}\) (c) \(0.002 \mathrm{~M} \mathrm{~HBr}\) (d) \(0.002 \mathrm{~M} \mathrm{~KOH}\)

Answer: (a) pH value of \(0.003 \mathrm{~M} \mathrm{~HCl}\) solution.
\(
\mathrm{HCl}(l)+\mathrm{H}_2 \mathrm{O}(l) \stackrel{(a q)}{\longrightarrow} \mathrm{H}_3 \mathrm{O}^{+}+\mathrm{Cl}^{-}(a q)
\)
\(
p \mathrm{H}=-\log \left[\mathrm{H}_3 \mathrm{O}^{+}\right]=-\log \left(3 \times 10^{-3}\right)=2.52
\)
(b) pH value of \(0.005 \mathrm{~M} \mathrm{~NaOH}\)
\(
\mathrm{NaOH}(s) \stackrel{(a q)}{\longrightarrow} \mathrm{Na}^{+}(a q)+\mathrm{OH}^{-}(a q)
\)
\(
\left[\mathrm{OH}^{-}\right]=0.005 \mathrm{M}=5 \times 10^{-3} \mathrm{~M}
\)
\(
\left[\mathrm{H}_3 \mathrm{O}^{+}\right]=\frac{\mathrm{K}_w}{\left[\mathrm{OH}^{-}\right]}=\frac{\left(10^{-14} \mathrm{M}^2\right)}{\left(5 \times 10^{-3} \mathrm{M}\right)}=2 \times 10^{-12} \mathrm{M}
\)
\(
p \mathrm{H}=-\log \left[\mathrm{H}_3 \mathrm{O}^{+}\right]=-\left(\log 2 \times 10^{-12}\right)=11.7
\)
(c) \(\mathrm{pH} \text { value of } 0.002 \mathrm{~M} \mathrm{~HBr}\)
\(
\mathrm{HBr}(l)+\mathrm{H}_2 \mathrm{O}(l) \stackrel{(a q)}{\longrightarrow} \mathrm{H}_3 \mathrm{O}^{+}(a q)+\mathrm{Br}^{-}(a q)
\)
\(
p \mathrm{H}=-\log \left[\mathrm{H}_3 \mathrm{O}^{+}\right]=-\log \left(2 \times 10^{-3}\right)=(3-\log 2)=3-0 \cdot 3010=\mathbf{2 \cdot 7}
\)
(d) \(\text { pH value of } 0.002 \mathrm{~M} \mathrm{~KOH}\)
\(
\mathrm{KOH}(s) \stackrel{(a q)}{\longrightarrow} \mathrm{K}^{+}(a q)+\mathrm{OH}^{-}(a q)
\)
\(
\left[\mathrm{OH}^{-}\right]=0.002 \mathrm{M}=2 \times 10^{-3} \mathrm{M}
\)
\(
\left[\mathrm{H}_3 \mathrm{O}^{+}\right]=\frac{\mathrm{K}_w}{\left[\mathrm{OH}^{-}\right]}=\frac{\left(10^{-14} \mathrm{M}^2\right)}{\left(2 \times 10^{-3} \mathrm{M}\right)}=5 \times 10^{-12} \mathrm{M}
\)
\(
\begin{aligned}
p \mathrm{H} & =-\log \left[\mathrm{H}_3 \mathrm{O}^{+}\right]=-\log \left(5 \times 10^{-12}\right) \\
& =(12-\log 5)=(12-0 \cdot 6989)=\mathbf{1 1} \cdot \mathbf{3 0}
\end{aligned}
\)

Q49. Calculate the \(\mathrm{pH}\) of the following solutions:
a) \(2 \mathrm{~g}\) of \(\mathrm{TlOH}\) dissolved in water to give 2 litre of solution.
b) \(\quad 0.3 \mathrm{~g}\) of \(\mathrm{Ca}(\mathrm{OH})_2\) dissolved in water to give \(500 \mathrm{~mL}\) of solution.
c) \(\quad 0.3 \mathrm{~g}\) of \(\mathrm{NaOH}\) dissolved in water to give \(200 \mathrm{~mL}\) of solution.
d) \(1 \mathrm{~mL}\) of \(13.6 \mathrm{M} \mathrm{HCl}\) is diluted with water to give 1 litre of solution.

Answer: (a) \(p H\) of \(2 \mathrm{~g}\) of TlOH in 2 litre of solution
\(
\text { Molarity of TlOH solution }=\frac{\text { Mass of } \mathrm{TlOH} / \text { Molar mass }}{\text { Volume of solution in litres }}=\frac{(2 \mathrm{~g}) /\left(204+17 \mathrm{~g} \mathrm{~mol}^{-1}\right)}{2 \mathrm{~L}}
\)
\(
=4.525 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1}=4.525 \times 10^{-3} \mathrm{M}
\)
\(
\mathrm{TlOH}(s) \stackrel{(a q)}{\longrightarrow} \mathrm{Tl}^{+}(a q)+\mathrm{OH}^{-}(a q)
\)
\(
\left[\mathrm{H}_3 \mathrm{O}^{+}\right]=\frac{\mathrm{K}_w}{\left[\mathrm{OH}^{-}\right]}=\frac{\left(10^{-14} \mathrm{M}^2\right)}{\left(4.525 \times 10^{-3} \mathrm{M}\right)}=2.21 \times 10^{-12} \mathrm{M}
\)
\(
p \mathrm{H}=-\log \left[\mathrm{H}_3 \mathrm{O}^{+}\right]=-\log \left(2 \cdot 21 \times 10^{-12}\right)=11.65
\)
(b) \(\mathrm{pH} \text { of } 0.3 \mathrm{~g} \text { of } \mathrm{Ca}(\mathrm{OH})_2 \text { in } 500 \mathrm{~mL} \text { of solution }\)
\(
\text { Molarity of } \mathrm{Ca}(\mathrm{OH})_2=\frac{\text { Mass of } \mathrm{Ca}(\mathrm{OH})_2 / \text { Molar mass }}{\text { Volume of solution in litres }}=\frac{(0.3 \mathrm{~g}) /\left(74 \mathrm{~g} \mathrm{~mol}^{-1}\right)}{0.5 \mathrm{~L}}
\)
\(
=8.1 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1}=8.1 \times 10^{-3} \mathrm{M}
\)
\(
\mathrm{Ca}(\mathrm{OH})_2 \stackrel{(a q)}{\longrightarrow} \mathrm{Ca}^{2+}(a q)+2 \mathrm{OH}^{-}(a q)
\)
\(
\quad \quad \quad \quad \quad  8.1 \times 10^{-3} \mathrm{~M}  \quad \quad 16.2 \times 10^{-3} \mathrm{~M}
\)
\(
\begin{aligned}
p \mathrm{OH} & =-\log \left[\mathrm{OH}^{-}\right]=-\log \left(16 \cdot 2 \times 10^{-3}\right) \\
& =-(\log 16 \cdot 2-3 \log 10)=(3-\log 16 \cdot 2) \\
& =(3-1 \cdot 209)=1 \cdot 791 \\
p \mathrm{H} & =14-p \mathrm{OH}=14-1 \cdot 791=\mathbf{1 2 \cdot 2 1}
\end{aligned}
\)
(c) \(\mathrm{pH} \text { of } 0.3 \mathrm{~g} \text { of } \mathrm{NaOH} \text { in } 200 \mathrm{~mL} \text { of solution }\)
\(
\text { Molarity of } \mathrm{NaOH} \text { solution }=\frac{\text { Mass of } \mathrm{NaOH} / \text { Molar mass }}{\text { Volume of solution in litres }}=\frac{(0.3 \mathrm{~g}) /\left(40 \mathrm{~g} \mathrm{~mol}^{-1}\right)}{0 \cdot 2 \mathrm{~L}}
\)
\(
=0.0375 \mathrm{~mol} \mathrm{~L}^{-1}=0.0375 \mathrm{~M}
\)
\(
\mathrm{NaOH} \stackrel{(a q)}{\longrightarrow} \mathrm{Na}^{+}(a q)+\mathrm{OH}^{-}(a q)
\)
\(
\quad \quad \quad \quad \quad 0.0375 \mathrm{M} \quad 0.0375 \mathrm{M}
\)
\(
\begin{aligned}
p \mathrm{OH} & =-\log \left[\mathrm{OH}^{-}\right]=-\log \left(3.75 \times 10^{-2}\right) \\
& =-(\log 3.75-2 \log 10)=(2-\log 3.75)=(2-0.574)=1.426
\end{aligned}
\)
\(
p \mathrm{H}=14-p \mathrm{OH}=14-1.426=\mathbf{1 2 . 5 7 4}
\)
(d) \(\mathrm{pH} \text { of } 1 \mathrm{~mL} \text { of } 13.6 \mathrm{~M} \mathrm{~HCl} \text { solution diluted to } 1 \mathrm{~L}\)
The molarity of diluted solution may be calculated as :
\(
\mathrm{M}_1 \mathrm{~V}_1 \equiv \mathrm{M}_2 \mathrm{~V}_2
\)
\(
\begin{aligned}
(13.6 \mathrm{M}) \times(1 \mathrm{~mL}) & =\mathrm{M}_2 \times(1000 \mathrm{~mL}) \text { or } \mathrm{M}_2=\frac{(13.6 \mathrm{M}) \times(1 \mathrm{~mL})}{(1000 \mathrm{~mL})} \\
& =1.36 \times 10^{-2} \mathrm{M}
\end{aligned}
\)
\(
\mathrm{HCl} \longrightarrow \mathrm{H}^{+}(a q)+\mathrm{Cl}^{-}(a q)
\)
\(
\quad \quad \quad 1.36 \times 10^{-2} \mathrm{~M} \quad \quad 1.36 \times 10^{-2} \mathrm{~M}
\)
\(
p \mathrm{H}=-\log \left[\mathrm{H}^{+}\right]=-\log \left(1.36 \times 10^{-2}\right)
\)
\(
\begin{aligned}
& =-(\log 1 \cdot 36-2 \log 10)=(2-\log 1.36) \\
& =(2-0.1335)=\mathbf{1 . 87}
\end{aligned}
\)

Q50. The degree of ionization of a \(0.1 \mathrm{~M}\) bromoacetic acid solution is 0.132 . Calculate the \(\mathrm{pH}\) of the solution and the \(p K_{\mathrm{a}}\) of bromoacetic acid.

Answer: \(\hspace { 3em} \mathrm{CH}_2(\mathrm{Br}) \mathrm{COOH} \rightleftharpoons \mathrm{CH}_2(\mathrm{Br}) \mathrm{COO}^{-}+\mathrm{H}^{+}\)
\(
\begin{array}{lccc}
\text { Initial conc. } & \mathrm{C} & & & & & & 0 & & & & & 0 \\
\text { Conc. at eqm. } & \mathrm{C}-\mathrm{C} \alpha & & & & & & \mathrm{C} \alpha & & & & & \mathrm{C} \alpha
\end{array}
\)
\(
K_a=\frac{C \alpha \cdot C \alpha}{C(1-\alpha)}=\frac{C \alpha^2}{1-\alpha} \simeq C \alpha^2=0.1 \times(0.132)^2=1.74 \times 10^{-3}
\)
\(
\begin{aligned}
p \mathrm{~K}_a & =-\log \left(1.74 \times 10^{-3}\right)=3-0.2405=2.76 \\
{\left[\mathrm{H}^{+}\right] } & =\mathrm{C} \alpha=0.1 \times 0.132=1.32 \times 10^{-2} \mathrm{M} \\
\mathrm{pH} & =-\log \left(1.32 \times 10^{-2}\right)=2-0.1206=1.88
\end{aligned}
\)

Q51. The pH of \(0.005 \mathrm{~M}\) codetne \(\left(\mathrm{C}_{18} \mathrm{H}_{21} \mathrm{NO}_3\right)\) solution 1s 9.95. Calculate 1ts ionization constant and \(\mathrm{p} K_{\mathrm{b}}\).

