Question

Phosphoric acid is a triprotic acid with Ka1 = 6.9 x 10-3, Ka2 = 6.2 x 10-8, and Ka3 = 4.8 x 10-13. Calculate the pH of a solution created by dissolving 23.4 g of KH2PO4(s) and 38.4 g of Na2HPO4(s) in 2 L of water.

Answer 1

The pH of the solution created is  7.21

Explanation:

Ka1 = 6.9 X 10⁻³

Ka2 = 6.2 X 10⁻⁸

Ka3 = 4.8 X 10⁻¹³

pKa = - log( Ka)

Thus,

pKa1 = - log ( 6.9 X 10⁻³) = 2.16

pKa2 = - log ( 6.2 X 10⁻⁸ ) = 7.21

pKa3 = - log ( 4.8 X 10⁻¹³ ) = 12.3

KH₂PO₄ + Na₂HPO₄ = KH₃PO₄ + Na₂PO₄

Ka2 = [ KH₃PO₄] [Na₂PO₄] / [KH₂PO₄] [Na₂HPO₄]

log (Ka2 ) = log [[ KH₃PO₄] [Na₂PO₄] / [KH₂PO₄] [Na₂HPO₄]]

pH = pKa2 + log [KH₂PO₄] /[Na₂HPO₄]  = 7.21

Therefore, the pH of the solution created is  7.21