What are the ${\mathbf{\text{[H}}}_{\mathbf{\text{3}}}{\mathbf{\text{O}}}^{\mathbf{\text{+}}}\mathbf{\text{]}}$and the pH of a benzoic acid–benzoate buffer that consists of 0.33M ${\mathbf{\text{C}}}_{\mathbf{\text{6}}}{\mathbf{\text{H}}}_{\mathbf{\text{5}}}\mathbf{\text{COOH}}$and 0.28M ${\mathbf{\text{C}}}_{\mathbf{\text{6}}}{\mathbf{\text{H}}}_{\mathbf{\text{5}}}\mathbf{\text{COONa}}$( ${\mathbf{\text{K}}}_{\mathbf{\text{a}}}$of benzoic acid=$\mathbf{6}\mathbf{.}\mathbf{3}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{5}}$) ?

An ionic equilibrium refers to an equilibrium that exists in the solution of weak electrolytes between unionized molecules and ions.

Write the 0.33 M${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}\text{COOH}$dissociation equation first.

${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{COOH +H}}_{\text{2}}\text{O}\rightleftharpoons {\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{COO}}^{\text{-}}{\text{+H}}_{\text{3}}{\text{O}}^{\text{+}}.$

Then, write the 0.28 M ${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}\text{COONa}$ dissociation equation.

${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}\text{COONa}\to {\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{COO}}^{\text{-}}{\text{+ Na}}^{\text{+}}.$

We now have the initial benzoic acid and benzonate concentrations.

$$\begin{gathered} {\text{C}}_{\text{6}}{\text{H}}_{\text{5}}\text{COOH = 0.33 M} \\ {\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{COO}}^{\text{-}}\text{= 0.28 M} \end{gathered}$$

To find the pH, we may utilize the Henderson-Hasselbalch equation. First, solve for the ${\text{pK}}_{\text{a}}$.

$$\begin{gathered} {\text{pK}}_{\text{a}}{\text{= - logK}}_{\text{a}} \\ {\text{= - log(6.3×10}}^{-5}\text{)} \\ =4.20 \\ \text{pH = pKa + log}\frac{\left[{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{COO}}^{\text{-}}\right]}{\left[{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}\text{COOH}\right]} \\ \text{= 4.20 + log}\frac{\text{0.28 M}}{\text{0.33 M}} \\ \text{= 4.13.} \end{gathered}$$

Then, solve for $\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]$ from the calculatedpHas:

$$\begin{gathered} \text{pH = - log}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right] \\ \left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]{\text{= 10}}^{\text{- pH}} \\ {\text{= 10}}^{\text{- 4.13}} \\ {\text{= 7.4×10}}^{-5} \end{gathered}$$

Therefore, the values obtained are:

$$\begin{gathered} \text{pH = 4.13.} \\ \left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]{\text{= 7.4×10}}^{-5} \end{gathered}$$