Answer: \(\mathrm{Cod}+\mathrm{H}_2 \mathrm{O} \rightleftharpoons \mathrm{Cod} \mathrm{H}^{+}+\mathrm{OH}^{-}\)
\(
\mathrm{pH}=9.95 \therefore \mathrm{pOH}=14-9.95=4.05, \text { i.e., }-\log \left[\mathrm{OH}^{-}\right]=4.05
\)
\(
\log \left[\mathrm{OH}^{-}\right]=-4.05=5.95 \text { or }\left[\mathrm{OH}^{-}\right]=8.913 \times 10^{-5} \mathrm{M}
\)
\(
\begin{aligned}
K_b & =\frac{\left[\mathrm{CodH}^{+}\right]\left[\mathrm{OH}^{-}\right]}{[\mathrm{Cod}]}=\frac{\left[\mathrm{OH}^{-}\right]^2}{[\mathrm{Cod}]}=\frac{\left(8.91 \times 10^{-5}\right)^2}{5 \times 10^{-3}}=\mathbf{1 . 6} \times 10^{-6} \\
\mathrm{pK}_b & =-\log \left(1.588 \times 10^{-6}\right)=6-0.1987=5.8
\end{aligned}
\)

Q52. What is the \(\mathrm{pH}\) of \(0.001 \mathrm{~M}\) aniline solution ? The ionization constant of aniline is \(4.27 \times 10^{-10}\). Calculate the degree of ionization of aniline in the solution. Also calculate the ionization constant of the conjugate acid of aniline.

Answer:

\(
\hspace { 7 em } \mathrm{C}_6 \mathrm{H}_5 \mathrm{NH}_2+\mathrm{H}_2 \mathrm{O} \rightleftharpoons \mathrm{C}_6 \mathrm{H}_5 \mathrm{NH}_3^{+}+\mathrm{OH}^{-}
\)
\(
\begin{array}{lccc}
\text { Initial conc. } & \mathrm{C} & & & & & & & 0 & & & 0 \\
\text { Eqm. conc. } & (\mathrm{C}-\mathrm{C} \alpha) & & & & & & &  \mathrm{C} \alpha & & & \mathrm{C} \alpha
\end{array}
\)
\(
\mathrm{K}_b=\frac{\mathrm{C} \alpha \cdot \mathrm{C} \alpha}{\mathrm{C}(1-\alpha)}=\frac{\mathrm{C} \alpha^2}{1-\alpha}=\mathrm{C} \alpha^2 \text { or } \alpha=\left(\mathrm{K}_b / \mathrm{C}\right)^{1 / 2}
\)
\(
\alpha=\left(4.27 \times 10^{-10} / 10^{-3}\right)^{1 / 2}=6.53 \times 10^{-4}
\)
\(
\begin{aligned}
& \text { Then, }\left[\text { anion] }=\mathrm{c} \alpha=.001 \times 65.34 \times 10^{-5}\right. \\
& =.065 \times 10^{-5} \\
& \text { pOH }=-\log \left(.065 \times 10^{-5}\right) \\
& =6.187 \\
& \text { pH }=7.813
\end{aligned}
\)
Now,
\(
\begin{aligned}
& K_a \times K_b=K_w \\
& \therefore 4.27 \times 10^{-10} \times K_a=K_w \\
& K_a=\frac{10^{-14}}{4.27 \times 10^{-10}} \\
& \quad=2.35 \times 10^{-5}
\end{aligned}
\)
Thus, the ionization constant of the conjugate acid of aniline is \(2.35 \times 10^{-5}\).

Q53. Calculate the degree of ionization of \(0.05 \mathrm{~M}\) acetic acid if its \(\mathrm{p} K_{\mathrm{a}}\) value is 4.74 . How is the degree of dissociation affected when its solution also contains (a) \(0.01 \mathrm{~M}\) (b) \(0.1 \mathrm{~M}\) in \(\mathrm{HCl}\)?

Answer: (i) Calculation of degree of ionization \((\alpha)\)
\(
\begin{aligned}
& \mathrm{c}=0.05 \mathrm{M} \\
& \mathrm{pK}_a=4.74 \\
& \mathrm{pK}_a=-\log \left(\mathrm{K}_a\right) \\
& \mathrm{K}_a=1.82 \times 10^{-5} \\
& \mathrm{~K}_a=\mathrm{C} \alpha^2 \quad \alpha=\sqrt{\frac{\mathrm{K}_a}{\mathrm{C}}} \\
& \alpha=\sqrt{\frac{1.82 \times 10^{-5}}{5 \times 10^{-2}}}=1.908 \times 10^{-2}
\end{aligned}
\)
In the presence of \(\mathrm{HCl}\), more of \(\mathrm{H}^{+}\)ions will be formed in solution. The equilibrium will be shifted to left. As a result, \(K_a\) for acetic acid will decrease.
(ii) Calculation of degree of dissociation ( \(\alpha\) ) in \(0.01 \mathrm{~M}\) solution
\(
\hspace { 9 em } \mathrm{CH}_3 \mathrm{COOH} \rightleftharpoons \mathrm{CH}_3 \mathrm{COO}^{-}+\mathrm{H}^{+}
\)
Initial conc. \(\quad \quad \quad \quad 0.05 \mathrm{~M} \hspace { 7 em } 0 \hspace { 3 em } 0.01\)
\(\begin{array}{llll}\text { Conc. after dissoc. } & 0.05-x \approx 0.05 & & x & & 0.01+x(\approx 0.01)\end{array}\)
\(
\begin{aligned}
\mathrm{K}_a & =\frac{\left[\mathrm{CH}_3 \mathrm{COO}^{-}\right]\left[\mathrm{H}^{+}\right]}{\left[\mathrm{CH}_3 \mathrm{COOH}\right]}=\frac{x(0.01)}{(0.05)} \text { or } x=\frac{\mathrm{K}_a \times(0.05)}{(0.01)} \\
x & =\frac{\left(1.82 \times 10^{-5}\right) \times(0.05)}{(0.01)}=\left(1.82 \times 10^{-5}\right) \times 5=9.1 \times 10^{-5}
\end{aligned}
\)
\(
\text { Degree of ionization }(\alpha)=\frac{x}{\mathrm{C}}=\frac{9 \cdot 1 \times 10^{-5}}{5 \cdot 0 \times 10^{-2}}=1.82 \times 10^{-3}=\mathbf{0 . 0 0 1 8}
\)
(iii) Calculation of degree of dissociation ( \(\alpha\) ) in \(0.1 M \mathrm{~HCl}\) solution
\(
\hspace { 9 em } \mathrm{CH}_3 \mathrm{COOH} \rightleftharpoons \mathrm{CH}_3 \mathrm{COO}^{-}+\mathrm{H}^{+}
\)
Initial conc. \(\quad \quad \quad \quad 0.05 \mathrm{~M} \hspace { 7 em } 0 \hspace { 3 em } 0.1\)
\(\begin{array}{llll}\text { Conc. after dissoc. } & 0.05-x \approx 0.05 & & x & & 0.1+x(\approx 0.1)\end{array}\)
\(
\begin{aligned}
\mathrm{K}_a & =\frac{\left[\mathrm{CH}_3 \mathrm{COO}^{-}\right]\left[\mathrm{H}^{+}\right]}{\left[\mathrm{CH}_3 \mathrm{COOH}\right]}=\frac{x(0.01)}{(0.05)} \\
x & =\frac{\left(1.82 \times 10^{-5}\right) \times(0.05)}{(0.1)}=9.1 \times 10^{-6}
\end{aligned}
\)
\(
\text { Degree of ionization }(\alpha)=\frac{x}{\mathrm{C}}=\frac{9.1 \times 10^{-6}}{5.0 \times 10^{-2}}=1.82 \times 10^{-4}=\mathbf{0 . 0 0 0 1 8}
\)

Q54. The ionization constant of dimethylamine is \(5.4 \times 10^{-4}\). Calculate its degree of 1onization in its \(0.02 \mathrm{~M}\) solution. What percentage of dimethylamine is ionized if the solution is also \(0.1 \mathrm{~M}\) in \(\mathrm{NaOH}\)?

Answer:

\(
\begin{aligned}
& K_a=5.4 \times 10^{-4} \\
& C=0.02 \mathrm{M}, K_a=C \alpha^2 \\
& \alpha^2=\frac{K_a}{C}=\frac{5.4 \times 10^{-4}}{0.02} \Rightarrow \alpha^2=270 \times 10^{-4} \\
& \alpha=0.164
\end{aligned}
\)
\(
\hspace { 7 em } \left(\mathrm{CH}_3\right)_2 \mathrm{NH}+\mathrm{H}_2 \mathrm{O} \rightleftharpoons\left(\mathrm{CH}_3\right)_2 \mathrm{NH}_2^{-}+\mathrm{OH}^{-}
\)
\(
\text { Initially: } \quad \quad \quad \quad C \quad \quad \quad \quad \quad \quad \quad \quad \quad 0 \quad \quad \quad \quad \quad 0
\)
\(
\text { At equili.: } \quad \quad \quad C(1-\alpha) \quad \quad \quad \quad \quad \quad C \alpha \quad \quad \quad C \alpha
\)
\(
\text { But }\left[\mathrm{OH}^{-}\right]=0.1 \mathrm{M}
\)
\(
K_n=\mathrm{C} \alpha \times\left[\mathrm{OH}^{-}\right] \Rightarrow 5.4 \times 10^{-4}=\frac{C \alpha \times 0.1}{C(1-\alpha)}
\)
But \(\alpha \ll<1\),
\(\therefore \quad \alpha\) is neglected in the denominator.
\(
\Rightarrow \quad \alpha=\frac{5.4 \times 10^{-4}}{0.1}=5.40 \times 10^{-3}=0.0054
\)
\(
\% \text { ionization }=100 \alpha=100 \times 0.0054=0.54
\)

Q55. Calculate the hydrogen ion concentration in the following biological fluids whose \(\mathrm{pH}\) are given below:
(a) Human muscle-fluid, 6.83
(b) Human stomach fluid, 1.2
(c) Human blood, 7.38
(d) Human saliva, 6.4.

Answer: (a)
\(
\begin{aligned}
& \mathrm{pH}=6.83 \\
& \therefore \quad\left[\mathrm{H}^{+}\right]=\operatorname{antilog}(-6.83)=1.48 \times 10^{-7} \mathrm{M}
\end{aligned}
\)
(b)
\(
\begin{aligned}
& \mathrm{pH}=1.2 \\
& \therefore \quad\left[\mathrm{H}^{+}\right]=\operatorname{antilog}(-1.2)=6.30 \times 10^{-2} \mathrm{M}
\end{aligned}
\)
(c)
\(
\begin{aligned}
& \mathrm{pH}=7.38 \\
& \therefore \quad\left[\mathrm{H}^{+}\right]=\operatorname{antilog}(-7.38)=4.17 \times 10^{-8} \mathrm{M}
\end{aligned}
\)
(d)
\(
\begin{aligned}
& \mathrm{pH}=6.4, \\
& \therefore \quad\left[\mathrm{H}^{+}\right]=\operatorname{antilog}(-6.4)=3.98 \times 10^{-7} \mathrm{M}
\end{aligned}
\)

Q56. The \(\mathrm{pH}\) 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.

