Question

in the laboratory, a general chemistry student measured the ph of a 0.501 m aqueous solution of hydrofluoric acid to be 1.739. use the information she obtained to determine the ( k_a ) for this acid. ( k_a(\text{experiment}) = )

Explanation:

Step1: Calculate $[H^+]$ from pH

$[H^+] = 10^{-\text{pH}} = 10^{-1.739}$
$[H^+] \approx 0.0182\ \text{M}$

Step2: Define dissociation equilibrium

Hydrofluoric acid dissociates: $\text{HF}
ightleftharpoons H^+ + F^-$
At equilibrium: $[H^+] = [F^-] = 0.0182\ \text{M}$
$[\text{HF}] = 0.501 - 0.0182 = 0.4828\ \text{M}$

Step3: Apply $K_a$ formula

$K_a = \frac{[H^+][F^-]}{[\text{HF}]}$
$K_a = \frac{(0.0182)^2}{0.4828}$

Step4: Compute final $K_a$

$K_a = \frac{3.3124 \times 10^{-4}}{0.4828} \approx 6.86 \times 10^{-4}$

Answer:

$6.9 \times 10^{-4}$ (or $6.86 \times 10^{-4}$ for more precision)