The pH of a 0.0338 M solution of a weak acid is 2.85. What is K _aof this acid?

Question

Quizplus · Accepted Answer

From the given data, we can write the expression for the ionization of the weak acid as follows:
HA ⇌ H+ + A-
Let's assume the initial concentration of the weak acid as 'x', then the concentration of H+ and A- ions will also be 'x' at equilibrium. Therefore, we can write the equilibrium concentration of the weak acid as (0.0338 - x).
The expression for the ionization constant (Ka) of the weak acid is given by:
Ka = [H+] [A-] / [HA]
At equilibrium, we have:
[H+] = x
[A-] = x
[HA] = (0.0338 - x)
Substituting these values in the expression for Ka, we get:
Ka = x * x / (0.0338 - x)
Given that pH = 2.85, we can write:
pH = -log[H+]
2.85 = -log(x)
x = 5.9 × 10^-3 M
Substituting this in the expression for Ka, we get:
Ka = (5.9 × 10^-3)^2 / (0.0338 - 5.9 × 10^-3)
Ka = 5.9 × 10^-5 / 0.0289
Ka = 2.04 × 10^-3
Therefore, the answer is (E) 5.9×10^-5.

The PH of a 0