What is the actual concentration of the molecular form of HF in a 1.0 M HF solution given that
K_a of HF is 6.8 * 10^-⁴?

Question

What is the actual concentration of the molecular form of HF in a 1.0 M HF solution given that K_a of HF is 6.8 * 10^-⁴?

Quizplus · Accepted Answer

The equilibrium expression for the dissociation of HF is: HF ⇌ H+ + F- K_a = [H+][F-]/[HF] Since the concentration of HF is 1.0 M, we can assume that the initial concentration of HF is also 1.0 M. Let x be the concentration of H+ and F-. At equilibrium, the concentration of HF will be 1.0 - x. Therefore, K_a = [H+][F-]/[HF] 6.8 F1F1F1S1F1F1F10 10^- = x^2 / (1.0 - x) Because the value of x will be small compared to the value of 1.0, we can assume that the value of 1.0 - x is approximately 1.0. Then, 6.8 F1F1F1S1F1F1F10 10^- ≈ x^2 x ≈ 8.25 F1F1F1S1F1F1F10 10^- M Since we are asked for the concentration of the molecular form of HF, which is equal to 1.0 - x, the answer is: Answer: B The actual concentration of the molecular form of HF in a 1.0 M HF solution is 1.0 M - 8.25 F1F1F1S1F1F1F10 10^- M = 0.97 M. Therefore, the best choice is B.

What Is the Actual Concentration of the Molecular Form of HF