Question

The number of moles of $Ca(OH{)}_2$ that is dissolved to produce $250mL$ of an aqueous solution of $pH10.0$, assuming complete dissociation, is $x\times {10}^{-5}$. Determine the value of $x$.

Answer 1

Let $a$ mole of $Ca(OH{)}_2$ be dissolved in $250mL$ solution to have $pH=10.0$
$\left[Ca\right(OH{)}_2]=\frac{a\times 1000}{250}=4aM$
On complete ionization,
$$\begin{array}{ccccc}& Ca(OH{)}_2& \to & C{a}^{2+}+& 2O{H}^{-}\\ \text{Initial moles}& 4a& & 0& 0\\ \text{Final moles}& 0& & 4a& 2\times 4a\end{array}$$
$\therefore \left[O{H}^{-}\right]=8aM\left[H^{+}\right]=\frac{{10}^{-14}}{8a}\because pH=-\log \left[H^{+}\right]\therefore 10.0=-\log \frac{{10}^{-14}}{8a}\frac{{10}^{-14}}{8a}={10}^{-10}\therefore a=0.125\times {10}^{-4}\text{moles}=1.25\times {10}^{-5}$