Q19.14P - Chemistry: Molecular Nature Of Matter And Change

Question

What are the $\mathbf{[}{\mathbf{\text{H}}}_{\mathbf{\text{3}}}{\mathbf{\text{O}}}^{\mathbf{\text{ + }}}\mathbf{\text{]}}$ and the pH of a buffer that consists of  0.20 M HF and  0.25 M KF ( ${\mathbf{\text{K}}}_{\mathbf{\text{a}}}$ of $\mathit{H}\mathit{F}\mathbf{=}\mathbf{6}\mathbf{.}\mathbf{8}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{4}}$ )?

Step-by-Step Solution

Answer

The pH is obtained is  3.26 and the $\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{ + }}\right]$  obtained is $5.4\times {10}^{-4}$.

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: Find the H 3 O  +  and pH

Write the 0.20 M HF dissociation equation first.

 ${\text{HF + H}}_{\text{2}}\text{O}\rightleftharpoons {\text{F}}^{\text{ - }}{\text{ + H}}_{\text{3}}{\text{O}}^{\text{ + }}.$

Then, write the  0.25 M KF dissociation equation.

 $\text{KF}\to {\text{F}}^{\text{ - }}{\text{ + K}}^{\text{ + }}.$

We now have initial hydrofluoric acid and fluoride.

 $$\begin{gathered} \text{ HF = 0.20 M} \\ {\text{F}}^{\text{ - }}\text{ = 0.25 M} \end{gathered}$$

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

  $$\begin{gathered} {\text{pK}}_{\text{a}}{\text{ = - logK}}_{\text{a}} \\ {\text{ = - log(6.8×10}}^{-4}\text{)} \\ =3.17 \\ \text{pH = pKa + log}\frac{\left[{\text{F}}^{\text{ - }}\right]}{\left[\text{HF}\right]} \\ \text{ = 3.17 + log}\frac{\text{0.25 M}}{\text{0.20 M}} \\ \text{ = 3.26.} \end{gathered}$$

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

  $$\begin{gathered} \text{ pH = - log}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{ + }}\right] \\ \left[{\text{H}}_{\text{3}}{\text{O}}^{\text{ + }}\right]{\text{ = 10}}^{\text{ - pH}} \\ {\text{ = 10}}^{\text{ - 3.26}} \\ {\text{ = 5.4×10}}^{-4} \end{gathered}$$

Therefore, the values obtained are:

 $$\begin{gathered} \text{ pH = 3.26.} \\ \left[{\text{H}}_{\text{3}}{\text{O}}^{\text{ + }}\right]{\text{ = 5.4×10}}^{-4} \end{gathered}$$