Question

ph of a saturated solution of $Ca{\left(OH\right)}_2$ is 9. The solubility product($K_{sp}$ ) of ${Ca\left(OH\right)}_2$ is

Answer 1

$Thereactionisgivenby$

$Ca{\left(OH\right)}_2⟺{Ca}^{2+}+{2OH}^{-}$

$GivenpH=9$

$HencepOH=14-9=5(pH+pOH=14)$

$\left[{OH}^{-}\right]={10}^{-5}M$

$Hence{Ca}^{2+}=\frac{{10}^{-5}}{2}$

$Thus{K}_{sp}=\left[C{a}^{2+}\right]{\left[{OH}^{-}\right]}^2$

$=\left(\frac{10^{-5}}{2}\right){\left({10}^{-5}\right)}^2$

$K_{sp}=0.5\times 10^{-15}$