Chapter 16.8, Problem 10PE

Interpretation Introduction

Interpretation: The change in $\text{pH}$ of $\text{1}\text{.00}\text{L}$ of a solution that contains $\text{1}\text{M}{\text{CH}}_{\text{3}}\text{COOH}$ and $\text{1}\text{M}{\text{CH}}_{\text{3}}\text{COONa}$ after the addition of $\text{10}\text{.00}\text{mL}$ of a $\text{0}\text{.100}\text{M}\text{NaOH}$ is to be calculated.

Concept introduction: The $\text{pH}$ of the solution is calculated by Henderson-Hasselbalch equation,

$\text{pH}=\text{p}{K}_{\text{a}}+\text{log}\left(\frac{\left[\text{Salt}\right]}{\left[\text{Acid}\right]}\right)$

Where,

• $\text{p}{K}_{\text{a}}$ is the acid dissociation constant.

The molarity of the solution is calculated by the formula,

$\begin{array}{c}\text{Molarity}=\frac{\text{Amount}\text{of}\text{solute}\left(\text{in}\text{moles}\right)}{\text{Volume}}\\ \text{Moles}=\text{Molarity}\times \text{Volume}\end{array}$

To determine: The change in $\text{pH}$ of $\text{1}\text{.00}\text{L}$ of a solution that contains $\text{1}\text{M}{\text{CH}}_{\text{3}}\text{COOH}$ and $\text{1}\text{M}{\text{CH}}_{\text{3}}\text{COONa}$ after the addition of $\text{10}\text{.00}\text{mL}$ of a $\text{0}\text{.100}\text{M}\text{NaOH}$.