Chapter 18, Problem 103A

Interpretation Introduction

Interpretation:

The $\text{p}\text{H}$ value of the given solution is to be calculated which is buffered by benzoic acid and sodium benzoate.

Concept introduction:

A buffer solution is a solution in which no change in $\text{p}\text{H}$ takes place on adding a small amount of acid and base.

The value of the acid ionization constant tells about the state of whether reactants or products are more favored in the reaction.

For a reaction,

$\text{H}\text{A}+{\text{H}}_2\text{O}\rightleftharpoons {\text{A}}^{-}+{\text{H}}_3{\text{O}}^{+}$

Acid ionization constant is given by:

$\begin{array}{l}{\text{k}}_{\text{a}}=\frac{\left[{\text{A}}^{-}\right]\times \left[{\text{H}}_3{\text{O}}^{+}\right]}{\left[\text{H}\text{A}\right]}\end{array}$

  $\begin{array}{c}\text{w}\text{h}\text{e}\text{r}\text{e},\end{array}$

  $\begin{array}{c}\left[{\text{A}}^{-}\right]\text{i}\text{s}\text{c}\text{o}\text{n}\text{c}\text{e}\text{n}\text{t}\text{r}\text{a}\text{t}\text{i}\text{o}\text{n}\text{o}\text{f}{\text{A}}^{-}\end{array}$

  $\begin{array}{c}\left[{\text{H}}_3{\text{O}}^{+}\right]\text{i}\text{s}\text{c}\text{o}\text{n}\text{c}\text{e}\text{n}\text{t}\text{r}\text{a}\text{t}\text{i}\text{o}\text{n}\text{o}\text{f}{\text{H}}_3{\text{O}}^{+}\end{array}$

  $\begin{array}{c}\left[\text{H}\text{A}\right]\text{i}\text{s}\text{c}\text{o}\text{n}\text{c}\text{e}\text{n}\text{t}\text{r}\text{a}\text{t}\text{i}\text{o}\text{n}\text{o}\text{f}\text{H}\text{A}\end{array}$

Answer to Problem 103A

$\text{p}\text{H}=4.19$

Explanation of Solution

Benzoic acid and sodium benzoate dissociate as:

$\begin{array}{l}{\text{C}}_6{\text{H}}_5\text{C}\text{O}\text{O}\text{H}\rightleftharpoons {\text{H}}^{+}+{\text{C}}_6{\text{H}}_5\text{C}\text{O}{\text{O}}^{-}\end{array}$

  $\begin{array}{c}{\text{C}}_6{\text{H}}_5\text{C}\text{O}\text{O}\text{N}\text{a}\to \text{N}{\text{a}}^{+}+{\text{C}}_6{\text{H}}_5\text{C}\text{O}{\text{O}}^{-}\end{array}$

  ${\text{k}}_{\text{a}}=\frac{\left[{\text{C}}_6{\text{H}}_5\text{C}\text{O}{\text{O}}^{-}\right]\times \left[{\text{H}}^{+}\right]}{\left[{\text{C}}_6{\text{H}}_5\text{C}\text{O}\text{O}\text{H}\right]}$

Given data:

  $\begin{array}{l}{\text{k}}_{\text{a}}=6.4\times {10}^{-5}\end{array}$

  $\begin{array}{c}\left[{\text{C}}_6{\text{H}}_5\text{C}\text{O}{\text{O}}^{-}\right]=0.0500\text{M}\end{array}$

  $\begin{array}{c}\left[{\text{C}}_6{\text{H}}_5\text{C}\text{O}\text{O}\text{H}\right]=0.0500\text{M}\end{array}$

Therefore,

$\begin{array}{c}\left[{\text{H}}^{+}\right]=\frac{6.4\times {10}^{-5}\times 0.05}{0.05}\end{array}$

  $\begin{array}{c}=6.4\times {10}^{-5}\text{M}\end{array}$

$\begin{array}{c}\text{p}\text{H}=-\text{log}(6.4\times {10}^{-5})\end{array}$

  $\begin{array}{c}=-\text{log}6.5-\text{log}{10}^{-5}\end{array}$

  $\begin{array}{c}=-0.81+5\end{array}$

  $\begin{array}{c}=4.19\end{array}$

Conclusion

Thus $\text{p}\text{H}$ value of given solution is $4.19$