Acid and bases/chemistry

Question

What is the pH of a 0.023M solution of a weak acid with Ka= 4.9x10^-10

Answer 1

Let the weak acid be HA

HA ==> H⁺ + A^-

Ka = 4.9x10^-10 = [H+][A-] / [HA] = (x)(x) / 0.023-x

assume x is small to avoid using quadratic…

4.9x10^-10 = x²/.023

x = 3.36x10^-6 M = [H⁺]

pH = - log 3.36x10^-6 = 5.47