Chapter 22, Problem 26QAP

When aniline, C₆H₅NH₂ $\left(K_b=7.4\times {10}^{-10}\right)$ , reacts with a strong acid, its conjugate acid, C₆H₅NH₃⁺, is formed. Calculate the pH of a 0.100 M solution of C₆H₅NH₃+ and compare it with the pH of acetic acid $\left(K_a=1.86\times {10}^{-5}\right)$ .

Interpretation Introduction

Interpretation:

The pH of 0.100 M solution of $C_6{H}_{4}N{H}_3^{+}$ is to be obtained and compared with the pH of acetic acid.

Concept introduction:

The relationship between the concentration of products and reactants at equilibrium for a general reaction:

$\text{aA}\text{+}\text{bB}\rightleftarrows \text{cC}\text{+}\text{dD}$

Where A, B, C, and D represents chemical species and a, b, c, and d are the coefficients for balanced reaction.

The equilibrium expression, K_c for reversible reaction is determined by multiplying the concentrations of products together and divided by the concentrations of the reactants. Every concentration term is raised to the power that is equal to the coefficient in the balanced reaction. So, the expression is:

${\text{K}}_{\text{c}}\text{=}\frac{{\text{[C]}}^{\text{c}}{\text{[D]}}^{\text{d}}}{{\text{[A]}}^{\text{a}}{\text{[B]}}^{\text{b}}}$

pH scale is a type of measurement of H+ ion concentration in the solution. It is taken as a negative log of H+ ion concentration as:

$pH=-{\log }_{1o}[H^{+}]$

For the equilibrium conditions:

$k_a\times k_b=k_w={10}^{-14}$

Where, k_a is acid dissociation constant, k_b is base dissociation constant and k_w is ionic product of water whose value is fixed as 10-14.

Answer to Problem 26QAP

The required $pH=2.93$ which is comparable to the pH of acetic acid $pH=2.87$.

Explanation of Solution

Given Information:

The aniline $C_6{H}_{4}N{H}_2(k_b=7.4\times {10}^{-10})$ reacts with a strong acid. Its conjugate acid is $C_6{H}_{4}N{H}_3^{+}$.

For, acetic acid $C{H}_{3}COOH(k_a=1.86\times {10}^{-5})$.

The conjugate acid of aniline is a weak acid which dissociates as:

$C_6{H}_{4}N{H}_3^{+}(aq)+H_{2}O\rightleftharpoons H_3{O}^{+}(aq)+C_6{H}_{4}N{H}_2(aq)$

Let’s discuss the main reaction in one direction and creating ICE table as:

$\begin{array}{l}{C}_6{H}_{4}N{H}_3^{+}(aq)+H_{2}O\rightleftharpoons H_3{O}^{+}(aq)+C_6{H}_{4}N{H}_2(aq)\\ I 0.100M 0 0\\ C -x x x\\ E (0.100M-x) x x\end{array}$

Hence, we have:

The equilibrium constant expression will be:

${\text{K}}_a\text{=}\frac{[H_3{O}^{+}][C_6{H}_{4}N{H}_2]}{[C_6{H}_{4}N{H}_3^{+}]}$

Putting the standard value of K_a :

$k_a=\frac{k_w}{k_b}=\frac{{10}^{-14}}{7.4\times {10}^{-10}}=1.35\times {10}^{-5}$

Now, the equilibrium expression is:

$\begin{array}{c}1.35\times {10}^{-5}=\frac{x\times x}{[0.100M-x]}\\ 0.100-x\approx 0.100\\ x^2=1.35\times {10}^{-5}\times 0.100=1.35\times {10}^{-6}\\ x=1.162\times {10}^{-3}\end{array}$

Further, pH is calculated as:

$\begin{array}{c}pH=-\log [H^{+}\text{] }\\ =-\log [1.162\times {10}^{-3}\text{]}\\ \text{=2}\text{.93}\end{array}$

Further the pH of acetic acid is calculated by its dissociation as:

$\begin{array}{l}C{H}_{3}COOH(aq)+H_{2}O(l)\rightleftharpoons H_3{O}^{+}(aq)+C{H}_{3}CO{O}^{-}(aq)\\ I 0.100M. 0 0\\ C -x x x\\ E (0.100M-x) x x\end{array}$

Hence, we have:

The equilibrium constant expression will be:

${\text{K}}_a\text{=}\frac{[H_3{O}^{+}][C{H}_{3}CO{O}^{-}]}{[C{H}_{3}COOH]}$

Putting the standard value of K_a the equilibrium expression is:

$\begin{array}{c}1.8\times {10}^{-5}=\frac{x\times x}{[0.100M-x]}\\ 0.100-x\approx 0.100\\ x^2=1.8\times {10}^{-5}\times 0.100=1.8\times {10}^{-6}\\ x=1.34\times {10}^{-3}\end{array}$

Further, pH is calculated as:

$\begin{array}{c}pH=-\log [H^{+}\text{] }\\ =-\log [1.34\times {10}^{-3}\text{]}\\ \text{=2}\text{.87}\end{array}$

Conclusion

The required $pH=2.93$ which is comparable to the pH of acetic acid $pH=2.87$.