Chapter 16, Problem 16.5QP

Determine the pH of (a) a 0.40 M CH₃COOH solution, (b) a solution that is 0.40 M CH₃COOH and 0.20 M CH₃COONa.

Interpretation Introduction

Interpretation:

By using the given concentration of acetic acid and sodium acetate, calculate the $\text{pH}$ values.

Concept introduction:

• $\text{pH}$ is the logarithm of the reciprocal of the hydrogen ion concentration in a solution.
•  $\text{pH}$ is used to determine the acidity or basicity of an aqueous solution.
• ${\text{pH = -log[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}$
• $\text{pH}\text{=}{\text{pK}}_{\text{a}}\text{+}\text{log}\frac{\text{[conjugate base]}}{\text{[acid]}}$

To calculate: the $\text{pH}$ of acetic acid

Answer to Problem 16.5QP

The $\text{pH}$ of acetic acid is $\text{2}\text{.57}$

Explanation of Solution

The given concentrations are recorded as shown.

The concentration of acetic acid and sodium acetate are $\text{0}\text{.40M}\text{and}\text{0}\text{.20M}$

Here acetic acid is a weak acid, the equilibrium table is as follows

$\begin{array}{l}{\text{CH}}_{\text{3}}{\text{COOH}}_{\text{(aq)}}\stackrel{}{\rightleftarrows }{\text{ H}}^{\text{+}}{}_{\text{(aq)}}{\text{ + CH}}_{\text{3}}{\text{COO}}^{\text{-}}{}_{\text{(aq)}}\\ \\ \text{Initial concentration(M): }\text{0}\text{.40 }\text{0}\text{.00 }\text{0}\text{.00}\\ \text{Change in concentration (M): }\text{-x }\text{+x }\text{+x}\\ \text{Equilibrium}\text{concentration (M): }\text{(0}\text{.40-x) }\text{x }\text{x}\end{array}$

${\text{K}}_{\text{a}}{\text{ value for CH}}_{\text{3}}\text{COOH is 1}\text{.8 }\times {\text{10}}^{\text{-5}}$

$\begin{array}{l}{\text{K}}_{\text{a}}\text{=}\frac{{\text{[H}}^{\text{+}}\text{]}{\text{[CH3COO}}^{\text{-}}\text{]}}{\text{[CH3COOH]}}\\ \text{1}\text{.8}{\text{×10}}^{\text{-5}}\text{=}\frac{{\text{x}}^{\text{2}}}{\text{(0}\text{.40-x)}}\approx \frac{{\text{x}}^{\text{2}}}{\text{0}\text{.40}}\\ \text{the}\text{value}\text{of}\text{x}\text{is}\text{very}\text{small}\text{and}\text{neglect}\text{it,}\text{because the ionization of }\\ {\text{CH}}_{\text{3}}{\text{COOH is reduced by the presence of CH}}_{\text{3}}{\text{COO}}^{\text{-}}\text{ ions}\\ \text{x}\text{=}{\text{[H}}^{\text{+}}\text{]}\text{=}\text{2}\text{.7 ×}{\text{10}}^{\text{-3}}\text{M}\\ \text{pH}\text{=}{\text{-log[H}}^{\text{+}}\text{]}\\ \text{=}\text{-log(2}\text{.7}\text{×}{\text{10}}^{\text{-3}}\text{M)}\\ \text{=}\text{2}\text{.57}\\ \end{array}$

The ratio of concentrations of acetate ion and hydrogen ion to concentration of acetic acid is ${\text{K}}_{\text{a}}$.  By substituting the concentrations from equilibrium table, we can find the hydrogen ion concentration.  The pH value can be calculated by taking negative logarithm of concentration of hydrogen ion.

Interpretation Introduction

Interpretation:

By using the given concentration of acetic acid and sodium acetate, calculate the $\text{pH}$ values.

Concept introduction:

• $\text{pH}$ is the logarithm of the reciprocal of the hydrogen ion concentration in a solution.
•  $\text{pH}$ is used to determine the acidity or basicity of an aqueous solution.
• ${\text{pH = -log[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}$
• $\text{pH}\text{=}{\text{pK}}_{\text{a}}\text{+}\text{log}\frac{\text{[conjugate base]}}{\text{[acid]}}$

To calculate: the $\text{pH}$ of solution contains acetic acid and sodium acetate.

