Question

The pH of a solution of $B{\left(OH\right)}_2$ is $10.6$. Calculate the solubility and solubility product of hydroxide.

Answer 1

$B(OH{)}_2\to B+2O{H}^{-}$

$S$ $2S$
$pH=10.6$

As $pH+pOH=14$
$pOH=3.4,\left[{OH}^{-}\right]=3.98\times {10}^{-4}mol$ $L^{-1}$
Solubility $=\frac{1}{2}\left[{OH}^{-}\right]=1.99\times {10}^{-4}mol$ $L^{-1}$

Solubility $=1.99\times {10}^{-4}mol{L}^{-1}$
Solubility product, $K_{sp}=\left[S\right][2S{]}^2=4S^3=3.15\times {10}^{-11}$