Q19.12P - Chemistry: Molecular Nature Of Matter And Change

Question

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 0.33 M  ${\mathbf{\text{C}}}_{\mathbf{\text{6}}}{\mathbf{\text{H}}}_{\mathbf{\text{5}}}\mathbf{\text{COOH}}$ and 0.28 M  ${\mathbf{\text{C}}}_{\mathbf{\text{6}}}{\mathbf{\text{H}}}_{\mathbf{\text{5}}}\mathbf{\text{COONa}}$ (  ${\mathbf{\text{K}}}_{\mathbf{\text{a}}}$of benzoic acid=$\mathbf{6}\mathbf{.}\mathbf{3}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{5}}$) ?

Step-by-Step Solution

Answer

The pH obtained is:  4.13 and the $\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{ + }}\right]$ obtained is: $\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{ + }}\right]=7.4\times {10}^{-5}$ .

Step 1: Define Ionic Equilibrium

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

Step 2: What is H 3 O  +  and pH of a benzoic acid?

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 calculated pH as:

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

Therefore, the values obtained are:

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