Question

When $100$ ml $Ca(OH{)}_2$ solution of $0.01$M is diluted with water, the pH of the resulting solution changes to $12$. What is the volume of water added?

Answer 1

$Ca{\left(OH\right)}_2⟶{Ca}^{2+}+2{OH}^{-}xx2x0.010.010.01\times 2$

Here $\left[O{H}^{-}\right]$=0.02 M

If $pH$ changes to 12, then $pOH=14-pH=14-12=2$

$⟹-\log \left[{OH}^{-}\right]=2⟹\left[{OH}^{-}\right]={10}^{-2}$

Then concentration of $Ca(OH{)}_2$ at $pH=12$ is $=\frac{{10}^{-2}}{2}=5\times {10}^{-3}$ M

By using $M_1{V}_1=M_2{V}_2$ we have,

${10}^{-2}\times 100$ ml=$5\times {10}^{-3}\times V$ ml (at $pH=12$)

$V$ ml $=200$ ml $⟹$ final volume

Thus, the volume of water added is=200 ml-100 ml=100 ml

Thus, $Ca(OH{)}_2$ of 0.01 M is diluted with 100 ml of water.