Question

What is the $pH$ of $Ca(OH{)}_2$ solution having a concentration of $5\times {10}^{-8}M?$
Take $log\left(6.18\right)=0.79$

• A. 8.99
• B. 7.21
• C. 5.69
• D. 6.01

Answer 1

The correct option is B 7.21

Here very dilute solution of $Ca(OH{)}_2$ is given, where water dissociation needs to be considered to calculate the total $\left[O{H}^{-}\right]$ ions present in the solution.
$\left[O{H}^{-}\right]fromCa(OH{)}_2=2\times 5\times {10}^{-8}M={10}^{-7}M.$

$H_{2}O\left(l\right)\rightleftharpoons H^{+}\left(aq\right)+O{H}^{-}\left(aq\right)$

From $H_{2}O$ ,
$\left[O{H}^{-}\right]=\left[H^{+}\right]$ = $x$
We know,
$\left[H^{+}\right]\left[O{H}^{-}\right]={10}^{-14}x({10}^{-7}+x)={10}^{-14}{x}^2+{10}^{-7}x-{10}^{-14}=0x=\frac{-{10}^{-7}\pm \sqrt{{10}^{-14}+4\times {10}^{-14}}}{2}x=6.18\times {10}^{-8}$

So, $\left[H^{+}\right]=6.18\times {10}^{-8}MpH=-log\left[H^{+}\right]pH=-log(6.18\times {10}^{-8})=(8-0.79)pH=7.21$