What are the \(\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\) and the \(\mathrm{pH}\) of a buffer that consists of \(0.20 M \mathrm{HF}\) and \(0.25 M \mathrm{KF}\left(K_{\mathrm{a}}\right.\) of \(\left.\mathrm{HF}=6.8 \times 10^{-4}\right) ?\)

The Henderson-Hasselbalch equation is a fundamental equation used to relate the pH of a buffer solution to the concentrations of its acid and conjugate base. The equation is written as: \[ \text{pH} = \text{p}K_\text{a} + \text{log} \frac{[\text{A}^-]}{[\text{HA}]} \] In this context:

• \( [\text{A}^-] \) represents the concentration of the conjugate base (here, fluoride ion \( \text{F}^- \)).
• \( [\text{HA}] \) represents the concentration of the weak acid (here, hydrofluoric acid \( \text{HF} \)).

The equation essentially combines the acid dissociation constant \( K_a \) and the concentrations of the acid and its conjugate base to determine the pH of the buffer solution.

The \( \text{p}K_\text{a} \) is a crucial component when using the Henderson-Hasselbalch equation. It represents the negative logarithm of the acid dissociation constant \( K_a \): \[ \text{p}K_\text{a} = -\text{log} K_\text{a} \] The \( K_a \) value is a measure of the strength of an acid in solution, indicating how easily the acid dissociates into its ions. In the problem, we are given \( K_a \) for HF is \( 6.8 \times 10^{-4} \). Using the equation, we calculate: \[ \text{p}K_\text{a} = -\text{log}(6.8 \times 10^{-4}) \] And find: \[ \text{p}K_\text{a} \rightarrow 3.17 \] This \( \text{p}K_\text{a} \) value is essential for determining the pH of the buffer solution. The smaller the \( \text{p}K_\text{a} \), the stronger the acid.

To find the pH of the buffer solution, we use the Henderson-Hasselbalch equation. For our specific problem, the concentrations of HF (acid) and \( \text{KF} \) (which provides the conjugate base \( \text{F}^- \)) are given as \( 0.20 \text{ M} \) and \( 0.25 \text{ M} \), respectively. Plugging these values into the equation, we get: \[ \text{pH} = 3.17 + \text{log} \frac{0.25}{0.20} \] Simplify the fraction to get 1.25, and calculate: \[ \text{pH} = 3.17 + \text{log}(1.25) \] Using a calculator for the log value, we find: \[ \text{pH} \rightarrow 3.17 + 0.1 = 3.27 \] Thus, the pH of the buffer solution is approximately 3.27. This calculated pH shows that the solution is slightly acidic, which makes sense given the starting materials.

The concentration of hydronium ions \( [\text{H}_3\text{O}^+] \) in a solution is directly connected to the pH. Once we know the pH, we can find \( [\text{H}_3\text{O}^+] \) using the formula: \[ [\text{H}_3\text{O}^+] = 10^{-\text{pH}} \] In our example, with a pH of 3.27, we calculate: \[ [\text{H}_3\text{O}^+] = 10^{-3.27} \] Using a calculator, we get: \[ [\text{H}_3\text{O}^+] = 5.37 \times 10^{-4} \text{ M} \] So, the concentration of hydronium ions in our buffer solution is approximately \( 5.37 \times 10^{-4} \text{ M} \). This value reflects the fact that the solution is slightly acidic, as indicated by the calculated pH.

Understanding acid dissociation is crucial for buffer solutions. When HF dissociates in water, it forms hydrogen ions (\( \text{H}^+ \)) and fluoride ions (\( \text{F}^- \)). This dissociation is represented by the equation: \[ \text{HF} \rightleftharpoons \text{H}^{+} + \text{F}^{-} \] The position of this equilibrium is determined by the acid dissociation constant \( K_a \). A high \( K_a \) value indicates a strong acid that easily dissociates, whereas a low \( K_a \) value indicates a weak acid. For HF, \( K_a = 6.8 \times 10^{-4} \), which shows it is a weak acid. Knowing how HF dissociates helps us calculate the pH of the buffer solution. When \( \text{KF} \) is added, it provides the conjugate base \( \text{F}^- \), shifting the equilibrium and affecting the pH.