Question

Choose the correct statement for $Ca(OH{)}_2$ when the $pH$ of a $5\times {10}^{-2}M$ solution of $Ca(N{O}_3{)}_2$​ is adjusted to 10.
[Given that: $K_{sp}\left(Ca\right(OH{)}_2)=5\times {10}^{-6}{M}^3]$

• A. $Ca(OH{)}_2$ precipitation will take place.
• B. $Ca(OH{)}_2$ precipitation will not take place.
• C. For $Ca(OH{)}_2,Q_{sp}=K_{sp}$
• D. Can not predicted.

Answer 1

The correct option is B $Ca(OH{)}_2$ precipitation will not take place.

At ${25}^{o}C$, $pH=10$ is given ,
So , $\left[H^{+}\right]={10}^{-10}M\left[O{H}^{-}\right]={10}^{-4}M$
$Ca(N{O}_3{)}_2$ is a strong electrolyte. The dissociation of $Ca(N{O}_3{)}_2$ is given below :
$Ca\left(N{O}_3{)}_2\right(s)\to C{a}^{2+}(aq)+2N{O}_3^{-}(aq)\therefore [C{a}^{2+}]=5\times {10}^{-2}M$

$Q_{sp}=\left[C{a}^{2+}\right][O{H}^{-}{]}^2{Q}_{sp}=(5\times {10}^{-2}\left)\right({10}^{-4}{)}^2{Q}_{sp}=5\times {10}^{-10}$
Given ,$K_{sp}\left(Ca\right(OH{)}_2=5\times {10}^{-6}{M}^3$
$\therefore Q_{sp}<K_{sp}$
So , precipitation will not take place.