H3PO4, please calculate the pH of the following solutions respectively: (a) 0.0250 M Na3PO4; (b) 0.000250 M of Na2HPO4; (c) 0.000250 M of NaH2PO4

Question

Ka1 = 7.11 × 103, Ka2 = 6.32 × 108, and Ka3 = 4.50 × 1013 for H3PO4,

Answer 1

PO₄^3- + H2O ==> HPO₄^2- + OH^- ... hydrolysis of PO₄^3-

Kb = 1x10^-14 / 4.50x10^-13 = 0.0222

Kb = 0.0222 = [HPO₄^2-][OH^-] / [PO₄^3-]

0.0222 = (x)(x) / 0.025 - x

x² = 0.000555 - 0.0222x

x² + 0.0222x - 0.000555 = 0

x = 0.0149 M = [OH^-]

pOH = -log 0.0149 = 1.83

pH = 14 - pOH = 14 - 1.83

pH = 12.17

HPO₄^2- + H2O ==> H₂PO₄^- + OH^- ... hydrolysis of HPO₄^2-

Kb = 1x10^-14 / 6.32x10^-8 = 1.6x10^-7

Kb = 1.60x10^-7 = [H₂PO₄^- ][ OH^- ] / [HPO₄^2-]

1.60x10^-7 = (x)(x) / 0.000250 - x and assume x is small relative to 0.000250 and ignore it

1.60x10^-7 = x² / 0.000250

x² = 3.96x10^-11

x = 6.29x10^-6 M = [OH^-] (note: this is less than 5% of 0.00025 so above assumption was valid)

pOH = -log 6.29x10^-6 = 5.20

pH = 14 - 5.20

pH = 8.80

H₂PO₄^- + H2O ==> H₃PO₄ + OH^- ... hydrolysis of H₂PO₄^-

Kb = 1x10^-14 / 7.11x10^-3 = 1.41x10^-12

Kb = 1.41x10^-12 = [H₃PO₄][ OH^-] / [H₂PO₄^-]

1.41x10^-12 = (x)(x) / 0.000250 - x and assume x is small relative to 0.000250 and ignore it

1.41x10^-12 = x² / 0.000250

x² = 3.52x10^-16

x = 2.0x10^-8 M = [OH^-] (note: this is less than 5% of 0.00025 M so above assumption was valid)

Since the [OH^-] is so low, we must now consider the contribution of H2O to the [OH^-], which is 1x10^-7 M

Total final [OH^-] = 2x10^-8 + 1x10^-7 = 1.20x10^-7 M

pOH = -log 1.20x10^-7

pOH = 6.92

pH = 14 - pOH

pH = 7.08