Answer:

\(
\begin{aligned}
\text { For milk, } \mathrm{pH}=6.8 & \\
\therefore \quad\left[\mathrm{H}^{+}\right]=\operatorname{antilog}(-\mathrm{pH}) & =\operatorname{antilog}(-6.8) =1.58 \times 10^{-7} \mathrm{M}
\end{aligned}
\)
\(
\begin{aligned}
& \text { For black coffee, } \mathrm{pH}=5.0 \\
& \begin{aligned}
\therefore \quad\left[\mathrm{H}^{+}\right]=\operatorname{antilog}(-\mathrm{pH}) & =\operatorname{antilog}(-5.0)=1 \times 10^{-5} \mathrm{M}
\end{aligned}
\end{aligned}
\)
For tomato juice, \(\mathrm{pH}=4.2\)
\(
\begin{aligned}
\therefore \quad\left[\mathrm{H}^{\mathrm{i}}\right]=\operatorname{antilog}(-\mathrm{pH}) & =\operatorname{antilog}(-4.2)=6.31 \times 10^{-5} \mathrm{M}
\end{aligned}
\)
For lemon juice, \(\mathrm{pH}=2.2\),
\(
\begin{aligned}
\therefore \quad\left[\mathrm{H}^{+}\right]=\operatorname{antilog}(-\mathrm{pH}) & =\operatorname{antilog}(-2.2)=6.31 \times 10^{-3} \mathrm{M}
\end{aligned}
\)
For egg white, \(\mathrm{pH}=7.8\)
\(
\begin{aligned}
\therefore \quad\left[\mathrm{H}^{+}\right]=\operatorname{antilog}(-\mathrm{pH}) & =\operatorname{antilog}(-7.8)=1.58 \times 10^{-8} \mathrm{M}
\end{aligned}
\)

Q57. If \(0.561 \mathrm{~g}\) of \(\mathrm{KOH}\) is dissolved in water to give \(200 \mathrm{~mL}\) of solution at \(298 \mathrm{~K}\). Calculate the concentrations of potassium, hydrogen and hydroxyl ions. What is its pH?

Answer: Molar mass of \(\mathrm{KOH}=56.0 \mathrm{~g} \mathrm{~mol}^{-1}\)
\(\begin{aligned} \therefore \quad & \text { No. of moles of KOH } \\ & =\frac{0.561 \mathrm{~g}}{56.0 \mathrm{~g} \mathrm{~mol}^{-1}}=0.01 \mathrm{~mol}\end{aligned}\)
and conc. of \(\mathrm{KOH}\)
\(
=\frac{0.01}{0.2 \mathrm{~L}} \mathrm{~mol}=0.05 \mathrm{~mol} / \mathrm{L}
\)
\(
\mathrm{KOH}_{(a q)} \rightarrow \mathrm{K}_{(a q)}^{+}+\mathrm{OH}_{(a q)}^{-}
\)
\(
\left[\mathrm{K}^{+}\right]=\left[\mathrm{OH}^{-}\right]=0.05 \mathrm{M}
\)
\(
\text { and }\left[\mathrm{H}^{+}\right]=\frac{K_w}{\left[\mathrm{OH}^{-}\right]}=\frac{1.0 \times 10^{-14}}{0.05}=2.0 \times 10^{-13}
\)
\(
\therefore \quad \mathrm{pH}=-\log \left(2.0 \times 10^{-13}\right)=12.70
\)

Q58. The solubility of \(\mathrm{Sr}(\mathrm{OH})_2\) at \(298 \mathrm{~K}\) is \(19.23 \mathrm{~g} / \mathrm{L}\) of solution. Calculate the concentrations of strontium and hydroxyl tons and the \(\mathrm{pH}\) of the solution.

Answer: Molar mass of
\(
\begin{aligned}
\mathrm{Sr}(\mathrm{OH})_2 & =87.6+34=121.6 \mathrm{~g} \mathrm{~mol}^{-1}
\end{aligned}
\)
\(
\text { Solubility of } \mathrm{Sr}(\mathrm{OH})_2 \text { in moles } \mathrm{L}^{-1}=\frac{19.23 \mathrm{~g} \mathrm{~L}^{-1}}{121.6 \mathrm{~g} \mathrm{~mol}^{-1}}=0.1581 \mathrm{M}
\)
Assuming complete dissociation,
\(
\mathrm{Sr}(\mathrm{OH})_2 \rightarrow \mathrm{Sr}^{2+}+2 \mathrm{OH}^{-}
\)
\(
\begin{array}{ll}
\therefore \quad & {\left[\mathrm{Sr}^{2+}\right]=0.1581 \mathrm{~M},} \\
& {\left[\mathrm{OH}^{-}\right]=2 \times 0.1581=0.3162 \mathrm{~M}} \\
& \mathrm{pOH}=-\log (0.3162)=0.5, \\
\therefore \quad & \mathrm{pH}=14-0.5=13.5
\end{array}
\)

Q59. The 1onization constant of propanoic acid is \(1.32 \times 10^{-5}\). Calculate the degree of 1onization of the acid in its \(0.05 \mathrm{~M}\) solution and also its \(\mathrm{pH}\). What will be its degree of ionization if the solution is \(0.01 \mathrm{~M} \mathrm{~in} \mathrm{~HCl}\) also?

Answer:

\(
\begin{gathered}
K_a=1.32 \times 10^{-5} ; C=0.05 M \\
\alpha \\
\alpha=\sqrt{\frac{K_a}{C}} \text { or } \alpha=\sqrt{\frac{1.32 \times 10^{-5}}{5 \times 10^{-2}}}=1.62 \times 10^{-2}
\end{gathered}
\)
\(
\begin{aligned}
& {\left[\mathrm{H}^{+}\right]=C \alpha=0.05 \times 1.62 \times 10^{-2}=8.1 \times 10^{-4}} \\
& \mathrm{pH}=-\log \left(8.1 \times 10^{-4}\right)=3.09
\end{aligned}
\)
In \(0.01 \mathrm{~M} \mathrm{~HCl},\left[\mathrm{H}^{+}\right]=0.01 \mathrm{M}\)
\(
\hspace { 7 em } \mathrm{C}_2 \mathrm{H}_5 \mathrm{COOH} \rightleftharpoons \mathrm{C}_2 \mathrm{H}_5 \mathrm{COO}^{-}+\mathrm{H}^{+}
\)
\(
\begin{array}{lccc}
\text { Initial molar conc. } & C & & & 0 & & & 0 \\
\text { Eqm. molar conc. } & C(1-\alpha) & & & C \alpha & & & C \alpha
\end{array}
\)
Applying the law of chemical equilibrium
\(
K_a=\frac{\left[\mathrm{C}_2 \mathrm{H}_5 \mathrm{COO}^{-}\right]\left[\mathrm{H}^{+}\right]}{\left[\mathrm{C}_2 \mathrm{H}_5 \mathrm{COOH}\right]}=\frac{C \alpha \times 0.01}{C(1-\alpha)},
\)
\(
\begin{aligned}
& K_{a}=\frac{C \alpha \times 0.01}{C} \\
& 1.32 \times 10^{-5}=0.01 \alpha \Rightarrow \alpha=1.32 \times 10^{-3}
\end{aligned}
\)

Q60. The \(\mathrm{pH}\) of \(0.1 \mathrm{~M}\) solution of cyanic acid \((\mathrm{HCNO})\) is 2.34. Calculate the ionization constant of the acid and its degree of ionization in the solution.

Answer:

\(
\mathrm{HCNO} \rightleftharpoons \mathrm{H}^{+}+\mathrm{CNO}^{-}
\)
\(
\begin{aligned}
& \mathrm{pH}=2.34 \text { or } \log \left[\mathrm{H}^{+}\right]=-2.34 \\
& \text { or, }\left[\mathrm{H}^{+}\right]=\operatorname{antilog}(-2.34)=4.57 \times 10^{-3} \mathrm{M} \\
& {\left[\mathrm{CNO}^{-}\right]=\left[\mathrm{H}^{+}\right]=4.57 \times 10^{-3} \mathrm{M}} \\
& K_a=\frac{\left(4.57 \times 10^{-3}\right)\left(4.57 \times 10^{-3}\right)}{0.1}=2.09 \times 10^{-4} \\
& \alpha=\sqrt{\frac{K_a}{\mathrm{C}}}=\sqrt{\frac{2.09 \times 10^{-4}}{0.1}}=0.0457
\end{aligned}
\)

Q61. The ionization constant of nitrous acid is \(4.5 \times 10^{-4}\). Calculate the \(\mathrm{pH}\) of \(0.04 \mathrm{~M}\) sodium nitrite solution and also its degree of hydrolysis.

Answer: Degree of hydrolysis, \(h=\sqrt{\frac{K_w}{K_a \cdot C}}\)
\(
=\sqrt{\frac{1 \times 10^{-14}}{4.5 \times 10^{-4} \times 0.04}}=2.36 \times 10^{-5}
\)
\(
\mathrm{pH}=\frac{1}{2} \mathrm{p} K_w+\frac{1}{2} \mathrm{p} K_a+\frac{1}{2} \log \mathrm{C}
\)
\(
=\frac{1}{2}\left[-\log 10^{-14}+\left(-\log 4.5 \times 10^{-4}\right)+\log \left(4 \times 10^{-2}\right)\right]=7.97
\)

Q62. A \(0.02 \mathrm{M}\) solution of pyridinium hydrochloride has \(\mathrm{pH}=3.44\). Calculate the ionization constant of pyridine.

Answer: Pyridinium hydrochloride is a salt of weak base and strong acid. Therefore,
\(
\mathrm{pH}=7-\frac{1}{2}\left(\log C+\mathrm{pK}_b\right)
\)
where \(K_b=\) dissociation constant of pyridine
\(
3.44=7-\frac{1}{2}\left(\log 0.02+p K_{b}\right)
\)
\(
-3.56=-\frac{1}{2}\left(-1.70+\mathrm{p} K_b\right) \Rightarrow-7.12=1.70-\mathrm{p} K_b
\)
\(
\begin{aligned}
& \mathrm{p} K_b=1.70+7.12=8.82 \\
& K_b=\operatorname{antilog}(-8.82)=1.513 \times 10^{-9}
\end{aligned}
\)

Q63. Predict if the solutions of the following salts are neutral, acidic or basic: \(\mathrm{NaCl}, \mathrm{KBr}, \mathrm{NaCN}, \mathrm{NH}_4 \mathrm{NO}_3, \mathrm{NaNO}_2\) and \(\mathrm{KF}\)

Answer: \(\mathrm{NaCN}, \mathrm{NaNO}_2, \mathrm{KF}\) solutions are basic, as they are salts of strong base and weak acid.
\(\mathrm{NaCl}, \mathrm{KBr}\) solutions are neutral, as they are salts of strong acid and strong base.
\(\mathrm{NH}_4 \mathrm{NO}_3\) solution is acidic, as it is a salt of strong acid and weak base.

