What is the pH of a 0.400 M solution of HF (Ka = 6.8 × 10^{-4})?

• A. 4.00
• B. 1.87
• C. 2.17
• D. 3.42

Answer: (C)

Verified step by step guidance

1. Write the dissociation equation for HF in water: \(\mathrm{HF} \rightleftharpoons \mathrm{H^+} + \mathrm{F^-}\).
2. Set up an ICE table (Initial, Change, Equilibrium) to express the concentrations: Initial HF = 0.400 M, initial \(\mathrm{H^+}\) and \(\mathrm{F^-}\) = 0, changes in concentration are \(-x\) for HF and \(+x\) for both \(\mathrm{H^+}\) and \(\mathrm{F^-}\), equilibrium concentrations are \$0.400 - x\( for HF and \)x$ for both ions.
3. Write the expression for the acid dissociation constant \(K_a\): \(K_a = \frac{[\mathrm{H^+}][\mathrm{F^-}]}{[\mathrm{HF}]} = \frac{x^2}{0.400 - x}\), where \(x\) represents the concentration of \(\mathrm{H^+}\) at equilibrium.
4. Assuming \(x\) is small compared to 0.400, approximate the denominator as 0.400 and solve for \(x\) using \(x = \sqrt{K_a \times 0.400}\).
5. Calculate the pH using the formula \(\mathrm{pH} = -\log_{10}(x)\), where \(x\) is the concentration of \(\mathrm{H^+}\) ions.