What is the H₃O⁺ concentration of a 0.250 M hydrofluoric acid (HF) solution, given that the acid dissociation constant (Ka) for HF is 6.8 x 10⁻⁴?

• A. 1.3 x 10⁻² M
• B. 6.8 x 10⁻⁴ M
• C. 4.1 x 10⁻³ M
• D. 2.5 x 10⁻² M

Answer: (C)

Verified step by step guidance

1. Start by writing the chemical equation for the dissociation of hydrofluoric acid (HF) in water: \( \text{HF} + \text{H}_2\text{O} \rightleftharpoons \text{H}_3\text{O}^+ + \text{F}^- \).
2. Use the expression for the acid dissociation constant \( K_a \) for HF: \( K_a = \frac{[\text{H}_3\text{O}^+][\text{F}^-]}{[\text{HF}]} \).
3. Assume that the initial concentration of HF is 0.250 M and that the change in concentration due to dissociation is \( x \). Therefore, \([\text{H}_3\text{O}^+] = x \), \([\text{F}^-] = x \), and \([\text{HF}] = 0.250 - x \).
4. Substitute these expressions into the \( K_a \) equation: \( 6.8 \times 10^{-4} = \frac{x^2}{0.250 - x} \).
5. Solve the equation for \( x \), which represents the \([\text{H}_3\text{O}^+] \) concentration. You can assume \( x \) is small compared to 0.250 M, simplifying the equation to \( 6.8 \times 10^{-4} = \frac{x^2}{0.250} \).