Question

question 12 of 32
what is the equation for the acid dissociation constant, $k_a$, of hf?

$\ce{hf(aq) + h2o(l) <=> h3o+(aq) + f-(aq)}$

$\boldsymbol{\text{o a. } k_a = \frac{\ce{h3o+}\ce{f-}}{\ce{hf}\ce{h2o}} }$
$\boldsymbol{\text{o b. } k_a = \frac{\ce{hf}}{\ce{h3o+}\ce{f-}} }$
$\boldsymbol{\text{o c. } k_a = \frac{\ce{h3o+}\ce{f-}}{\ce{hf}} }$
$\boldsymbol{\text{o d. } k_a = \frac{\ce{hf}\ce{h2o}}{\ce{h3o+}\ce{f-}} }$

Explanation:

Step1: Recall acid dissociation constant formula

For a reaction \( aA + bB
ightleftharpoons cC + dD \), the equilibrium constant \( K = \frac{[C]^c[D]^d}{[A]^a[B]^b} \). For acid dissociation, pure liquids (like \( H_2O(l) \)) have activity 1, so their concentration is not included.

Step2: Apply to HF dissociation

Reaction: \( \text{HF}(aq) + \text{H}_2\text{O}(l)
ightleftharpoons \text{H}_3\text{O}^+(aq) + \text{F}^-(aq) \). Exclude \( [\text{H}_2\text{O}] \). So \( K_a = \frac{[\text{H}_3\text{O}^+][\text{F}^-]}{[\text{HF}]} \).

Answer:

C. \( K_{a} = \frac{[\text{H}_3\text{O}^+][\text{F}^-]}{[\text{HF}]} \)