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 $\mathit{H}\mathit{F}\mathbf{=}\mathbf{6}\mathbf{.}\mathbf{8}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{4}}$ )?

Short 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-by-step solution

Ionic Equilibrium

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

Find the H3O + 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 MKFdissociation 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${\text{pK}}_{\text{a}}$.

role="math" localid="1663405063668" $$\begin{gathered} {\text{pK}}_{\text{a}}{\text{= - logK}}_{\text{a}} \\ {\text{= - log(6.8×10}}^{-4}\text{)} \\ =3.17 \\ \text{pH = pKa + log}\frac{\left[{\text{F}}^{\text{-}}\right]}{\left[\text{HF}\right]} \\ \text{= 3.17 + log}\frac{\text{0.25 M}}{\text{0.20 M}} \\ \text{= 3.26.} \end{gathered}$$

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

role="math" localid="1663405281683" $$\begin{gathered} \text{pH = - log}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right] \\ \left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]{\text{= 10}}^{\text{- pH}} \\ {\text{= 10}}^{\text{- 3.26}} \\ {\text{= 5.4×10}}^{-4} \end{gathered}$$

Therefore, the values obtained are:

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