Chapter 22, Problem 60QAP

Write an equation for the reaction of chloroacetic acid $\left(K_{\text{a}}=1.5\times {10}^{-3}\right)$ with trimethylamine $\left(K_{\text{b}}=5.9\times {10}^{-5}\right)$ . Calculate the equilibrium constant for the reaction. If 0.10 M solutions of these two species are mixed, what will be their concentrations at equilibrium?

Interpretation Introduction

Interpretation:

The equation for the reaction between chloroacetic acid and trimethylamine, the equilibrium constant and concentration at equilibrium for the reaction should be determined.

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. Each concentration is raised to the power that is equal to the coefficient in the balanced reaction. So, the expression is:

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

Square brackets represent the concentration.

Answer to Problem 60QAP

The reaction between chloroacetic acid and trimethylamine is:

${\text{ClCH}}_{\text{2}}{\text{COOH+(CH}}_{\text{3}}{\text{)}}_{\text{3}}\text{N}\to {\text{ClCH}}_{\text{2}}{\text{COO}}^{\text{-}}{\text{+(CH}}_{\text{3}}{\text{)}}_{\text{3}}{\text{NH}}^{\text{+}}$, the equilibrium constant, K = $8.85\times {10}^6$.

The concentration is $\text{3}{\text{.4 × 10}}^{\text{-5}}\text{ M}$.

Explanation of Solution

The reaction between chloroacetic acid and trimethylamine is neutralization reaction.

The neutralization reaction between chloroacetic acid and trimethylamine is:

${\text{ClCH}}_{\text{2}}{\text{COOH+(CH}}_{\text{3}}{\text{)}}_{\text{3}}\text{N}\to {\text{ClCH}}_{\text{2}}{\text{COO}}^{\text{-}}{\text{+(CH}}_{\text{3}}{\text{)}}_{\text{3}}{\text{NH}}^{\text{+}}$

In order to determine the value of equilibrium constant of the reaction, the dissociation of chloroacetic acid and trimethylamine in accordance to Arrhenius view may be written as:

${{\text{(CH}}_{\text{3}}\text{)}}_{\text{3}}{\text{N(aq) + H}}_2\text{O}\to {{\text{(CH}}_{\text{3}}\text{)}}_{\text{3}}{\text{NH}}^{\text{+}}{\text{(aq) + OH}}^{-}{\text{(aq) k}}_{\text{b}}\text{= 5}\text{.9}\times {\text{10}}^{-5}$ -(1)

${\text{ClCH}}_{\text{2}}{\text{COOH(aq)+H}}_2\text{O}\to {\text{ClCH}}_{\text{2}}{\text{COO}}^{\text{-}}{\text{+H}}_3{\text{O}}^{+}{\text{ k}}_{\text{a}}\text{=1}{\text{.5×10}}^{\text{-3}}$ -(2)

Since, ${\text{H}}_2\text{O }\to {\text{H}}^{\text{+}}{\text{+OH}}^{\text{-}}{\text{ k}}_{\text{w}}{\text{=10}}^{\text{-14}}$

So,

${\text{H}}^{\text{+}}{\text{+OH}}^{\text{-}}\to {\text{ H}}_2{\text{O k}}^{\text{'}}\text{=}\frac{1}{{\text{10}}^{\text{-14}}}$ -(3)

As on reversing the reaction, the value of constant of reversed reaction is the invert value of K.

When the reaction is expressed as the sum of two or more reactions the value of equilibrium constant, K for the final reaction is determined by the product of the equilibrium constant of individual reactions.

Adding equation (1), (2), and (3):

$\begin{array}{l}{{\text{(CH}}_{\text{3}}}_{\text{)}}{\text{N(aq) + H}}_2\text{O}\rightleftarrows {{\text{(CH}}_{\text{3}}}_{\text{)}}{\text{NH}}^{\text{+}}{\text{(aq) + OH}}^{-}{\text{(aq) k}}_{\text{b}}\text{= 5}\text{.9}\times {\text{10}}^{-5}\\ {\text{ClCH}}_{\text{2}}{\text{COOH(aq)+H}}_2\text{O}\rightleftarrows {\text{ClCH}}_{\text{2}}{\text{COO}}^{\text{-}}{\text{(aq)+H}}_3{\text{O}}^{+}{\text{ k}}_{\text{a}}\text{=1}{\text{.5×10}}^{\text{-3}}\\ \underset{\_}{{\text{ H}}^{\text{+}}{\text{+OH}}^{\text{-}}\to {\text{ H}}_2{\text{O k}}^{\text{'}}\text{=}\frac{1}{{\text{10}}^{\text{-14}}}}\\ {{\text{(CH}}_{\text{3}}}_{\text{)}}{\text{N(aq)+ ClCH}}_{\text{2}}\text{COOH(aq)}\to {\text{ClCH}}_{\text{2}}{\text{COO}}^{\text{-}}{\text{(aq)+(CH}}_{\text{3}}{\text{)}}_{\text{3}}{\text{NH}}^{\text{+}}\text{(aq) K = }\frac{\text{5}\text{.9}\times {\text{10}}^{-5}\times \text{1}{\text{.5×10}}^{\text{-3}}}{{\text{10}}^{\text{-14}}}\end{array}$

Thus, K = $8.85\times {10}^6$

The concentration at equilibrium is determined by constructing the ICE table for the reaction:

$\begin{array}{l}{\text{ ClCH}}_{\text{2}}{\text{COOH+(CH}}_{\text{3}}{\text{)}}_{\text{3}}\text{N}\rightleftarrows {\text{ClCH}}_{\text{2}}{\text{COO}}^{\text{-}}{\text{+(CH}}_{\text{3}}{\text{)}}_{\text{3}}{\text{NH}}^{\text{+}}\\ \text{I 0}\text{.1 M 0}\text{.1 M 0 0}\\ \text{C -x -x +x +x}\\ \text{E 0}\text{.1-x 0}\text{.1-x +x +x}\end{array}$

The equilibrium constant for the above reaction is:

$\text{K}=\frac{[{\text{ClCH}}_{\text{2}}{\text{COO}}^{\text{-}}][{{\text{(CH}}_{\text{3}}\text{)}}_{\text{3}}{\text{NH}}^{\text{+}}]}{[{\text{ClCH}}_{\text{2}}\text{COOH}][{{\text{(CH}}_{\text{3}}\text{)}}_{\text{3}}\text{N}]}$

Substituting the values:

$\text{8}{\text{.85×10}}^{\text{6}}\text{=}\frac{{\text{x}}^{\text{2}}}{{\text{(0}\text{.1-x)}}^{\text{2}}}$

On solving the quadratic equation:

$\text{x = 3}{\text{.4 × 10}}^{\text{-5}}$

Thus, the concentration is $\text{3}{\text{.4 × 10}}^{\text{-5}}\text{ M}$.

Conclusion

The reaction between chloroacetic acid and trimethylamine is:

${\text{ClCH}}_{\text{2}}{\text{COOH+(CH}}_{\text{3}}{\text{)}}_{\text{3}}\text{N}\to {\text{ClCH}}_{\text{2}}{\text{COO}}^{\text{-}}{\text{+(CH}}_{\text{3}}{\text{)}}_{\text{3}}{\text{NH}}^{\text{+}}$, the equilibrium constant, K = $8.85\times {10}^6$.

The concentration is $\text{3}{\text{.4 × 10}}^{\text{-5}}\text{ M}$.