A 0.15 M solution of an unknown monoprotic acid has a pH of 4.1, calculate the Ka for this unknown acid.

Question

I got 4.2e-8 for the answer I just don't think I solved it right.

Answer 1

Let's let the dissociation of the acid be represented by HA ↔ H⁺ + A^-.

The equilibrium expression is

K_a = [H⁺] [A^-] / [HA] . As [H⁺] = [A^-] and assuming the degree of dissociation is small. The equation can be simplified to

K_a = [H⁺]² / [HA] where [HA] ≈ 0.15 M

pH = - log [H⁺], take antilog of - pH to get [H⁺] = 10^-pH = 7.9 x 10^-5. Put this into the last expression for K_a and solve.

K_a = (7.9 x 10^-5)² / 0.15 = 4.2 x 10^-8

We are either both right or both wrong!