Q64. The ionization constant of chloroacetic acid is \(1.35 \times 10^{-3}\). What will be the \(\mathrm{pH}\) of \(0.1 \mathrm{~M}\) acid and its \(0.1 \mathrm{~M}\) sodium salt solution?

Answer: \(p \mathrm{H}\) of acid solution
\(
\begin{aligned}
& {\left[\mathrm{H}^{+}\right]=\sqrt{K_a \times C}} \\
& K_a=1.35 \times 10^{-3}, C=0.1 \mathrm{~M}
\end{aligned}
\)
\(
\begin{aligned}
{\left[\mathrm{H}^{+}\right]=\sqrt{1.35 \times 10^{-3} \times 0.1} } & =\sqrt{1.35 \times 10^{-4}}=1.16 \times 10^{-2}
\end{aligned}
\)
\(
\begin{aligned}
\mathrm{pH}=-\log \left[\mathrm{H}^{+}\right] & =-\log \left(1.16 \times 10^{-2}\right)=2-0.064=1.936
\end{aligned}
\)
\(\mathrm{pH}\) of \(0.1 \mathrm{~M}\) sodium salt solution.
Sodium salt of chloroacetic acid is salt of weak acid and strong base. Hence,
\(
\mathrm{pH}=\frac{1}{2}\left[\mathrm{p} K_{w}+\mathrm{pK}_a+\log \mathrm{C}\right]
\)
\(
K_a=1.35 \times 10^{-3}
\)
\(
\begin{aligned}
p K_a= & -\log K_a=-\log \left(1.35 \times 10^{-3}\right) \\
& =-(0.1303-3)=2.8697
\end{aligned}
\)
\(
\begin{aligned}
& \mathrm{p} K_w=-\log 10^{-14}=14 \\
& \quad \mathrm{C}=0.1 \mathrm{M}, \log \mathrm{C}=\log (0.1)=-1
\end{aligned}
\)
\(
\Rightarrow \quad \mathrm{pH}=\frac{1}{2}[14+2.8697-1]=7.935
\)

Q65. Ionic product of water at \(310 \mathrm{~K}\) is \(2.7 \times 10^{-14}\). What is the \(\mathrm{pH}\) of neutral water at this temperature?

Answer: We know that
\(
\begin{aligned}
& {\left[\mathrm{H}^{+}\right]=\sqrt{K_w}=\sqrt{2.7 \times 10^{-14}}=1.643 \times 10^{-7} \mathrm{M}} \\
& \therefore \quad \mathrm{pH}=-\log \left[\mathrm{H}^{+}\right]=-\log \left(1.643 \times 10^{-7}\right) \\
& =7-0.2156=6.78
\end{aligned}
\)

Q66. Calculate the \(\mathrm{pH}\) of the resultant mixtures:
a) \(10 \mathrm{~mL}\) of \(0.2 \mathrm{~M} \mathrm{~Ca}(\mathrm{OH})_2+25 \mathrm{~mL}\) of \(0.1 \mathrm{~M} \mathrm{~HCl}\)
b) \(10 \mathrm{~mL}\) of \(0.01 \mathrm{~M} \mathrm{~H}_2 \mathrm{SO}_4+10 \mathrm{~mL}\) of \(0.01 \mathrm{~M} \mathrm{~Ca}(\mathrm{OH})_2\)
c) \(10 \mathrm{~mL}\) of \(0.1 \mathrm{~M} \mathrm{~H}_2 \mathrm{SO}_4+10 \mathrm{~mL}\) of \(0.1 \mathrm{~M} \mathrm{~KOH}\)

Answer: (a)
\(
\begin{array}{rl}
10 \mathrm{~mL} \text { of } 0.2 \mathrm{~V} & \mathrm{Ca}(\mathrm{OH})_2=10 \times 0.2 \text { millimoles }=2 \text { millimoles of } \mathrm{Ca}(\mathrm{OH})_2
\end{array}
\)
\(
\begin{aligned}
25 \mathrm{~mL} \text { of } 0.1 \mathrm{~M} \mathrm{~HCl} & =25 \times 0.1 \text { millimoles }=2.5 \text { millimoles of } \mathrm{HCl}
\end{aligned}
\)
\(
\mathrm{Ca}(\mathrm{OH})_2+2 \mathrm{HCl} \longrightarrow \mathrm{CaCl}_2+2 \mathrm{H}_2 \mathrm{O}
\)
1 millimole of \(\mathrm{Ca}(\mathrm{OH})_2\) reacts with 2 millimoles of \(\mathrm{HCl}\)
\(\therefore \quad 2.5\) millimoles of \(\mathrm{HCl}\) will react with 1.25 millimoles of \(\mathrm{Ca}(\mathrm{OH})_2\)
\(\therefore \quad \mathrm{Ca}(\mathrm{OH})_2\) left \(=2-1.25=0.75\) millimoles
( \(\mathrm{HCl}\) is the limiting reactant)
\(
\begin{aligned}
\text { Total volume of the solution } & =10+25 \mathrm{~mL}=35 \mathrm{~mL}
\end{aligned}
\)
\(\therefore\) Molarity of \(\mathrm{Ca}(\mathrm{OH})_2\) in the mixture
\(
\begin{aligned}
& \text { solution }=\frac{0.75}{35} \mathrm{M}=0.0214 \mathrm{M} \\
& \begin{aligned}
\therefore \quad\left[\mathrm{OH}^{-}\right]=2 \times 0.0214 \mathrm{M} & =0.0428 \mathrm{M} \\
& =4.28 \times 10^{-2} \mathrm{M} \\
\text { pOH }=-\log \left(4.28 \times 10^{-2}\right) & =2-0.6314 \\
& =1.3686 \approx 1.37
\end{aligned}
\end{aligned}
\)
\(
\therefore \quad \mathrm{pH}=14-1.37=12.63
\)

(b)

\(
\begin{aligned}
& 10 \mathrm{~mL} \text { of } 0.01 \mathrm{~M} \mathrm{~H}_2 \mathrm{SO}_4=10 \times 0.01 \mathrm{millimole}=0.1 \text { millimole }
\end{aligned}
\)
\(
\begin{aligned}
& 10 \mathrm{~mL} \text { of } 0.01 \mathrm{M} \mathrm{Ca}(\mathrm{OH})_2=10 \times 0.01 \text { millimole }=0.1 \text { millimole }
\end{aligned}
\)
\(
\mathrm{Ca}(\mathrm{OH})_2+\mathrm{H}_2 \mathrm{SO}_4 \rightarrow \mathrm{CaSO}_4+2 \mathrm{H}_2 \mathrm{O}
\)
1 mole of \(\mathrm{Ca}(\mathrm{OH})_2\) reacts with 1 mole of \(\mathrm{H}_2 \mathrm{SO}_4\)
\(\therefore \quad 0.1\) millimole of \(\mathrm{Ca}(\mathrm{OH})_2\) will react completely with 0.1 millimole of \(\mathrm{H}_2 \mathrm{SO}_4\). hence, solution will be neutral with \(\mathrm{pH}=7.0\)

(c) \(10 \mathrm{~mL}\) of \(0.1 \mathrm{~M} \mathrm{~H}_2 \mathrm{SO}_4=1\) millimole
\(10 \mathrm{~mL}\) of \(0.1 \mathrm{~M} \mathrm{~KOH}=1 \mathrm{millimole}\)
\(
2 \mathrm{KOH}+\mathrm{H}_2 \mathrm{SO}_4 \rightarrow \mathrm{K}_2 \mathrm{SO}_4+2 \mathrm{H}_2 \mathrm{O}
\)
1 millimole of \(\mathrm{KOH}\) will react with 0.5 millimole of \(\mathrm{H}_2 \mathrm{SO}_4\)
\(
\begin{aligned}
& \therefore \quad \mathrm{H}_2 \mathrm{SO}_4 \text { left }=1-0.5=0.5 \text { millimole } \\
& \text { Volume of reaction mixture }=10+10=20 \mathrm{~mL} \\
&
\end{aligned}
\)
\(\therefore\) Molarity of \(\mathrm{H}_2 \mathrm{SO}_4\) in the mixture
\(
\begin{aligned}
& \text { solution }=\frac{0.5}{20}=2.5 \times 10^{-2} \mathrm{M} \\
& {\left[\mathrm{H}^{+}\right]=2 \times 2.5 \times 10^{-2}=5 \times 10^{-2}} \\
& \mathrm{pH}=-\log \left(5 \times 10^{-2}\right)=2-0.699=1.3
\end{aligned}
\)

Q67. Determine the solubility of silver chromate, barium chromate and ferric hydroxide at \(289 \mathrm{~K}\) from their solubility product constants. Determine also the molarities of the individual ions.
\(
\begin{array}{ll}
\quad \text { Salt } & K_{\text {sp }} \text { value } \\
\text { Silver chromate } & 1.1 \times 10^{-12} \\
\text { Barium chromate } & 1.2 \times 10^{-10} \\
\text { Ferric hydroxide } & 1.0 \times 10^{-38} \\
\text { Lead chloride } & 1.6 \times 10^{-5} \\
\text { Mercurous iodide } & 4.5 \times 10^{-29}
\end{array}
\)

Answer: Silver chromate, \(\mathrm{Ag}_2 \mathrm{CrO}_4\)
\(
\mathrm{Ag}_2 \mathrm{CrO}_{4(\mathrm{~s})} \rightleftharpoons 2 \mathrm{Ag}_{(a q)}^{+}+\mathrm{CrO}_{4(a q)}^{2-}
\)
\(
K_{s p}=4 S^3=1.1 \times 10^{-12} \therefore S=6.50 \times 10^{-5} \mathrm{M}
\)
\(
\therefore \quad\left[\mathrm{Ag}^{+}\right]=2 \times 6.50 \times 10^{-5}=1.30 \times 10^{-4} \mathrm{M} \text { and } \left[\mathrm{CrO}_4^{2-}\right]=6.50 \times 10^{-5} \mathrm{M}
\)
Barium chromate, \(\mathrm{BaCrO}_4\)
\(
\mathrm{BaCrO}_{4(s)} \rightleftharpoons \mathrm{Ba}_{(a q)}^{2+}+\mathrm{CrO}_{4(a q)}^{2-}
\)
\(
K_{s p}=S^2=1.2 \times 10^{-10} \quad \therefore S=1.09 \times 10^{-5} \mathrm{M}
\)
\(
\left[\mathrm{Ba}^{2+}\right]=\left[\mathrm{CrO}_4^{2-}\right]=1.09 \times 10^{-5} \mathrm{M}
\)
Ferric hydroxide, \(\mathrm{Fe}(\mathrm{OH})_3\)
\(
\mathrm{Fe}(\mathrm{OH})_{3(\mathrm{s})} \rightleftharpoons \mathrm{Fe}_{(a q)}^{3+}+3 \mathrm{OH}_{(a q)}^{-}
\)
\(
K_{s p}=27 S^4=1.0 \times 10^{-38} \quad \therefore S=1.38 \times 10^{-10} \mathrm{M}
\)
\(
\left[\mathrm{Fe}^{3+}\right]=1.38 \times 10^{-10} \mathrm{M} \text { and }\left[\mathrm{OH}^{-}\right]=3 \times 1.38 \times 10^{-10}=4.16 \times 10^{-10} \mathrm{M}
\)
Lead chloride, \(\mathrm{PbCl}_2\)
\(\mathrm{PbCl}_{2(s)} \rightleftharpoons \mathrm{Pb}_{(a q)}^{2+}+2 \mathrm{Cl}_{(a q)}^{-}\)
\(
\begin{aligned}
& K_{s p}=4 S^3=1.6 \times 10^{-3} \\
& \therefore \quad S=1.59 \times 10^{-2} \mathrm{M}
\end{aligned}
\)
\(
\left[\mathrm{Pb}^{2+}\right]=1.59 \times 10^{-2} \mathrm{M} \text { and }\left[\mathrm{Cl}^{-}\right]=2 \times 1.59 \times 10^{-2}=3.18 \times 10^{-2} \mathrm{M}
\)
Mercurous iodide, \(\mathrm{Hg}_2 \mathrm{I}_2\)
\(
\mathrm{Hg}_2 \mathrm{I}_{2(s)} \rightleftharpoons \mathrm{Hg}_{2(a q)}^{2+}+2 \mathrm{I}_{(\mathrm{a} q)}^{-}
\)
\(
K_{s p}=4 S^3=4.5 \times 10^{-29} \text { and } S=2.24 \times 10^{-10} \mathrm{M}
\)
\(
\begin{aligned}
{\left[\mathrm{Hg}_2^{2+}\right]=2.24 \times 10^{-10} \mathrm{M} \text { and }\left[\mathrm{I}^{-}\right] } & =2 \times 2.24 \times 10^{-10}=4.48 \times 10^{-10} \mathrm{M}
\end{aligned}
\)

