Question

Molar solubility of $\text{Ca}{\left(\text{OH}\right)}_2$ in a solution that has a pH of 12.$\left[K_{SP}\left[\text{Ca}{\left(\text{OH}\right)}_2\right]=5.6\times {10}^{-12}\right]$

• A. $5.6\times {10}^{-10}M$
• B. $5.6\times {10}^{-8}M$
• C. $4\times {10}^{-4}M$
• D. None

Answer 1

The correct option is B $5.6\times {10}^{-8}M$

We know that, for a solution, $pH+pOH$ = 14

$pH$ of the solution is 14, hence $pOH$ = 2

$-log\left[O{H}^{-}\right]=2$

$log\left[O{H}^{-}\right]=-2$

taking antilog

$O{H}^{-}={10}^{-2}$

Consider, the molar solubility of $Ca(OH{)}_2=x$

Now,

$K_{s}p=\left[Ca\right][2O{H}^{-}{]}^2$

$5.6\times {10}^{-12}=\left[x\right][2x+{10}^{-2}{]}^2$

$5.6\times {10}^{-12}=\left[x\right][4x^2+2x{10}^{-2}+{10}^{-4}]$

Now, given that $x<<1$

We can ignore the terms $4x^2$ and $2x{10}^{-2}$

So,

$5.6\times {10}^{-12}=\left[x\right]\left[{10}^{-4}\right]$

$x=5.6\times {10}^{-8}$