The pH of a solution of $B {(O H)}_{2}$ is $10.6$ . Calculate the solubility and solubility product of hydroxide.

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Solution

$B (O H)_{2} \to B + 2 O H^{-}$
$S$ $2 S$
$p H = 10.6$
As $p H + p O H = 14$
$p O H = 3.4, [{O H}^{-}] = 3.98 \times 10^{- 4} m o l$ $L^{- 1}$
Solubility $= \frac{1}{2} [{O H}^{-}] = 1.99 \times 10^{- 4} m o l$ $L^{- 1}$
Solubility $= 1.99 \times 10^{- 4} m o l L^{- 1}$
Solubility product, $K_{s p} = [S] [2 S]^{2} = 4 S^{3} = 3.15 \times 10^{- 11}$

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