Answer to Problem 16.5QP

The $\text{pH}$ of a solution contains acetic acid and sodium acetate is $\text{4}\text{.44}$

Explanation of Solution

$\begin{array}{l}{\text{CH}}_{\text{3}}{\text{COONa}}_{\text{(aq)}}\to {\text{Na}}^{\text{+}}{}_{\text{(aq)}}{\text{ + CH}}_{\text{3}}{\text{COO}}^{\text{-}}{}_{\text{(aq)}}\\ {\text{CH}}_{\text{3}}{\text{COOH}}_{\text{(aq)}}\stackrel{}{\rightleftarrows }{\text{ H}}^{\text{+}}{}_{\text{(aq)}}{\text{ + CH}}_{\text{3}}{\text{COO}}^{\text{-}}{}_{\text{(aq)}}\\ \\ \text{Initial concentration(M): }\text{0}\text{.40 }\text{0}\text{.00 }\text{0}\text{.20}\\ \text{Change in concentration (M): }\text{-x }\text{+x }\text{+x}\\ \text{Equilibrium}\text{concentration (M): }\text{(0}\text{.40-x) }\text{x }(0.20+\text{x)}\\ \text{The concentration of acetate ion should be added, because sodium acetate fully ionizes }\\ \text{and forms equlibrium with acetic acid}\end{array}$

$\begin{array}{l}{\text{K}}_{\text{a}}\text{=}\frac{{\text{[H}}^{\text{+}}\text{]}{\text{[CH3COO}}^{\text{-}}\text{]}}{\text{[CH3COOH]}}\\ \text{1}\text{.8}{\text{×10}}^{\text{-5}}\text{=}\frac{\text{(x)(0}\text{.20 + x)}}{\text{(0}\text{.40 - x)}}\approx \frac{\text{x(0}\text{.20)}}{\text{0}\text{.40}}\\ \text{the}\text{value}\text{of}\text{x}\text{is}\text{very}\text{small}\text{and}\text{neglect}\text{it,}\text{because the ionization of }\\ {\text{CH}}_{\text{3}}{\text{COOH is reduced by the presence of CH}}_{\text{3}}{\text{COO}}^{\text{-}}\text{ ions}\\ \text{x}\text{=}{\text{[H}}^{\text{+}}\text{]}\text{=}3.6\text{ ×}{\text{10}}^{\text{-5}}\text{M}\\ \text{pH}\text{=}{\text{-log[H}}^{\text{+}}\text{]}\\ \text{=}\text{-log(}3.6\text{ ×}{\text{10}}^{\text{-5}}\text{M)}\\ \text{=}4.44\\ \end{array}$

Another method to calculate $\text{pH}$, by using Henderson-Hasselbalch equation

$\begin{array}{l}\text{pH}\text{=}{\text{pK}}_{\text{a}}\text{+}\text{log}\frac{{\text{[CH}}_{\text{3}}{\text{COO}}^{\text{-}}\text{]}}{{\text{[CH}}_{\text{3}}\text{COOH]}}\\ \text{pH}\text{=}\text{-log(1}\text{.8}\text{×}{\text{10}}^{\text{-5}}\text{)}\text{+}\text{log}\frac{\text{0}\text{.20}\text{M}}{\text{0}\text{.40}\text{M}}\\ \text{=}\text{-log(1}\text{.8}\text{×}{\text{10}}^{\text{-5}}\text{)}\text{+}\text{log}\text{0}\text{.50}\\ \text{=}\text{4}\text{.74}\text{-}\text{0}\text{.30}\\ \text{=}\text{4}\text{.44}\end{array}$

The ratio of concentrations of acetate ion and hydrogen ion to concentration of acetic acid is ${\text{K}}_{\text{a}}$.  By substituting the concentrations from equilibrium table, we can find the hydrogen ion concentration.  The pH value can be calculated by taking negative logarithm of concentration of hydrogen ion.  Alternative way to find $\text{pH}$ value is by Henderson-Hasselbalch equation.  By substituting the ${\text{pK}}_{\text{a}}$ value and logarithm of concentrations of acetate and acetic acid, we can find the $\text{pH}$ value.