Question

pH of a saturated solution of $Ca(OH{)}_2$ is 9. The solubility product (Ksp) of $Ca(OH{)}_2$ is:

• A. $0.5\times {10}^{-15}$
• B. $0.25\times {10}^{-10}$
  
• C. $0.125\times {10}^{-15}$
• D. $0.5\times {10}^{-10}$

Answer 1

The correct option is A $0.5\times {10}^{-15}$

$Ca(OH{)}_2$ $\rightleftharpoons$ $C{a}^{2+}$ + $2O{H}^{-}$
pH = 9, Hence pOH = 14 – 9 = 5
$[OH{]}^{-}$= ${10}^{-5}M$

Hence $\left[C{a}^{2+}\right]$= $\frac{{10}^{-5}}{2}$
Thus Ksp = $[Ca{]}^{2+}$ + $[O{H}^{-}{]}^2$
= ( $\frac{{10}^{-5}}{2}$) (${10}^{-5}{)}^2$
= $0.5\times {10}^{-15}$