Question

What is the pH of H2PO4-, HPO42-, and PO43-. All listed are 0.1M. Please show work. I know the last one is 12, but why?

Answer 1

H₂PO₄^- <--> HPO₄^-2 + H⁺
Ka= 6.2 x 10^-8

Ka = [HPO₄^-2] [H⁺] / [H₂PO₄^-1]
6.2 x 10^-8 = [X] [X] / [0.1]
X² = 6.2 X 10-7
X = [H⁺] = 7.87 X 10-4
pH = - log[H⁺] = 3.10

HPO₄^-2 <---> PO₄^-3 + H⁺
Ka= 4.8 x 10^-13
Ka = [PO₄^-3] [H⁺] / [HPO₄^-2]
4.8 X 10^-13 = [X] [X] / [0.1]
X² =4.8 X 10^-12
X = [H⁺] = 2.19 X 10^-6
pH = -log[H⁺] = 5.66

The conjugate base PO₄^-3 does a hydrolysis
PO₄^-3 + H₂O <---> HPO₄^-2 + OH^-
since conjugate pair's have......... Ka times Kb = K water
Kb = (1 X 10^-14) / 4.8 X 10^-13
Kb = 2.08 X 10^-2

Kb = [HPO₄^-2] [OH^-] / PO₄^-3]

2.08 X 10^-2 = [X] [X] / [0.1 -X]

2.08 X 10^-2 [0.1 -X] = X²

(0.00208) - (0.0208X) = X²

X² + (0.0208X) - (0.00208) = 0
using the quadratic equation to calculate root
X = = [OH^-] = 0.0364
pOH = -log[OH^-] = 1.44
so, since pH + pOH = 14
pH = 14-1.44 = 12.56