Question

A 0.425-M aqueous solution of a weak acid has a pH of 3.6. Calculate K_a for the acid.

K_a =

Answer 1

Let the acid be HA

then HA + H₂O <-----> H₃O⁺ + A^-

Intial   0.425                  0        0

Change x                       x        x

Equilibrium 0.425-x         x        x

pH = 3.6

pH = -log[H₃O⁺]

[H₃O⁺] = 10 ^-pH = 10^-3.6 = 0.00025

[HA] = 0.425 - x = 0.425 - 0.00025 = 0.4247

Ka = [H₃O⁺][A^-] / [HA] = (0.00025)(0.00025)/(0.4247) = 1.47 x 10^-7