Q68. The solubility product constant of \(\mathrm{Ag}_2 \mathrm{CrO}_4\) and \(\mathrm{AgBr}\) are \(1.1 \times 10^{-12}\) and \(5.0 \times 10^{-13}\) respectively. Calculate the ratio of the molaritles of their saturated solutions.

Answer:

\(
K_{s p}\left(\mathrm{Ag}_2 \mathrm{CrO}_4\right)>K_{s p}(\mathrm{AgBr})
\)
\(\therefore \quad \mathrm{Ag}_2 \mathrm{CrO}_4\) is more soluble.
\(\mathrm{Ag}_2 \mathrm{CrO}_{4(s)} \rightleftharpoons 2 \mathrm{Ag}_{(a q)}^{-}+\mathrm{CrO}_{4(a q))}^{2-}\)
\(
K_{s p}=4 S^3 \text { and }
\)
\(
\therefore \quad S=\left(\frac{1.1 \times 10^{-12}}{4}\right)^{1 / 3}=6.50 \times 10^{-5} \mathrm{M}
\)
\(
\mathrm{AgBr}_{(s)} \rightleftharpoons \mathrm{Ag}_{(a q)}^{-}+\mathrm{Br}_{(a q)}^{-} ; K_{s p}=S^{\prime 2}
\)
\(
\therefore \quad S^{\prime}=\left(5 \times 10^{-13}\right)^{1 / 2}=7.07 \times 10^{-7} \mathrm{M}
\)
\(
\begin{aligned}
\text { The ratio of the molarities } & =\frac{S}{S^{\prime}}=\frac{6.50 \times 10^{-5}}{7.07 \times 10^{-7}}=91.94
\end{aligned}
\)
Silver chromate is more soluble and the ratio of their molarities = 91.9.

Q69. Equal volumes of \(0.002 \mathrm{~M}\) solutions of sodium iodate and cupric chlorate are mixed together. Will it lead to precipitation of copper 1odate? (For cupric 1odate \(K_{\mathrm{sp}}=7.4 \times 10^{-8})\).

Answer: When equal volumes of sodium iodate and copper chlorate are mixed, the molar concentrations of both the solutes would be reduced to half i.e., \(0.001 \mathrm{M}\)
\(
\mathrm{NaIO}_3 \longrightarrow \mathrm{Na}^{+}+\mathrm{IO}_3^{-}
\)
\(
0.001 \mathrm{M} \quad \quad \quad \quad 0.001 \mathrm{M}
\)
\(
\mathrm{Cu}\left(\mathrm{ClO}_3\right)_2 \longrightarrow \mathrm{Cu}^{2+}+2 \mathrm{ClO}_3^{-}
\)
\(
0.001 \mathrm{M} \quad \quad \quad \quad 0.001 \mathrm{M}
\)
\(\therefore \quad\) After mixing, \(\left[\mathrm{IO}_3{ }^{-}\right]=\left[\mathrm{NaIO}_3\right]=0.001 \mathrm{M}\)
\(
\left[\mathrm{Cu}^{2+}\right]=\left[\mathrm{Cu}\left(\mathrm{IO}_3\right)_2\right]=0.001 \mathrm{M}
\)
Solubility equilibrium for copper iodate may be written as,
\(\mathrm{Cu}\left(\mathrm{IO}_3\right)_{2(s)} \rightleftharpoons \mathrm{Cu}_{(a q)}^{2+}+2 \mathrm{IO}_3^{-}{ }_{(a q)}\)
Ionic product of copper iodate \(=\left[\mathrm{Cu}^{2+}\right]\left[\mathrm{IO}_3\right]^2=(0.001)(0.001)^2=1 \times 10^{-9}\)
Since ionic product \(\left(1 \times 10^{-9}\right)\) is less than \(K_{s p}\left(7.4 \times 10^{-8}\right)\), therefore, no precipitation will take place.

Q70. The ionization constant of benzoic acid is \(6.46 \times 10^{-5}\) and \(K_{\mathrm{sp}}\) for silver benzoate is \(2.5 \times 10^{-13}\). How many times is silver benzoate more soluble in a buffer of \(\mathrm{pH}\) 3.19 compared to its solubility in pure water?

Answer: Suppose \(S\) is the molar solubility of silver benzoate in water, then
\(
\mathrm{C}_6 \mathrm{H}_5 \mathrm{COOAg}_{(\mathrm{s})} \rightleftharpoons \mathrm{C}_6 \mathrm{H}_5 \mathrm{COO}_{(a q)}^{-}+\mathrm{Ag}_{(\mathrm{aq})}^{+}
\)
\(
K_{s p}=S^2 \therefore S=\sqrt{2.5 \times 10^{-13}}=5.0 \times 10^{-7} \mathrm{M}
\)
If the solubility of salt of weak acid of ionization constant \(K_a\) is \(S^{\prime}\), then \(K_{s p}, K_a\) and \(S^{\prime}\) are related to each other at \(\mathrm{pH}=3.19\).
\(
\therefore \quad\left[\mathrm{H}^{+}\right]=6.46 \times 10^{-4} \mathrm{M}
\)
\(
K_{s p}=S^{\prime 2}\left[\frac{K_a}{K_a+\left[\mathrm{H}^{+}\right]}\right]
\)
\(
S^{\prime}=\left\{\frac{2.5 \times 10^{-13}}{\left[\frac{6.46 \times 10^{-5}}{6.46 \times 10^{-5}+6.46 \times 10^{-4}}\right]}\right\}^{1 / 2}
\)
\(
S^{\prime}=\left\{\frac{2.5 \times 10^{-13} \times 7.106 \times 10^{-4}}{6.46 \times 10^{-5}}\right\}^{1 / 2}
\)
\(
=\left(2.75 \times 10^{-12}\right)^{1 / 2}=1.658 \times 10^{-6} \mathrm{M}
\)
\(\therefore \quad\) The ratio \(\frac{S^{\prime}}{S}=\frac{1.658 \times 10^{-6}}{5.0 \times 10^{-7}}=3.32\) Silver benzoate is 3.32 times more soluble in buffer of \(\mathrm{pH} 3.19\) than in pure water.

Q71. 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 tron sulphide, \(K_{\mathrm{sp}}=6.3 \times 10^{-18}\) ).

Answer: Let the concentration of each of \(\mathrm{FeSO}_4\) and \(\mathrm{Na}_2 \mathrm{S}\) is \(x\) mol \(\mathrm{L}^{-1}\). Then after mixing equal volumes:
\(
\left[\mathrm{FeSO}_4\right]=\frac{x}{2} \mathrm{M},\left[\mathrm{Na}_2 \mathrm{~S}\right]=\frac{x}{2} \mathrm{M}
\)
\(
\text { or }\left[\mathrm{Fe}^{2+}\right]=\frac{x}{2} \mathrm{M} \text { and }\left[\mathrm{S}^{2-}\right]=\frac{x}{2} \mathrm{M}
\)
\(
\mathrm{FeS} \rightleftharpoons \mathrm{Fe}^{2+}+\mathrm{S}^{2-}, K_{s p}=\left[\mathrm{Fe}^{2+}\right]\left[\mathrm{S}^{2-}\right]
\)
\(
6.3 \times 10^{-18}=\frac{x}{2} \times \frac{x}{2}
\)
or \(x^2=4 \times 6.3 \times 10^{-18}=25.2 \times 10^{-18}\) or \(x=5.02 \times 10^{-9} \mathrm{~mol} \mathrm{~L}^{-\mathrm{1}}\)
Maximum concentration of \(\mathrm{FeSO}_4\) and \(\mathrm{Na}_2 \mathrm{S}\) which will not precipitate iron sulphide \(=5.02 \times 10^{-9} \mathrm{~mol} \mathrm{~L}^{-1}\)

Q72. What is the minimum volume of water required to dissolve \(1 \mathrm{~g}\) of calcium sulphate at \(298 \mathrm{~K}\)? (For calclum sulphate, \(K_{\mathrm{sp}}\) is \(9.1 \times 10^{-6}\) ).

Answer:

\(
\mathrm{CaSO}_{4(s)} \rightleftharpoons \mathrm{Ca}_{(a q)}^{2+}+\mathrm{SO}_{4(\mathrm{aq})}^{2-}
\)
If \(S\) is the solubility of \(\mathrm{CaSO}_4\) in \(\mathrm{mol} \mathrm{~L}^{-1}\), then
\(
K_{s p}=\left[\mathrm{Ca}^{2+}\right] \times\left[\mathrm{SO}_4^{2-}\right]=\mathrm{S}^2
\)
\(
\begin{aligned}
S & =\sqrt{K_{s p}}=\sqrt{9.1 \times 10^{-6}}=3.02 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} \\
& =3.02 \times 10^{-3} \times 136 \mathrm{~g} \mathrm{~L}^{-1}=0.411 \mathrm{~g} \mathrm{~L}^{-1}
\end{aligned}
\)
\(\left(\right.\) Molar mass of \(\left.\mathrm{CaSO}_4=136 \mathrm{~g} \mathrm{~mol}^{-1}\right)\). Thus, for dissolving \(0.411 \mathrm{~g}\), water required \(=1 \mathrm{~L}\)
\(\therefore \quad\) For dissolving \(1 \mathrm{~g}\),
\(
\text { water required }=\frac{1}{0.411} \mathrm{~L}=2.43 \mathrm{~L}
\)

