Calculate pH of $1 M P O_{4}^{3 -}$ (aq) solution $p^{k_{b}} (P O_{4}^{3 -}) = 1.62$

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Solution

$P O_{4}^{3 -}$ in aqueous solution gives $O H^{-}$ ion hence from Ostwaid's dilution law.
$[O H^{-}] = \sqrt{K_{b} . C} P O H = \frac{1}{2} [P k b - log C] = \frac{1}{2} [1.62 - log 1] = 0.81 p H = 14 - p O H = 14 - 0.81 = 13.19$

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