Phosphoric acid, H 3 P O 4 (aq), is a triprotic acid, meaning that one molecule of the acid has three acidic protons. Estimate the pH and the concentrations of all species in a 0.100 M phosphoric acid solution.

H3PO4 = 0.30 M

H2PO4 - = 4.59 x 10^-2M

HPO2^4 - = 6.17 x 10^-8 M

PO3^4 - = 6.44 x 10^-19 M

H + = 4.59 x 10^-2 M

Phosphoric acid can be classified as a weak acid which dissciates to give three acid where the products of one dissociation equilibrium become components of the subsequent dissociation equilibrium. Below are their equilibrium equations:

H3PO4 (aq) ⇌ H2PO4 - (aq) + H + (aq) H2PO4 - (aq) ⇌ HPO2^4 - (aq) + H + (aq)

HPO2^4 - (aq) ⇌ PO3^4 - (aq) + H + (aq)

The pH for the concentrations are:

Ka = 10^-pKa,

Ka1 = 10-2.16

= 6.17 x 10^-3

Ka2 = 10-7.21

= 6.17 * 10^-8

Ka3 = 10-12.32

= 4.79 * 10

The following concentrations are:

Solving for [H2PO4^-] and [H+],

Ka1 = {[H2PO4^-]*[H+]}/[H3PO4]

6.92 x 10^-3 = ((x) * (x)) / 0.35 - x

2.42 x 10^-3 - 6.92 x 10^-3 = x2

Solving the above using general quadratic formula,

x = 4.59 x 10^-2M

x = [H2PO4^-] = [H+]

Solving for [HPO4^2-],

Ka2 = {[HPO4^2-]*[H+]}/[H2PO4^-]

6.17 x 10^8 - = ((x) * (4.59 x 10-2)) / 4.59 x 10^-2

x = 6.17 x 10^-7

x = [HPO4^2-]

Solving for [PO4^3-],

Ka3 = [PO4^3-]*[H+]}/[HPO4^2-]

4.79 x 10^-13 = ((x) * (4.5 x 10^-*)) / 6.17 x 10^-8

x = 6.44 x 10^-19M

x = [PO4^3-]