When 0.20 mole HF is dissolved in water to a volume of 1.00 L,5.8% of the HF dissociates to form F^-(aq).What is the equilibrium constant for the reaction?

Question

Quizplus · Accepted Answer

The balanced equation for the dissociation of HF is:

HF (aq) ⇌ H+ (aq) + F- (aq)

According to the problem, 5.8% of the HF dissociates, which means that the concentration of H+ is 5.8% of the initial concentration of HF, and the concentration of F- is also 5.8% of the initial concentration of HF.

Therefore, at equilibrium, the concentrations are:

[H+] = 0.058 mol/L
[F-] = 0.058 mol/L
[HF] = 0.20 - 0.058 = 0.142 mol/L

The equilibrium constant expression is:

Kc = [H+][F-]/[HF]

Substituting the values obtained above, we get:

Kc = (0.058)(0.058)/(0.142) = 7.1 x 10^-4

Using scientific notation and rounding to one significant figure, the answer is B: 7.1 x 10^-4.

When 020 Mole HF Is Dissolved in Water to a Volume