What is the pH of a 0.1 M solution of HF, given that the acid dissociation constant (Ka) for hydrofluoric acid is 7.1 x 10^-4?

• A. pH = 4.0
• B. pH = 3.5
• C. pH = 1.0
• D. pH = 2.1

Answer: (D)

Verified step by step guidance

1. Start by writing the balanced chemical equation for the dissociation of hydrofluoric acid (HF) in water: \( \text{HF} \rightleftharpoons \text{H}^+ + \text{F}^- \).
2. Use the expression for the acid dissociation constant \( K_a \) for HF: \( K_a = \frac{[\text{H}^+][\text{F}^-]}{[\text{HF}]} \). Given \( K_a = 7.1 \times 10^{-4} \).
3. Assume that the initial concentration of HF is 0.1 M and that the change in concentration of HF upon dissociation is \( x \). Therefore, at equilibrium, \([\text{H}^+] = x\), \([\text{F}^-] = x\), and \([\text{HF}] = 0.1 - x\).
4. Substitute these equilibrium concentrations into the \( K_a \) expression: \( 7.1 \times 10^{-4} = \frac{x^2}{0.1 - x} \).
5. Assume \( x \) is small compared to 0.1, so \( 0.1 - x \approx 0.1 \). This simplifies the equation to \( 7.1 \times 10^{-4} = \frac{x^2}{0.1} \). Solve for \( x \) to find \([\text{H}^+]\), and then calculate the pH using \( \text{pH} = -\log[\text{H}^+] \).