What is the pH of a solution of 0.800 M KH2PO4, potassium dihydrogen phosphate?

KH₂PO₄ hydrolyzes as;

H₂PO₄⁻ + H₂O ↔ H₃PO₄ + OH⁻

Let x amount of H₂PO₄⁻ has reacted with water then,

Kb₁ = [H₃PO₄][OH⁻] / [H₂PO₄⁻]

[H₂PO₄⁻] = 0.8-x M

Kb₁ = x² / (0.8 - x)

Given Ka₁ = 7.5 x 10⁻³

so Kb₁ = 1 x 10⁻¹⁴ / (7.5 x 10⁻³) = 1.33 x 10⁻¹²

From this information:

1.33 x 10⁻¹² = x² / 0.8

x = [OH⁻] = 1.03 x 10⁻⁶ M

pOH = - log (1.03 x 10⁻⁶) = 5.99

pH = 14 - pOH = 14 - 5.99 = 8.01