An unknown weak acid with a concentration of 0.093 M has a pH of 1.80. What is the Ka of the weak acid?

Answer 1

Let HA represent the weak acid.

HA ==> H⁺ + A^-

Ka = [H⁺][A^-] / [HA] = (x)(x) / 0.093 - x

From pH = 1.80, we find the [H⁺] which is also the [A^-] and plug it into the equation.

[H⁺] = [A^-] = 1x10^-1.80 = 1.58x10^-2 M

Ka = (1.58x10^-2)² / 0.093 - 0.0158 = 2.51x10^-4 / 0.0772

Ka = 3.25x10^-3