Chapter 16, Problem 16.12QP

Calculate the pH of the following two buffer solutions: (a) 2.0 M CH₃COONa/2.0 M CH₃COOH, (b) 0.20 M CH₃COONa/0.20 M CH₃COOH. Which is the more effective buffer? Why?

Interpretation Introduction

Interpretation:

From the given two different solution concentrations of acetic acid and sodium acetate, $\text{pH}$ of the buffer solution has to be calculated and effective buffer between them has to be explained.

Concept introduction:

• Buffer solution is defined as a solution that oppose changes in $\text{pH}$ while adding little amount of either an acid or a base.
• Buffer solution is a combination of a weak acid and its conjugate base or vice-versa.
• $\text{pH}$ in a solution is defined as the logarithm of the reciprocal of concentration of the hydrogen ion.
•  $\text{pH}$ is used to determine the acidity or basicity of an aqueous solution.
• $\text{pH}\text{=}{\text{pK}}_{\text{a}}\text{+}\text{log}\frac{\text{[conjugate base]}}{\text{[acid]}}$

Answer to Problem 16.12QP

The $\text{pH}$ of the solution containing $\text{2}\text{.0M}$ acetic acid and $\text{2}\text{.0M}$ sodium acetate is $4.74$

Explanation of Solution

To calculate the $\text{pH}$ of buffer solution using the concentrations of components at equilibrium

$\begin{array}{l}{\text{CH}}_{\text{3}}{\text{COOH}}_{\text{(aq)}}\stackrel{}{\rightleftarrows }{\text{ H}}^{\text{+}}{}_{\text{(aq)}}{\text{ + CH}}_{\text{3}}{\text{COO}}^{\text{-}}{}_{\text{(aq)}}\\ \\ \text{Initial concentration(M): }\text{2}\text{.0 }\text{0}\text{.00 }\text{2}\text{.0}\\ \text{Change in concentration (M): }\text{-x }\text{+x }\text{+x}\\ \text{Equilibrium}\text{concentration (M): }\text{(2}\text{.0-x) }\text{x }\text{2}\text{.0}\text{+}\text{x}\end{array}$

$\begin{array}{l}{\text{K}}_{\text{a}}\text{=}\frac{{\text{[H}}^{\text{+}}\text{]}{\text{[CH3COO}}^{\text{-}}\text{]}}{\text{[CH3COOH]}}\\ {\text{K}}_{\text{a}}\text{=}\frac{{\text{[H}}^{\text{+}}\text{](2}\text{.0}\text{+}\text{x)}}{\text{(2}\text{.0-x)}}\approx \frac{{\text{[H}}^{\text{+}}\text{](2}\text{.0)}}{\text{2}\text{.0}}\\ \text{the}\text{value}\text{of}\text{x}\text{is}\text{very}\text{small}\text{and}\text{neglect}\text{it,}\text{because the ionization of }\\ {\text{CH}}_{\text{3}}{\text{COOH is reduced by the presence of CH}}_{\text{3}}{\text{COO}}^{\text{-}}\text{ ions}\\ {\text{K}}_{\text{a}}\text{=}{\text{[H}}^{\text{+}}\text{]}\\ {\text{-logK}}_{\text{a}}\text{=}{\text{-log[H}}^{\text{+}}\text{]}\\ \text{By taking -log on both sides we get,}\\ {\text{pK}}_{\text{a}}\text{ = pH}\\ {\text{K}}_{\text{a}}{\text{ value for CH}}_{\text{3}}\text{COOH is 1}\text{.8 ×}{\text{10}}^{\text{-5}}\\ \text{pH}\text{=}{\text{-logK}}_{\text{a}}\\ \text{=}\text{-log(1}\text{.8}\text{×}{\text{10}}^{\text{-5}}\text{M)}\\ \text{=}\text{4}\text{.74}\end{array}$

The $\text{pH}$ of the buffer solution can be calculated using concentrations of components at equilibrium.  In this solution, the concentration of conjugate base and acid are same, then Henderson–Hasselbalch equation becomes ${\text{pK}}_{\text{a}}\text{ = pH}$.  By substituting the values of ${\text{K}}_{\text{a}}$ the $\text{pH}$ value can be calculated.

Interpretation Introduction

Interpretation:

From the given two different solution concentrations of acetic acid and sodium acetate, $\text{pH}$ of the buffer solution has to be calculated and effective buffer between them has to be explained.

