Chemistry Math Question?

Question

The acid ionization constant of Fe(H₂O)₆²⁺(aq) is 3E-6. Compute the pH of a 0.04 M aqueous solution of Fe(NO₃)₂.

Answer 1

[Fe(H₂O)₆]²⁺ <===> [Fe(H₂O)₅(OH)]⁺ + H⁺ Ka = 3x10^-6

Ka = [H⁺] [Fe(H₂O)₅(OH)]⁺ / [Fe(H2O)₆²⁺] - x

3x10^-6 = (x)(x)/0.04-x

3x10^-6 = (x²)/0.04-x

x = [H⁺] = 3.4x10^-4

pH = -log 3.4x10^-4

pH = 3.47