Determine the pH for sodium hydrogen phosphate.

Question

pK_a = 12.38

pK_b = 6.80

Concentration = 9.44×10^-3 M

If the 5% approximation is valid, use the assumption to find the pH.

Answer 1

Na₂HPO₄ = sodium hydrogen phosphate (aka di-sodium hydrogen phosphate)

HPO₄^2- + H₂O ==> H₂PO₄^- + OH^-

Kb = 1.6x10^-7 (obtained from the pKb of 6.80)

Kb = 1.6x10^-7 = [H₂PO₄^-] [OH^-] / [HPO₄^2-] - x and assume x is small relative to [HPO₄^2-] and ignore it.

1.6x10^-7 = [H₂PO₄^-] [OH^-] / [HPO₄^2-]

1.6x10^-7 = (x)(x) / 9.44x10^-3

x² = 1.496x10^-9

x = [OH^-] = 3.868x10^-5 M note: this is less than 5% of 9.44x10^-3 so above assumption was valid.

pOH = -log [OH^-] = -log 3.868x10^-5

pOH = 4.413

pH = 14 - 4.413

pH = 9.587 (3 sig.figs.)