Q73. The concentration of sulphide ion in \(0.1 \mathrm{~M} \mathrm{~HCl}\) solution saturated with hydrogen sulphide is \(1.0 \times 10^{-19} \mathrm{M}\). If \(10 \mathrm{~mL}\) of this is added to \(5 \mathrm{~mL}\) of \(0.04 \mathrm{~M}\) solution of the following: \(\mathrm{FeSO}_4, \mathrm{MnCl}_2, \mathrm{ZnCl}_2\) and \(\mathrm{CdCl}_2\). In which of these solutions precipitation will take place?
Given \(\mathrm{K}_{\mathrm{sp}}\) for
\(
\begin{aligned}
& \mathrm{FeS}=6.3 \times 10^{-18} \\
& \mathrm{MnS}=2.5 \times 10^{-13} \\
& \mathrm{ZnS}=1.6 \times 10^{-24} \text { and } \\
& \mathrm{CdS}=8.0 \times 10^{-27}
\end{aligned}
\)

Answer: Precipitation will take place in the solution for which the ionic product is greater than the solubility product. As \(10 \mathrm{~mL}\) of solution containing \(\mathrm{S}^{2-}\) ion is mixed with \(5 \mathrm{~mL}\) of metal salt solution, after mixing
\(
\left[\mathrm{S}^{2-}\right]=1.0 \times 10^{-19} \times \frac{10}{15}=6.67 \times 10^{-20}
\)
\(
\left[\mathrm{Fe}^{2+}\right]=\left[\mathrm{Mn}^{2+}\right]=\left[\mathrm{Zn}^{2+}\right]=\left[\mathrm{Cd}^{2+}\right]=\frac{5}{15} \times 0.04=1.33 \times 10^{-2} \mathrm{M}
\)
Hence the ionic product \(\left[M^{2+}\right]\left[\mathrm{S}^{2-}\right]\)
\(
=1.33 \times 10^{-2} \times 6.67 \times 10^{-20}=8.87 \times 10^{-22}
\)
Therefore in \(\mathrm{ZnS}\left(K_{s p}=1.6 \times 10^{-24}\right)\) and \(\mathrm{CdS}\) \(\left(K_{s p}=8.0 \times 10^{-27}\right)\), ionic product exceeds solubility product. Hence, precipitation will take place.

Exemplar

Q74. The ionisation of hydrochloric in water is given below:
\(
\mathrm{HCl}(\mathrm{aq})+\mathrm{H}_2 \mathrm{O}(l) \rightleftharpoons \mathrm{H}_3 \mathrm{O}^{+}(\mathrm{aq})+\mathrm{Cl}^{-}(\mathrm{aq})
\)
Label two conjugate acid-base pairs in this ionisation.

Answer:

\(
\begin{array}{lr}
\mathrm{HCl} \text { (Acid) } & \mathrm{Cl}^{-} \text {(Conjugate base) } \\
\mathrm{H}_2 \mathrm{O} \text { (Base) } & \mathrm{H}_3 \mathrm{O}^{+} \text {(Conjugate acid }
\end{array}
\)

Q75. The aqueous solution of sugar does not conduct electricity. However, when sodium chloride is added to water, it conducts electricity. How will you explain this statement on the basis of ionisation and how is it affected by concentration of sodium chloride?

Answer: (i) Sugar being a non-electrolyte does not ionise in water whereas \(\mathrm{NaCl}\) ionises completely in water and produces \(\mathrm{Na}^{+}\)and \(\mathrm{Cl}^{-}\)ion which help in the conduction of electricity.
(ii) When concentration of \(\mathrm{NaCl}\) is increased, more \(\mathrm{Na}^{+}\)and \(\mathrm{Cl}^{-}\)ions will be produced. Hence, conductance or conductivity of the solution increases.

Q76. \(\mathrm{BF}_3\) does not have proton but still acts as an acid and reacts with \(\mathrm{NH}_3\). Why is it so? What type of bond is formed between the two?

Answer: \(\mathrm{BF}_3\) is an electron deficient compound and hence acts as Lewis acid. \(\mathrm{NH}_3\) has one lone pair which it can donate to \(\mathrm{BF}_3\) and form a coordinate bond. Hence, \(\mathrm{NH}_3\) acts as a Lewis base.
\(
\mathrm{H}_3 \mathrm{~N}: \rightarrow \mathrm{BF}_3
\)

Q77. Ionisation constant of a weak base \(\mathrm{MOH}\), is given by the expression
\(
K_b=\frac{\left[\mathrm{M}^{+}\right]\left[\mathrm{OH}^{-}\right]}{[\mathrm{MOH}]}
\)
Values of ionisation constant of some weak bases at a particular temperature are given below:
\(
\begin{array}{lllll}
\text { Base } & \text { Dimethylamine } & \text { Urea } & \text { Pyridine } & \text { Ammonia } \\
K_{\mathrm{b}} & 5.4 \times 10^{-4} & 1.3 \times 10^{-14} & 1.77 \times 10^{-9} & 1.77 \times 10^{-5}
\end{array}
\)
Arrange the bases in decreasing order of the extent of their ionisation at equilibrium. Which of the above base is the strongest?

Answer: Greater is the ionization constant \(\left(\mathrm{K}_{\mathrm{b}}\right)\) of a base, greater is the ionization of the base. Order of extent of ionization at equilibrium is dimethylamine > ammonia > pyridine > urea. Dimethylamine is the strongest base due to maximum value of \(K_b\) (Since dimethylamine will ionise to the maximum extent it is the strongest base out of the four given bases).

Q78. Conjugate acid of a weak base is always stronger. What will be the decreasing order of basic strength of the following conjugate bases?
\(
\mathrm{OH}^{-}, \mathrm{RO}^{-}, \mathrm{CH}_3 \mathrm{COO}^{-}, \mathrm{Cl}^{-}
\)

Answer: Conjugate acids of given bases are \(\mathrm{H}_2 \mathrm{O}, \mathrm{ROH}, \mathrm{CH}_3 \mathrm{COOH}, \mathrm{HCl}\) Their acidic strength is in the order \(\mathrm{HCl}>\mathrm{CH}_3 \mathrm{COOH}>\mathrm{H}_2 \mathrm{O}>\mathrm{ROH}\). Hence, basic strength is in the order
\(
\mathrm{RO}^{-}>\mathrm{OH}^{-}>\mathrm{CH}_3 \mathrm{COO}^{-}>\mathrm{Cl}^{-}
\)

Q79. Arrange the following in increasing order of \(\mathrm{pH}\).
\(
\mathrm{KNO}_3(a q), \mathrm{CH}_3 \mathrm{COONa}(a q), \mathrm{NH}_4 \mathrm{Cl}(a q), \mathrm{C}_6 \mathrm{H}_5 \mathrm{COONH}_4(a q)
\)

Answer: (i) \(\mathrm{KNO}_3\) is a salt of strong acid \(\left(\mathrm{HNO}_3\right)\) strong base \((\mathrm{KOH})\), hence its aqueous solution is neutral; \(\mathrm{pH}=7\).
(ii) \(\mathrm{CH}_3 \mathrm{COONa}\) is a salt of weak acid \(\left(\mathrm{CH}_3 \mathrm{COOH}\right)\) and strong base \((\mathrm{NaOH})\), hence, its aqueous solution is basic; \(\mathrm{pH}>7\).
(iii) \(\mathrm{NH}_4 \mathrm{Cl}\) is a salt of strong acid \((\mathrm{HCl})\) and weak base \(\left(\mathrm{NH}_4 \mathrm{OH}\right)\) hence, its aqueous solution is acidic; \(\mathrm{pH}<7\).
(iv) \(\mathrm{C}_6 \mathrm{H}_5 \mathrm{COONH}_4\) is a salt of weak acid, \(\mathrm{C}_6 \mathrm{H}_5 \mathrm{COOH}\) and weak base, \(\mathrm{NH}_4 \mathrm{OH}\). But \(\mathrm{NH}_4 \mathrm{OH}\) is slightly stronger than \(\mathrm{C}_6 \mathrm{H}_5 \mathrm{COOH}\). \(\mathrm{Hence}, \mathrm{pH}\) is slightly \(>7\).
Therefore, increasing order of \(\mathrm{pH}\) of the given salts is,
\(
\mathrm{NH}_4 \mathrm{Cl}<\mathrm{C}_6 \mathrm{H}_5 \mathrm{COONH}_4<\mathrm{KNO}_3<\mathrm{CH}_3 \mathrm{COONa}
\)

Q80. The value of \(K_{\mathrm{c}}\) for the reaction \(2 \mathrm{HI}(\mathrm{g}) \rightleftharpoons \mathrm{H}_2(\mathrm{~g})+\mathrm{I}_2(\mathrm{~g})\) is \(1 \times 10^{-4}\) At a given time, the composition of reaction mixture is \([\mathrm{HI}]=2 \times 10^{-5} \mathrm{~mol}, \quad\left[\mathrm{H}_2\right]=1 \times 10^{-5} \mathrm{~mol}\) and \(\left[\mathrm{I}_2\right]=1 \times 10^{-5} \mathrm{~mol}\) In which direction will the reaction proceed?

Answer: At a given time the reaction quotient \(Q\) for the reaction will be given by the expression.
\(
\begin{aligned}
Q & =\frac{\left[\mathrm{H}_2\right]\left[\mathrm{I}_2\right]}{[\mathrm{HI}]^2} \\
& =\frac{1 \times 10^{-5} \times 1 \times 10^{-5}}{\left(2 \times 10^{-5}\right)^2}=\frac{1}{4} \\
& =0.25=2.5 \times 10^{-1}
\end{aligned}
\)
As the value of reaction quotient is greater than the value of \(K_c\) i.e. \(1 \times 10^{-4}\) the reaction will proceed in the reverse direction.

Q81. On the basis of the equation \(\mathrm{pH}=-\log \left[\mathrm{H}^{+}\right]\), the \(\mathrm{pH}\) of \(10^{-8} \mathrm{~mol} \mathrm{dm}{ }^{-3}\) solution of \(\mathrm{HCl}\) should be 8. However, it is observed to be less than 7.0. Explain the reason.

Answer: Concentration of \(10^{-8} \mathrm{~mol} \mathrm{dm}^{-3}\) indicates that the solution is very dilute. Hence, the contribution of \(\mathrm{H}_3 \mathrm{O}^{+}\)concentration from water is significant and should also be included for the calculation of \(\mathrm{pH}\). So, we cannot neglect the contribution of \(\mathrm{H}_3 \mathrm{O}^{+}\)ions produced from \(\mathrm{H}_2 \mathrm{O}\) in the solution. Total \(\left[\mathrm{H}_3 \mathrm{O}^{+}\right]=10^{-8}+10^{-7} \mathrm{M}\). From this we get the value of \(\mathrm{pH}\) close to 7 but less than 7 because the solution is acidic.
From calculation, it is found that \(\mathrm{pH}\) of \(10^{-8} \mathrm{~mol} \mathrm{dm}^{-3}\) solution of \(\mathrm{HCl}\) is equal to 6.96 .

Q82. \(\mathrm{pH}\) of a solution of a strong acid is 5.0 . What will be the \(\mathrm{pH}\) of the solution obtained after diluting the given solution a 100 times?

