What is the concentration (in units of molarity) of HCrO−4 in a 0.073 M solution of chromic acid, H2CrO4?

Question

What is the concentration (in units of molarity) of HCrO−4 in a 0.073 M solution of chromic acid, H2CrO4?

H2CrO4(aq)+H2O(l)↽−−⇀H3O+(aq)+HCrO−4(aq)    Ka1=9.6

HCrO−4(aq)+H2O(l)↽−−⇀H3O+(aq)+CrO2−4(aq)    Ka2=3.2×10−7

Answer 1

H₂CrO₄ <==> H⁺ + HCrO₄^- Ka = 9.6

HCrO₄^- <==> H⁺ + CrO₄^2- Ka = 3.2x10^-7

Like sulfuric acid (H₂SO₄) chromic acid is a strong acid but only for the first ionization. Since both Ka values are given, apparently they want you to consider both ionizations.

For the first ionization:

9.6 = (x)(x)/0.073-x

x² = 0.70 - 9.6x and x² + 9.6x - 0.70 = 0 and x = [HCrO4^-] = 0.072 M

For the 2nd ionization:

HCrO₄^- <==> H⁺ + CrO₄^2-

3.2x10^-7 = x²/0.072-x

x² + 3.2x10^-7x - 2.3x10^-8 = 0 and x = 1.5x10^-4

[HCrO₄^-] = 0.072 - 1.5x10^-4 = 0.07185 M = 0.072 M (to 2 sig. figs.) NOTE: this is the same as just using the first ionization.