At 25∘C phosphoric acid, H3PO4, has the following equilibrium constants:

Question

H3PO4(aq) + H2O(l) ⇌ H3O+(aq) + H2PO4-(aq) K_a1 = 7.5 x 10^-3

H2PO4-(aq) + H2O(l) ⇌ H3O+(aq) + HPO42-(aq) K_a2 = 6.2 x 10^-8

HPO42-(aq) + H2O(l) ⇌ H3O+(aq) + PO43-(aq) K_a3 = 4.2 x 10^-13

1. What is the pH of a solution of 0.25 M K3PO4, potassium phosphate?
2. What is the pH of a solution of 0.450 M K2HPO4, potassium hydrogen phosphate?
3. What is the pH of a solution of 0.550 M KH2PO4, potassium dihydrogen phosphate?

Answer 1

Since the Ka2 and Ka3 are more than 1000 x the Ka1, we don't really need to consider them when finding the pH of a solution of H₃PO₄.

1). 0.25 M H₃PO₄ ==> H⁺ + H₂PO₄^-

Ka = 7.5x10^-3 = [H⁺][H₂PO₄^-] / [H₃PO₄]

7.5x10^-3 = (x)(x) / 0.25 - x

x² + 7.5x10^-3x - 1.875x10^-3 = 0

x = [H+] = 0.0397 M

pH = -log 0.0397

pH = 1.40

2). 0.450 M HPO4^2- ==> H⁺ + PO4^3- (use Ka3 for this problem)

Ka3 = 4.2x10^-13 = [H⁺][PO₄^3-] / [HPO₄^2-]

4.2x10^-13 = (x)(x) / 0.450 - x (ignore x in denominator as is small relative to 0.45 and no need for quadratic)

x2 = 1.89x10^-13

x = [H⁺] = 4.35x10^-7 M

pH = -log 4.35x10^-7

pH = 6.36 (but this ignores the contribution of H⁺ from the autoionization of H2O. So we need to add 1x10^-7 M H⁺. Thus we have a final H⁺ of 5.35x10^-7 M and pH = 6.27

3). Do the same way as above but use Ka2 to solve