Chapter 21, Problem 21.1P

Interpretation Introduction

Interpretation:

$\text{pH}$ of a mixture of $\text{0}\text{.050 M}$ in ${\text{CH}}_{\text{3}}{\text{COOH}}_{\left(\text{aq}\right)}$ and $\text{0}\text{.050 M}$ in ${\text{CH}}_{\text{3}}{\text{COONa}}_{\left(\text{aq}\right)}$ solution has to be calculated.

Concept Introduction:

Henderson-Hasselbalch equation is given by

    $\text{pH}\text{=}{\text{pK}}_{\text{a}}\text{+}\text{log}\frac{\text{[Conjugate}\text{base]}}{\text{[Acid]}}$

Explanation of Solution

Given data,

$\text{0}\text{.050 M}$${\text{CH}}_{\text{3}}{\text{COOH}}_{\left(\text{aq}\right)}$ and $\text{0}\text{.050 M}$${\text{CH}}_{\text{3}}{\text{COONa}}_{\left(\text{aq}\right)}$

From the standard data, $\text{pKa}$ of acetic acid is $4.8$

$\text{pH}$ of given buffer solution can be calculated as

  $\begin{array}{l}{\text{pH=pK}}_{\text{a}}\text{+log}\frac{\text{0}\text{.050M}}{\text{0}\text{.050M}}\\ \\ \text{pH=4}\text{.8+log}\frac{\text{0}\text{.050M}}{\text{0}\text{.050M}}\\ \\ \text{pH=4}\text{.8+log1}\\ \\ \text{pH=4}\text{.8+0=4}\text{.8}\text{.}\end{array}$

Hence, the $\text{pH}$ of a mixture of $\text{0}\text{.050 M}$ in ${\text{CH}}_{\text{3}}{\text{COOH}}_{\left(\text{aq}\right)}$ and $\text{0}\text{.050 M}$ in ${\text{CH}}_{\text{3}}{\text{COONa}}_{\left(\text{aq}\right)}$ is $\text{4}\text{.8}$.