Calculate $p H$ of $1.0 M P O_{4}^{3 -}$ in aqueous solution. Given $p K_{b} o f P O_{4}^{3 -} = 1.62$ .

13.19

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6.5

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11.5

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Solution

The correct option is A 13.19
Given, $[P O_{4}^{3 -}] = 1 M$
$p K_{b} (P O_{4}^{3 -}) = 1.62$
$P O_{4}^{3 -}$ in aqueous solution gives $O H^{-}$ ion, hence from ostwald's dilution law
$[O H^{-}] = \sqrt{K_{b} . C}$
the relation between $P O H$ and $P k_{b}$ is given by
$P O H = \frac{1}{2} [P k_{b} - log C]$
putting the value of $P k_{b}$ and C we get,
$\Rightarrow P O H = \frac{1}{2} [1.62 - log 1] = 0.81$
Also,
$p H = 14 - p O H = 14 - 0.81 = 13.19$
Hence, pH of $P O_{4}^{3 -}$ in aqueous solution is 13.19

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