Question

How many moles of calcium hydroxide must be dissolved to produce 250.0 ml of an aqueous solution of pH 10.65 ? Assume complete dissociation.
Given ${10}^{-3.35}=4.47\times {10}^{-4}$

• A. $8.9\times {10}^{-4}\text{mol}$
• B. $5.6\times {10}^{-6}\text{mol}$
• C. $2.8\times {10}^{-5}\text{mol}$
• D. $2.8\times {10}^{-6}\text{mol}$

Answer 1

The correct option is A $8.9\times {10}^{-4}\text{mol}$

$pH+pOH=14$
$pOH=14-pH$
$-\text{log}\left[O{H}^{-}\right]=3.35$
$\text{log}\left[O{H}^{-}\right]=-3.35$
$\left[O{H}^{-}\right]={10}^{-3.35}=4.47\times {10}^{-4}M$
$=4.47\times {10}^{-4}{\text{mol L}}^{-1}$
Moles of $O{H}^{-}$ required for 250 ml
$=\frac{4.47\times {10}^{-4}{\text{mol L}}^{-1}}{\left(\frac{250}{1000}\right)L}=1.78\times {10}^{-3}\text{mol}$
As 1 mol $Ca(OH{)}_2$ produces 2 mol $O{H}^{-}$ ions.
So moles of $Ca(OH{)}_2$ dissolved
$=\frac{1.78\times {10}^{-3}\text{mol}}{2}=8.94\times {10}^{-4}\text{mol}Ca(OH{)}_2$