Concept introduction:

• Buffer solution is defined as a solution that oppose changes in $\text{pH}$ while adding little amount of either an acid or a base.
• Buffer solution is a combination of a weak acid and its conjugate base or vice-versa.
• $\text{pH}$ in a solution is defined as the logarithm of the reciprocal of concentration of the hydrogen ion.
•  $\text{pH}$ is used to determine the acidity or basicity of an aqueous solution.
• $\text{pH}\text{=}{\text{pK}}_{\text{a}}\text{+}\text{log}\frac{\text{[conjugate base]}}{\text{[acid]}}$

Answer to Problem 16.12QP

The $\text{pH}$ of the solution containing $\text{0}\text{.2M}$ acetic acid and $\text{0}\text{.2M}$ sodium acetate is $4.74$

Buffer solution (a) is more effective than buffer (b).  Because, the concentrations of spices in buffer (a) is ten times more than buffer(b).  The neutralisation of buffer (a) by an acid or base can be ten times to buffer (b).

Explanation of Solution

To calculate the $\text{pH}$ of buffer solution using the concentrations of components at equilibrium

$\begin{array}{l}{\text{CH}}_{\text{3}}{\text{COOH}}_{\text{(aq)}}\stackrel{}{\rightleftarrows }{\text{ H}}^{\text{+}}{}_{\text{(aq)}}{\text{ + CH}}_{\text{3}}{\text{COO}}^{\text{-}}{}_{\text{(aq)}}\\ \\ \text{Initial concentration(M): }\text{0}\text{.20 }\text{0}\text{.00 }\text{0}\text{.20}\\ \text{Change in concentration (M): }\text{-x }\text{+x }\text{+x}\\ \text{Equilibrium}\text{concentration (M): }\text{(0}\text{.20}\text{-}\text{x) }\text{x }\text{0}\text{.20}\text{+}\text{x}\end{array}$

$\begin{array}{l}{\text{K}}_{\text{a}}\text{=}\frac{{\text{[H}}^{\text{+}}\text{]}{\text{[CH3COO}}^{\text{-}}\text{]}}{\text{[CH3COOH]}}\\ {\text{K}}_{\text{a}}\text{=}\frac{{\text{[H}}^{\text{+}}\text{](0}\text{.20}\text{+}\text{x)}}{\text{(0}\text{.20-x)}}\approx \frac{{\text{[H}}^{\text{+}}\text{](0}\text{.20)}}{\text{0}\text{.20}}\\ \text{the}\text{value}\text{of}\text{x}\text{is}\text{very}\text{small}\text{and}\text{neglect}\text{it,}\text{because the ionization of }\\ {\text{CH}}_{\text{3}}{\text{COOH is reduced by the presence of CH}}_{\text{3}}{\text{COO}}^{\text{-}}\text{ ions}\\ {\text{K}}_{\text{a}}\text{=}{\text{[H}}^{\text{+}}\text{]}\\ {\text{-logK}}_{\text{a}}\text{=}{\text{-log[H}}^{\text{+}}\text{]}\\ \text{By taking -log on both sides we get,}\\ {\text{pK}}_{\text{a}}\text{ = pH}\\ {\text{K}}_{\text{a}}{\text{ value for CH}}_{\text{3}}\text{COOH is 1}\text{.8 ×}{\text{10}}^{\text{-5}}\\ \text{pH}\text{=}{\text{-logK}}_{\text{a}}\\ \text{=}\text{-log(1}\text{.8}\text{×}{\text{10}}^{\text{-5}}\text{M)}\\ \text{=}\text{4}\text{.74}\end{array}$

The $\text{pH}$ of the buffer solution can be calculated using concentrations of components at equilibrium.  In this solution the concentration of conjugate base and acid are same, then Henderson–Hasselbalch equation becomes ${\text{pK}}_{\text{a}}\text{ = pH}$.  By substituting the values of ${\text{K}}_{\text{a}}$ the $\text{pH}$ value can be calculated.

To compare: The effectiveness of buffer (a) and buffer (b)

Buffer solution (a) is more effective than buffer (b).  Because, the concentrations of spices in buffer (a) is ten times more than buffer(b).  The neutralisation of buffer (a) by an acid or base can be ten times to buffer (b).