Answer: 
\(
\begin{aligned}
& \mathrm{pH}=5 \\
& {\left[\mathrm{H}^{+}\right]=10^{-5} \mathrm{~mol} \mathrm{~L}^{-1}}
\end{aligned}
\)
On 100 times dilution
\(
\left[\mathrm{H}^{+}\right]=10^{-7} \mathrm{~mol} \mathrm{~L}^{-1}
\)
On calculating the \(\mathrm{pH}\) using the equation \(\mathrm{pH}=-\log \left[\mathrm{H}^{+}\right]\), value of \(\mathrm{pH}\) comes out to be 7 . It is not possible. This indicates that solution is very dilute. Hence,
\(
\begin{aligned}
& \text { Total hydrogen ion concentration} \\
\end{aligned}=\left[\mathrm{H}^{+}\right]
\)
=Contribution of \(\mathrm{H}_3 \mathrm{O}^{+} \text {ion concentration of acid}+\)Contribution of \(\mathrm{H}_3 \mathrm{O}^{+}\)ion concentration of water
\(
=10^{-7}+10^{-7} \text {. }
\)
\(
\mathrm{pH}=2 \times 10^{-7}=7-\log 2=7-0.3010=6.6990
\)

Q83. A sparingly soluble salt gets precipitated only when the product of concentration of its ions in the solution \(\left(\mathrm{Q}_{\mathrm{sp}}\right)\) becomes greater than its solubility product. If the solubility of \(\mathrm{BaSO}_4\) in water is \(8 \times 10^{-4} \mathrm{~mol} \mathrm{dm}^{-3}\). Calculate its solubility in \(0.01 \mathrm{~mol} \mathrm{dm}^{-3}\) of \(\mathrm{H}_2 \mathrm{SO}_4\).

Answer: 

\(
\begin{array}{llcc}
& \mathrm{BaSO}_4(\mathbf{s}) \rightleftharpoons \mathrm{Ba}^{2+}(a q) & +\mathrm{SO}_4^{2-}(a q) \\
\text { At } \mathrm{t}=0 & 1 \quad \quad \quad \quad \quad \quad 0 & 0 \\
\text { At equilibrium in water } & 1-\mathrm{S} \quad \quad \quad \quad \quad \mathrm{S} & \mathrm{S} \\
\begin{array}{l}
\text { At equilibrium in the presence } \\
\text { of sulphuric acid }
\end{array} & 1-\mathrm{S}\quad \quad \quad \quad \quad \mathrm{S} & (\mathrm{S}+0.01)
\end{array}
\)
\(
K_{\mathrm{sp}} \text { for } \mathrm{BaSO}_4 \text { in water }=\left[\mathrm{Ba}^{2+}\right]\left[\mathrm{SO}_4^{2-}\right]=(\mathrm{S})(\mathrm{S})=\mathrm{S}^2
\)
\(
\text { But } \mathrm{S}=8 \times 10^{-4} \mathrm{~mol} \mathrm{~dm}^{-3}
\)
\(
\therefore K_{\mathrm{sp}}=\left(8 \times 10^{-4}\right)^2=64 \times 10^{-8} \dots(1)
\)
The expression for \(K_{\mathrm{sp}}\) in the presence of sulphuric acid will be as follows :
\(
K_{\mathrm{sp}}=(\mathrm{S})(\mathrm{S}+0.01) \dots(2)
\)
Since value of \(K_{\mathrm{sp}}\) will not change in the presence of sulphuric acid, therefore from (1) and (2)
\(
\begin{aligned}
& (\mathrm{S})(\mathrm{S}+0.01)=64 \times 10^{-8} \\
& \mathrm{~S}^2+0.01 \mathrm{~S}=64 \times 10^{-8} \\
& \mathrm{~S}^2+0.01 \mathrm{~S}-64 \times 10^{-8}=0
\end{aligned}
\)
\(
\begin{aligned}
S & =\frac{-0.01 \pm \sqrt{(0.01)^2+\left(4 \times 64 \times 10^{-8}\right)}}{2} \\
& =\frac{-0.01 \pm \sqrt{10^{-4}+\left(256 \times 10^{-8}\right)}}{2} \\
& =\frac{-0.01 \pm \sqrt{10^{-4}\left(1+256 \times 10^{-2}\right)}}{2} \\
& =\frac{-0.01 \pm 10^{-2} \sqrt{1+0.256}}{2} \\
& =\frac{-0.01 \pm 10^{-2} \sqrt{1.256}}{2}
\end{aligned}
\)
\(
=6 \times 10^{-4} \mathrm{~mol} \mathrm{~dm}^{-3}
\)

Q84. \(\mathrm{pH}\) of \(0.08 \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{HOCl}\) solution is 2.85 . Calculate its ionisation constant.

Answer: \(\mathrm{pH}\) of \(\mathrm{HOCl}=2.85\)
But, \(-\mathrm{pH}=\log \left[\mathrm{H}^{+}\right]\)
\(
\begin{aligned}
\therefore-2.85 & =\log \left[\mathrm{H}^{+}\right] \\
3.15 & =\log \left[\mathrm{H}^{+}\right] \\
{\left[\mathrm{H}^{+}\right] } & =1.413 \times 10^{-3}
\end{aligned}
\)
\(
\text { For weak mono basic acid }\left[\mathrm{H}^{+}\right]=\sqrt{K_{\mathrm{a}} \times \mathrm{C}}
\)
\(
\begin{aligned}
K_{\mathrm{a}} & =\frac{\left[\mathrm{H}^{+}\right]^2}{C}=\frac{\left(1.413 \times 10^{-3}\right)^2}{0.08} \\
& =24.957 \times 10^{-6}=2.4957 \times 10^{-5}
\end{aligned}
\)

Q85. Calculate the \(\mathrm{pH}\) of a solution formed by mixing equal volumes of two solutions \(\mathrm{A}\) and \(\mathrm{B}\) of a strong acid having \(\mathrm{pH}=6\) and \(\mathrm{pH}=4\) respectively.

Answer: \(\mathrm{pH}\) of Solution \(\mathrm{A}=6\)
Therefore, concentration of \(\left[\mathrm{H}^{+}\right]\)ion in solution \(\mathrm{A}=10^{-6} \mathrm{~mol} \mathrm{~L}^{-1}\) \(\mathrm{pH}\) of Solution \(\mathrm{B}=4\)
Therefore, Concentration of \(\left[\mathrm{H}^{+}\right]\)ion concentration of solution \(\mathrm{B}=10^{-4} \mathrm{~mol} \mathrm{~L}^{-1}\) On mixing one litre of each solution, total volume \(=1 \mathrm{~L}+1 \mathrm{~L}=2 \mathrm{~L}\)
Amount of \(\mathrm{H}^{+}\)ions in \(\mathrm{lL}\) of Solution \(\mathrm{A}=\) Concentration \(\times\) volume \(V\)
\(
=10^{-6} \mathrm{~mol} \times 1 \mathrm{~L}
\)
Amount of \(\mathrm{H}^{+}\)ions in \(1 \mathrm{~L}\) of solution \(\mathrm{B}=10^{-4} \mathrm{~mol} \times 1 \mathrm{~L}\)
\(\therefore\) Total amount of \(\mathrm{H}^{+}\)ions in the solution formed by mixing solutions \(\mathrm{A}\) and \(\mathrm{B}\) is \(\left(10^{-6} \mathrm{~mol}+10^{-4} \mathrm{~mol}\right)\)
This amount is present in \(2 \mathrm{~L}\) solution.
\(
\begin{aligned}
\therefore \text { Total }\left[\mathrm{H}^{+}\right]=\frac{10^{-4}(1+0.01)}{2}=\frac{1.01 \times 10^{-4}}{2} \mathrm{~mol} \mathrm{~L}^{-1} & =\frac{1.01 \times 10^{-4}}{2} \mathrm{~mol} \mathrm{~L}^{-1} \\
& =0.5 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \\
& =5 \times 10^{-5} \mathrm{~mol} \mathrm{~L}^{-1}
\end{aligned}
\)
\(
\begin{aligned}
\mathrm{pH}=-\log \left[\mathrm{H}^{+}\right] & =-\log \left(5 \times 10^{-5}\right) \\
& =-[\log 5+(-5 \log 10)] \\
& =-\log 5+5 \\
& =5-\log 5 \\
& =5-0.6990 \\
& =4.3010=4.3
\end{aligned}
\)

Q86. The solubility product of \(\mathrm{Al}(\mathrm{OH})_3\) is \(2.7 \times 10^{-11}\). Calculate its solubility in \(\mathrm{gL}^{-1}\) and also find out \(\mathrm{pH}\) of this solution. (Atomic mass of \(\mathrm{Al}=27 \mathrm{u}\) ).

Answer:Let \(\mathrm{S}\) be the solubility of \(\mathrm{Al}(\mathrm{OH})_3\).

\(
\begin{aligned}
& K_{\mathrm{sp}}=\left[\mathrm{Al}^{3+}\right][\mathrm{OH}]^3=(\mathrm{S})(3 \mathrm{~S})^3=27 \mathrm{~S}^4 \\
& \mathrm{~S}^4=\frac{K_{s p}}{27}=\frac{27 \times 10^{-11}}{27 \times 10}=1 \times 10^{-12} \\
& \mathrm{~S}=1 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1}
\end{aligned}
\)
(i) Solubility of \(\mathrm{Al}(\mathrm{OH})_3\)
Molar mass of \(\mathrm{Al}(\mathrm{OH})_3\) is \(78 \mathrm{~g}\). Therefore,
Solubility of \(\mathrm{Al}(\mathrm{OH})_3\) in \(\mathrm{g} \mathrm{L}^{-1}=1 \times 10^{-3} \times 78 \mathrm{~g} \mathrm{~L}^{-1}=78 \times 10^{-3} \mathrm{~g} \mathrm{~L}^{-1}\)
\(
=7.8 \times 10^{-2} \mathrm{~g} \mathrm{~L}^{-1}
\)
(ii) \(\mathrm{pH}\) of the solution
\(
\mathrm{S}=1 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1}
\)
\(
\begin{aligned}
& {\left[\mathrm{OH}^{-}\right]=3 \mathrm{~S}=3 \times 1 \times 10^{-3}=3 \times 10^{-3}} \\
& \mathrm{pOH}=3-\log 3 \\
& \mathrm{pH}=14-\mathrm{pOH}=11+\log 3=11.4771
\end{aligned}
\)

Q87. Calculate the volume of water required to dissolve \(0.1 \mathrm{~g}\) lead (II) chloride to get a saturated solution. \(\left(K_{\mathrm{sp}}\right.\) of \(\mathrm{PbCl}_2=3.2 \times 10^{-8}\), atomic mass of \(\left.\mathrm{Pb}=207 \mathrm{u}\right)\).

