Q19.11P - Chemistry: Molecular Nature Of Matter And Change

Question

What are the $\mathbf{\left[}{\mathbf{\text{H}}}_{\mathbf{\text{3}}}{\mathbf{\text{O}}}^{\mathbf{\text{ + }}}\mathbf{\right]}$ and the pH of a propanoic acid– propanoate buffer that consists of  0.35 M   ${\mathbf{\text{CH}}}_{\mathbf{\text{3}}}{\mathbf{\text{CH}}}_{\mathbf{\text{2}}}\mathbf{\text{COONa }}$and 0.15 M ${\mathbf{\text{CH}}}_{\mathbf{\text{3}}}{\mathbf{\text{CH}}}_{\mathbf{\text{2}}}\mathbf{\text{COOH}}$ ( ${\mathbf{\text{K}}}_{\mathbf{\text{a}}}$ of propanoic acid=$\mathbf{1}\mathbf{.}\mathbf{3}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{5}}$)?

Step-by-Step Solution

Answer

The pH of propanoic acid is: 5.25 and $\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{ + }}\right]$ is obtained as: $\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{ + }}\right]=5.6\times {10}^{-6}$.

Step 1: Ionic Equilibrium

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

Step 2: The H 3 O  +  and the pH of a propanoic acid

Write the  0.15 M  ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{COOH}$ dissociation equation first.

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

Then write the 0.35 M  ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{COONa}$ dissociation equation.H

 ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{COONa}\to {\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{COO}}^{\text{ - }}{\text{ + Na}}^{\text{ + }}.$

We now have the initial propanoic acid and propanoate concentrations.

 $$\begin{gathered} {\text{ CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{COOH = 0.15M} \\ {\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{COO}}^{\text{ - }}\text{ = 0.35M} \end{gathered}$$

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

 $$\begin{gathered} {\text{pK}}_{\text{a}}{\text{ = - logK}}_{\text{a}} \\ {\text{ = - log(1.3×10}}^{-5}\text{)} \\ =4.89 \\ \text{pH = pKa + log}\frac{\left[{\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{COO}}^{\text{ - }}\right]}{\left[{\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{COOH}\right]} \\ \text{ = 4.89 + log}\frac{\text{0.35M}}{\text{0.15M}} \\ \text{ = 5.25.} \end{gathered}$$

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

 $$\begin{gathered} \text{ pH = - log}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{ + }}\right] \\ \left[{\text{H}}_{\text{3}}{\text{O}}^{\text{ + }}\right]{\text{ = 10}}^{\text{ - pH}} \\ {\text{ = 10}}^{\text{ - 5.25}} \\ \text{ = 5.6} \end{gathered}$$

Therefore, the values we obtained are:

  $$\begin{gathered} \text{ pH = 5.25.} \\ \left[{\text{H}}_{\text{3}}{\text{O}}^{\text{ + }}\right]{\text{ = 5.6×10}}^{-6} \end{gathered}$$