Question

Ph of a saturated solution of ca(oh)2 is 9. The solubility product (ksp) of ca(oh)2 is

Answer 1

Answer:

The solubility product of the calcium hydroxide is $2.0\times 10^{-15}$.

Explanation:

The pH of the saturated solution of calcium hydroxide = 9

$pH+pOH=14$

pOH = 14 - pH = 14 - 9 = 5

$pOH=-\log[OH^-]$

$5=-\log[OH^-]$

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

$Ca(OH)_2(aq)\rightarrow Ca^{2+}(aq)+2OH^-(aq)$

1 mol of dissociates into 1 mol of calcium ion and 2 moles of hydroxide ions.

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

$[Ca^{2+}]=\frac{[OH^-]}{2}=0.5\times 10^{-5} M$

$Ca(OH)_2(aq)\rightarrow Ca^{2+}(aq)+2OH^-(aq)$

                                       S      2S

$K_{sp}=[Ca^{2+}][[OH^-]]^2=S\times (2S)^2=4S^3$

$K_{sp}=4\times (0.5\times 10^{-5})^3$

$K_{sp}=2.0\times 10^{-15}$

The solubility product of the calcium hydroxide is $2.0\times 10^{-15}$.