Chapter 22, Problem 28QAP

When the conjugate acid of aniline, C₆H₅NH₃⁺, reacts with the acetate ion, the following reaction takes place:
${\text{C}}_6{\text{H}}_5{\text{NH}}_3{}^{+}\left(aq\right)+{\text{CH}}_3{\text{COO}}^{-}\left(aq\right)\rightleftharpoons {\text{C}}_6{\text{H}}_5{\text{NH}}_2\left(aq\right)+{\text{CH}}_3\text{COOH}\left(aq\right)$
If K_afor C₆H₅NH₃+ is $1.35\times {10}^{-5}$ and K_afor CH₃COOH is $1.86\times {10}^{-5}$ , what is K for the reaction?

Interpretation Introduction

Interpretation:

The K of the given reaction is to be calculated.

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 28QAP

The required $K=0.73$.

Explanation of Solution

Given Information:

The conjugate acid of aniline $C_6{H}_{5}N{H}_3^{+}(k_a=1.35\times {10}^{-5})$ reacts with acetate ion. For, acetic acid $C{H}_{3}COOH(k_a=1.86\times {10}^{-5})$

The reaction is:

$C_6{H}_{5}N{H}_3^{+}(aq)+C{H}_{3}CO{O}^{-}(aq)\rightleftharpoons C_6{H}_{5}N{H}_2(aq)+C{H}_{3}COOH(aq)$

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

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

The equilibrium constant expression will be:

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

Further acetic acid is dissociated as:

$C{H}_{3}COOH(aq)+H_{2}O(l)\rightleftharpoons H_3{O}^{+}(aq)+C{H}_{3}CO{O}^{-}(aq)$

Hence, we have:

The equilibrium constant expression will be:

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

Further when the reactions are added:

$C_6{H}_{4}N{H}_3^{+}(aq)+C{H}_{3}CO{O}^{-}(aq)\rightleftharpoons C_6{H}_{4}N{H}_2(aq)+C{H}_{3}COOH(aq)$

The expression is:

$\text{K=}\frac{[C_6{H}_{4}N{H}_2][C{H}_{3}COOH]}{[C_6{H}_{4}N{H}_3^{+}][C{H}_{3}CO{O}^{-}]}$

The expression can be rearranged as:

$\begin{array}{l}\text{K=}\frac{[C_6{H}_{4}N{H}_2][C{H}_{3}COOH][H_3{O}^{+}]}{[C_6{H}_{4}N{H}_3^{+}][C{H}_{3}CO{O}^{-}][H_3{O}^{+}]}=\frac{[C{H}_{3}COOH]}{[C{H}_{3}O{O}^{-}][H_3{O}^{+}]}\times \frac{[C_6{H}_{4}N{H}_2][H_3{O}^{+}]}{[C_6{H}_{4}N{H}_3^{+}]}=\frac{k_{a(C_6{H}_{4}N{H}_3^{+})}}{k_{a(C{H}_{3}COOH)}}\\ \text{K=}\frac{k_{a(C_6{H}_{4}N{H}_3^{+})}}{k_{a(C{H}_{3}COOH)}}\end{array}$

Putting the values we get:

$\begin{array}{c}\text{K=}\frac{k_{a(C_6{H}_{4}N{H}_3^{+})}}{k_{a(C{H}_{3}COOH)}}=\frac{1.35\times {10}^{-5}}{1.86\times {10}^{-5}}\\ K=0.73\end{array}$

Conclusion

The required $K=0.73$.