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=$\mathbf{1}\mathbf{.}\mathbf{3}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{5}}$)?

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

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 calculatedpHas:

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

Therefore, the values we obtained are:

role="math" localid="1663409335423" $$\begin{gathered} \text{pH = 5.25.} \\ \left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]{\text{= 5.6×10}}^{-6} \end{gathered}$$