Calculate the pH for this buffer using Ka1, Ka2, and Ka3

Question

A buffer consists of 0.35 M Na₂HPO₄ and 0.35 M Na₃PO₄. Given that the K values for H₃PO₄ are, K_a1 = 7.2 x 10^-3, K_a2 = 6.3 x 10^-8, and K_a3 = 4.2 x 10^-13, calculate the pH for this buffer.

the answer is: 12.38

I am completely lost with this question, please show steps so I can follow along. Thank you!

Answer 1

Hi Janel,

1) The first key is to recognize that the solution described in the problem (0.35 M Na₂HPO₄ and 0.35 M Na₃PO₄ ) is a buffer solution because it contains HPO₄^2- and PO₄^3- ions floating around in the beaker. The HPO₄^2- and PO₄^3- ions are conjugates of each other, thus they form a buffer solution. The Na⁺ ions are just spectator ions and don't affect the solution.

2) The second key is to recognize that H₃PO₄ is a triprotic acid that will undergo three successive dissociations, as follows:

K_a1 refers to the equilibrium reaction for the 1st dissocation: H₃PO₄ + H₂O <=> H₂PO₄^- + H₃O⁺

K_a2 refers to the equilibrium reaction for the 2nd dissocation: H₂PO₄^- + H₂O <=> HPO₄^2- + H₃O⁺

K_a3 refers to the equilibrium reaction for the 3rd dissocation: HPO₄^2-+ H₂O <=> PO₄^3- + H₃O⁺

3) The problem asks about a buffer with HPO₄^2- and PO₄^3- ions, and the equilibrium reaction for K_a3 contains the HPO₄^2- and PO₄^3- ions. Therefore, you use K_a3 in the Henderson-Hasselbalch equation:

pH = pK_a3 + log([conjugate base] / [acid])

pH = -log(4.2 x 10^-13) + log([PO₄^3- ] / [HPO₄^2-])

pH = -log(4.2 x 10^-13) + log([.35 ] / [.35])

pH = 12.38 + 0

pH = 12.38

Hope this helps!