Answer: \(K_{\text {sp }} \text { of } \mathrm{PbCl}_2=3.2 \times 10^{-8}\)
Let \(\mathrm{S}\) be the solubility of \(\mathrm{PbCl}_2\).
\(
\begin{array}{llcc}
& \mathbf{P b C l}_2(\mathbf{s}) \rightleftharpoons \mathbf{P b}^{2+}(\mathbf{a q})+2 \mathbf{C l}^{-}(\mathbf{a q}) \\
\begin{array}{l}
\text { Concentration of } \\
\text { species at } \mathrm{t}=0
\end{array} & 1 \quad \quad \quad \quad \quad \quad \quad 0 \quad \quad \quad \quad \quad 0 \\
\begin{array}{l}
\text { Concentration of various } \\
\text { species at equilibrium }
\end{array} & 1-\mathrm{S} \quad \quad \quad \quad \quad \mathrm{S}  \quad \quad \quad \quad \quad 2 \mathrm{~S}
\end{array}
\)
\(
\begin{aligned}
& K_{\mathrm{sp}}=\left[\mathrm{Pb}^{2+}\right]\left[\mathrm{Cl}^{-}\right]^2=(\mathrm{S})(2 \mathrm{~S})^2=4 \mathrm{~S}^3 \\
& K_{\mathrm{sp}}=4 \mathrm{~S}^3 \\
& \mathrm{~S}^3=\frac{K_{\mathrm{sp}}}{4}=\frac{3.2 \times 10^{-8}}{4} \mathrm{~mol} \mathrm{~L}^{-1}=8 \times 10^{-9} \mathrm{~mol} \mathrm{~L}^{-1} \\
& \mathrm{~S}=\sqrt[3]{8 \times 10^{-9}}=2 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} \quad \therefore \mathrm{S}=2 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1}
\end{aligned}
\)
Molar mass of \(\mathrm{PbCl}_2=278\)
\(\therefore\) Solubility of \(\mathrm{PbCl}_2\) in \(\mathrm{g} \mathrm{L}^{-1}=2 \times 10^{-3} \times 278 \mathrm{~g} \mathrm{~L}^{-1}\)
\(
\begin{aligned}
& =556 \times 10^{-3} \mathrm{~g} \mathrm{~L}^{-1} \\
& =0.556 \mathrm{~g} \mathrm{~L}^{-1}
\end{aligned}
\)
To get saturated solution, \(0.556 \mathrm{~g}\) of \(\mathrm{PbCl}_2\) is dissolved in \(1 \mathrm{~L}\) water.
\(0.1 \mathrm{~g} \mathrm{PbCl}_2\) is dissolved in \(\frac{0.1}{0.556} \mathrm{~L}=0.1798 \mathrm{~L}\) water.
To make a saturated solution, dissolution of \(0.1 \mathrm{~g} \mathrm{PbCl}_2\) in \(0.1798 \mathrm{~L} \approx 0.2 \mathrm{~L}\) of water will be required.

Q88. A reaction between ammonia and boron trifluoride is given below:
\(
: \mathrm{NH}_3+\mathrm{BF}_3 \longrightarrow \mathrm{H}_3 \mathrm{~N}: \mathrm{BF}_3
\)
Identify the acid and base in this reaction. Which theory explains it? What is the hybridisation of \(\mathrm{B}\) and \(\mathrm{N}\) in the reactants?

Answer: Although \(\mathrm{BF}_3\) does not have a proton but acts as Lewis acid as it is an electron deficient compound. It reacts with \(\mathrm{NH}_3\) by accepting the lone pair of electrons from \(\mathrm{NH}_3\) and completes its octet. The reaction can be represented by
\(
\mathrm{BF}_3+: \mathrm{NH}_3 \rightarrow \mathrm{BF}_3 \leftarrow: \mathrm{NH}_3
\)
Lewis electronic theory of acids and bases can explain it. Boron in \(\mathrm{BF}_3\) is \(\mathrm{sp}^2\) hybridised, whereas \(\mathrm{N}_{\text {in }} \mathrm{NH}_3\) is \(\mathrm{sp}^3\) hybridised.

Q89. Following data is given for the reaction: \(\mathrm{CaCO}_3(\mathrm{~s}) \rightarrow \mathrm{CaO}(\mathrm{s})+\mathrm{CO}_2(\mathrm{~g})\)
\(
\begin{aligned}
& \Delta_f H^{\ominus}[\mathrm{CaO}(\mathrm{s})]=-635.1 \mathrm{~kJ} \mathrm{~mol}^{-1} \\
& \Delta_f H^{\ominus}\left[\mathrm{CO}_2(\mathrm{~g})\right]=-393.5 \mathrm{~kJ} \mathrm{~mol}^{-1} \\
& \Delta_f H^{\ominus}\left[\mathrm{CaCO}_3(\mathrm{~s})\right]=-1206.9 \mathrm{~kJ} \mathrm{~mol}^{-1}
\end{aligned}
\)
Predict the effect of temperature on the equilibrium constant of the above reaction.

Answer:

\(
\begin{aligned}
& \Delta_{\mathrm{r}} H^{\ominus}=\Delta_f H^{\ominus}[\mathrm{CaO}(\mathrm{s})]+\Delta_f H^{\ominus}\left[\mathrm{CO}_2(\mathrm{~g})\right]-\Delta_f H^{\ominus}\left[\mathrm{CaCO}_3(\mathrm{~s})\right] \\
& \therefore \Delta_{\mathrm{r}} H^{\ominus}=178.3 \mathrm{~kJ} \mathrm{~mol}^{-1}
\end{aligned}
\)
The reaction is endothermic. Hence, according to Le-Chatelier’s principle, reaction will proceed in forward direction on increasing temperature.

Q90. How can you predict the following stages of a reaction by comparing the value of \(K_c\) and \(Q_c\)?
(i) Net reaction proceeds in the forward direction.
(ii) Net reaction proceeds in the backward direction.
(iii) No net reaction occurs.

Answer: (i) \(Q_{\mathrm{c}}<K_{\mathrm{c}}\), the reaction will proceed in the direction of the products (forward reaction).
(ii) \(Q_{\mathrm{c}}>K_{\mathrm{c}}\), the reaction will proceed in the direction of reactants (reverse reaction).
(iii) \(Q_{\mathrm{c}}=K_{\mathrm{c}}\), the reaction mixture is already at equilibrium.
where, \(Q_{\mathrm{c}}\) is reaction quotient in terms of concentration and \(K_{\mathrm{c}}\) is equilibrium constant.

Q91. On the basis of Le Chatelier principle explain how temperature and pressure can be adjusted to increase the yield of ammonia in the following reaction.
\(
\mathrm{N}_2(\mathrm{~g})+3 \mathrm{H}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_3(\mathrm{~g}) \quad \Delta \mathrm{H}=-92.38 \mathrm{~kJ} \mathrm{~mol}^{-1}
\)
What will be the effect of addition of argon to the above reaction mixture at constant volume?

Answer: It is an exothermic process as \(\Delta \mathrm{H}\) is negative.
Effect of temperature According to Le-Chatelier’s principle, low temperature is favourable for high yield of ammonia, but practically very low temperatures slow down the reaction. So, optimum temperature \(700 \mathrm{~K}\) is favourable in attainment of equilibrium.
Effect of pressure Similarly, high pressure about 200 atm is favourable for high yield of ammonia. On increasing pressure, reaction goes in the forward direction because the number of moles decreases in the forward direction.
Addition of argon At constant volume addition of argon does not affect the equilibrium because it does not change the partial pressures of the reactants or products involved in the reaction and the equilibrium remains undisturbed.

Q92. A sparingly soluble salt having general formula \(\mathrm{A}_x^{p+} \mathrm{B}_y^{q-}\) and molar solubility \(\mathrm{S}\) is in equilibrium with its saturated solution. Derive a relationship between the solubility and solubility product for such salt.

Answer: A sparingly soluble salt having general formula \(\mathrm{A}_{\mathrm{x}}^{\mathrm{p}+} \mathrm{B}_{\mathrm{y}}^{\mathrm{q}-}\). Its molar solubility is \(\mathrm{S} \mathrm{~mol} \mathrm{~L}^{-1}\).
\(
\mathrm{A}_x^{p+} \mathrm{B}_y^{q-} \rightleftharpoons x \mathrm{~A}^{p+}(\mathrm{aq})+y \mathrm{~B}^{q-}(\mathrm{aq})
\)
\(
\mathrm{S} \text { moles of } \mathrm{A}_x \mathrm{~B}_y \text { dissolve to give } x \mathrm{~S} \text { ~moles of } \mathrm{A}^{\mathrm{p}+} \text { and } y \mathrm{~S} \text { ~moles of } \mathrm{B}^{\mathrm{q}-} \text {. }
\)
Therefore, solubility product \(\left(\mathrm{K}_{\mathrm{sp}}\right)=\left[\mathrm{A}^{\mathrm{p}+}\right]^{\mathrm{x}}\left[\mathrm{B}^{\mathrm{q}-}\right]^{\mathrm{y}}\)
\(
\begin{aligned}
& =[\mathrm{xS}]^{\mathrm{x}}[\mathrm{yS}]^{\mathrm{y}} \\
& =\mathrm{x}^{\mathrm{x}} \mathrm{y}^{\mathrm{y}} \mathrm{S}^{\mathrm{x}+\mathrm{y}}
\end{aligned}
\)

Q93. Write a relation between \(\Delta G\) and \(Q\) and define the meaning of each term and answer the following:
(a) Why a reaction proceeds forward when \(Q<K\) and no net reaction occurs when \(Q=K\).
(b) Explain the effect of increase in pressure in terms of reaction quotient \(Q\). for the reaction: \(\mathrm{CO}(\mathrm{g})+3 \mathrm{H}_2(\mathrm{~g}) \rightleftharpoons \mathrm{CH}_4(\mathrm{~g})+\mathrm{H}_2 \mathrm{O}(\mathrm{g})\)

Answer:

\(
\Delta G=\Delta G^{\ominus}+R T \ln Q
\)
\(\Delta G^{\ominus}=\) Change in free energy as the reaction proceeds
\(\Delta G=\) Standard free energy change
\(Q=\) Reaction quotient
\(\mathrm{R}=\) Gas constant
\(T=\) Absolute temperature
Since \(\Delta G^{\ominus}=-\mathrm{R} T \ln K\)
\(
\therefore \quad \Delta G=-R T \ln K+\mathrm{R} T \ln Q=\mathrm{RT} \ln \frac{Q}{K}
\)
If \(Q<K, \Delta G\) will be negative. Reaction proceeds in the forward direction. If \(Q=K, \Delta G=0\), no net reaction.
(b) \(\text { (b) } \mathrm{CO}(\mathrm{g})+3 \mathrm{H}_2(\mathrm{~g}) \rightleftharpoons \mathrm{CH}_4(\mathrm{~g})+\mathrm{H}_2 \mathrm{O}(\mathrm{g})\)
\(
\mathrm{K}_{\mathrm{c}}=\frac{\left[\mathrm{CH}_4\right]\left[\mathrm{H}_2 \mathrm{O}\right]}{[\mathrm{CO}]\left[\mathrm{H}_2\right]^3}
\)
On increasing pressure, volume decreases. If we doubled the pressure, volume will be halved but the molar concentrations will be doubled. Then,
\(
\mathrm{Q}_{\mathrm{c}}=\frac{2\left[\mathrm{CH}_4\right] \cdot 2\left[\mathrm{H}_2 \mathrm{O}\right]}{2[\mathrm{CO}]\left\{2\left[\mathrm{H}_2\right]\right\}^3}=\frac{1}{4} \frac{\left[\mathrm{CH}_4\right]\left[\mathrm{H}_2 \mathrm{O}\right]}{[\mathrm{CO}]\left[\mathrm{H}_2\right]^3}=\frac{1}{4} \mathrm{~K}_{\mathrm{c}}
\)
Therefore, \(Q_c\) is less than \(K_c\), so \(Q_c\) will tend to increase to re-establish equilibrium and the reaction will go in forward direction.
[Hint: Next relate \(Q\) with concentration of \(\mathrm{CO}, \mathrm{H}_2, \mathrm{CH}_4\) and \(\mathrm{H}_2 \mathrm{O}\) in view of reduced volume (increased pressure). Show that \(Q^4<\mathrm{K}\) and hence the reaction proceeds in forward direction.]