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20 June, 17:12

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

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  1. 20 June, 17:25
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    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
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