Question

Calculate the pH of 0.103 M phosphoric acid (H₃PO₄, a triprotic acid). K_a1 = 7.5 x 10^-3, K_a2 = 6.2 x 10^-8, and K_a3 = 4.8 x 10^-13.

Hint, if you are doing much work, you are making the problem harder than it needs to be.

Answer 1

When finding the pH of a solution of a polyprotic acid, almost all of the H₃O⁺ produced comes from the first ionization step (Ka1). The H₃O⁺ obtained from Ka2 and Ka3 are negligible.

The overall reaction:

r: H₃PO₄ + H₂O <---------> H₂PO₄^- + H₃O⁺   Ka1 = 7.5x10^-3

i: 0.103 0 0

e: 0.103-x x x

7.5x10^-3 = x² / 0.103-x as Ka is small, we can neglect the substract of 0.103-x, to 0.103 only so:

7.5x10^-3 * 0.103 = x²

x = 0.0278 M = [H₃O⁺]

pH = -log(0.0278)

pH = 1.56

Hope this helps.