Question

Calculate the concentrations of the molecular and ionic species and the pH in aqueous solution that has a formal composition of 0.2350 M KH2PO4 + 0.4300 M K2HPO4.

ph of 7.46

What is the concentration of the molecular species H₃PO₄?

Answer 1

Solution.

The phosphoric acid is a triprotic acid with the pKa values of

pK_a1 = 2.15; pK_a2 =7.20; pK_a3 = 12.35;

The pH of the solution can be calculated as follows:

$pH = pK_{a2}+\log\frac{[HPO_4^{2-}]}{[H_2PO_4]^{-}} =7.20+\log\frac{0.430}{0.235} = 7.462;$

Using the first dissociation constant expression we can derive that

$\frac{[H_2PO_4^{-}][H^{+}]}{[H_3PO_4]} =K_{a1};$

$[H_3PO_4] = \frac{[H_2PO_4^{-}][H^{+}]}{K_{a1}} = \frac{[H_2PO_4^{-}]10^{-pH}}{10^{-pK_{a1}}} = \frac{0.235\times 10^{-7.462}}{10^{-2.15}} =1.146\times 10^{-6} \ M.$

Using the third dissociation constant expression we can derive that

$\frac{[PO_4^{3-}][H^{+}]}{[HPO_4^{2-}]} =K_{a3};$

$[PO_4^{3-}] = \frac{K_{a3}[HPO_4^{2-}]}{[H^{+}]} =\frac{10^{-pK_{a3}}[HPO_4^{2-}]}{10^{-pH}} = \frac{0.43\times 10^{-12.35}}{10^{-7.462}} =5.565\times 10^{-6} \ M.$