Question

A 0.10 M solution of HF is 8.0% dissocaited What is the K_(a)

Answer 1

The $K_a$ is $7.0 \times 10^{-4}$.

Answer 2

Explanation of the solution:

1. Define the equilibrium: Hydrofluoric acid (HF) is a weak acid that dissociates in water according to the equilibrium: $HF(aq) \rightleftharpoons H^+(aq) + F^-(aq)$

2. Initial concentrations: Initial $[HF] = C = 0.10$ M Initial $[H^+] = 0$ Initial $[F^-] = 0$

3. Degree of dissociation ($\alpha$): The solution is 8.0% dissociated, so $\alpha = 8.0\% = 0.08$.

4. Equilibrium concentrations: At equilibrium, the change in concentration of HF is $-C\alpha$, and the change in concentrations of $H^+$ and $F^-$ is $+C\alpha$. $[HF]_{eq} = C - C\alpha = C(1-\alpha)$ $[H^+]_{eq} = C\alpha$ $[F^-]_{eq} = C\alpha$

   Substitute the given values: $[HF]_{eq} = 0.10 (1 - 0.08) = 0.10 \times 0.92 = 0.092$ M $[H^+]_{eq} = 0.10 \times 0.08 = 0.008$ M $[F^-]_{eq} = 0.10 \times 0.08 = 0.008$ M

5. Expression for $K_a$: The acid dissociation constant ($K_a$) is given by: $K_a = \frac{[H^+][F^-]}{[HF]}$

6. Calculate $K_a$: Substitute the equilibrium concentrations into the $K_a$ expression: $K_a = \frac{(0.008)(0.008)}{0.092}$ $K_a = \frac{0.000064}{0.092}$ $K_a \approx 0.00069565$

   Rounding to two significant figures (consistent with the given data 0.10 M and 8.0%): $K_a \approx 7.0 \times 10^{-4}$