The pH of a 0.013 M weak acid solution is 5.27. Calculate K_a for this acid. A) 1.9 *10^-⁹ B) 2.2 * 10^-⁹ C) 5.4* 10^-⁶ D) 2.6*10^-¹⁶ E) 4.2*10^-⁸

Question

Quizplus · Accepted Answer

The equation for the ionization of a weak acid is:

HA ⇌ H+ + A-

The equilibrium expression for this reaction is:

Ka = [H+][A-]/[HA]

Since the acid is weak, we assume that [HA] is approximately equal to the initial concentration (0.013 M).

We are given the pH of the solution, which we can use to calculate [H+]:

pH = -log[H+]

5.27 = -log[H+]

[H+] = 1.9 × 10^-5 M

Since the acid is monoprotic, [H+] = [A-], so we can substitute into the equilibrium expression:

Ka = (1.9 × 10^-5)^2 / 0.013

Ka = 2.2 × 10^-10

Therefore, the answer is B, 2.2 × 10^-10.

The PH of a 0