Question

How many moles of calcium hydroxide must be dissolved to produce $250mL$ of an aqueous solution of $pH$ $10.65$. Assume the complete ionisation.

Answer 1

We know that
$pH+pOH=14$
So, $pOH=(14-10.65)=3.35$
$\left[{OH}^{-}\right]={10}^{-3.35}={10}^{-4}\times {10}^{0.65}=4.47\times {10}^{-4}M$
One molecule of $Ca{\left(OH\right)}_2$ furnishes $2{OH}^{-}$ ions
Hence concentration of
$Ca{\left(OH\right)}_2=\frac{4.47\times {10}^{-4}}{2}=2.235\times {10}^{-4}M$
No. of moles in $250mL=\frac{2.235\times {10}^{-4}}{4}=5.58\times {10}^{-5}$