Chapter 7, Problem 7.63P

Interpretation Introduction

(a)

Interpretation:

The equilibrium concentrations of H+ and CH₃ COO- have to be determined given the initial concentration and final concentration of the weak acid, CH₃ COOH.

Concept Introduction:

${\text{CH}}_{\text{3}}\text{COOH}$ is a weak acid which partially ionizes in solution. The products of ionization, CH₃ COO- and H+ exist in equilibrium with the weak acid. The change in concentration of the weak acid when the system reaches equilibrium will be equal to the concentration of the ionization products.

Answer to Problem 7.63P

${{\text{[H}}^{+}]}_{\text{equilbrium}}={[{\text{CH}}_{\text{3}}{\text{COO}}^{-}]}_{\text{equilibrium}}=0.0020M$.

Explanation of Solution

The equilibrium reaction representing the ionization of weak acid, ${\text{CH}}_{\text{3}}\text{COOH}$ is,

${\text{CH}}_3\text{COOH(}\text{a}q\text{)}\rightleftharpoons {\text{H}}^{+}(\text{a}q)+{\text{CH}}_{\text{3}}{\text{COO}}^{-}(\text{a}q)$

The initial concentration of the acid = 0.10 M.

Concentration of acid at equilibrium = 0.098 M.

Change in concentration of ${\text{CH}}_{\text{3}}\text{COOH}$

$=(0.10-0.098)\text{M}=0.0020\text{M}$

This change in concentration of CH₃ COOH will be equal to the concentration of ${\text{[H}}^{+}]$ and $[{\text{CH}}_{\text{3}}{\text{COO}}^{-}]$ formed during the ionization of CH₃ COOH.

Therefore, ${{\text{[H}}^{+}]}_{\text{equilbrium}}={[{\text{CH}}_{\text{3}}{\text{COO}}^{-}]}_{\text{equilibrium}}=0.0020M$.

Interpretation Introduction

(b)

Interpretation:

The expected equilibrium constant for the above reaction has to be calculated.

Concept Introduction:

Equilibrium constant is the ratio of the concentrations of products to the reactants concentrations, each term raised to the power equal to the stoichiometric coefficient.

${\text{CH}}_3\text{COOH(}\text{a}q\text{)}\rightleftharpoons {\text{H}}^{+}(\text{a}q)+{\text{CH}}_{\text{3}}{\text{COO}}^{-}(\text{a}q)K=\frac{[{\text{CH}}_{\text{3}}{\text{COO}}^{-}]{\text{[H}}^{+}]}{{\text{[CH}}_3\text{COOH}]}$.

Answer to Problem 7.63P

The expected equilibrium constant for the reaction is $4.1\times {10}^{-5}$.

Explanation of Solution

The given equilibrium ionization reaction is,

$\begin{array}{l}{\text{CH}}_3\text{COOH(}\text{a}q\text{)}\rightleftharpoons {\text{H}}^{+}(\text{a}q)+{\text{CH}}_{\text{3}}{\text{COO}}^{-}(\text{a}q)\\ K=\frac{[{\text{CH}}_{\text{3}}{\text{COO}}^{-}]{\text{[H}}^{+}]}{{\text{[CH}}_3\text{COOH}]}\end{array}$

Plugging in the equilibrium concentrations to calculate the K value,

$\begin{array}{l}K=\frac{[{\text{CH}}_{\text{3}}{\text{COO}}^{-}]{\text{[H}}^{+}]}{{\text{[CH}}_3\text{COOH}]}\\ =\frac{(0.0020)(0.0020)}{(0.098)}\\ =4.08\times {10}^{-5}\\ \simeq 4.1\times {10}^{-5}(\text{On}\text{rounding}\text{to two}\text{significant}\text{figures)}\end{array}$.