Question

The value of $\left[M{g}^{2+}\right][O{H}^{⊝}{]}^2$ in a solution of 0.001 M $Mg(OH{)}_2$ in $Mg(N{O}_3{)}_2$ if the pH of solution is adjusted to 9 is:

$K_{sp}$ of $Mg(OH{)}_2=8.9\times {10}^{-12}$

• A. lesser than $K_{sp}$ and $Mg(OH{)}_2$ won't precipitate
  
• B. greater than $K_{sp}$ and $Mg(OH{)}_2$ won't precipitate
  
• C. lesser than $K_{sp}$ and $Mg(OH{)}_2$ will precipitate
  
• D. greater than $K_{sp}$ and $Mg(OH{)}_2$ will precipitate
  

Answer 1

The correct option is A lesser than $K_{sp}$ and $Mg(OH{)}_2$ won't precipitate

$pH=9\therefore \left[H^{\oplus }\right]={10}^{-9}Mor\left[O{H}^{⊝}\right]={10}^{-5}M$

Now if $Mg(N{O}_3{)}_2$ is present in a solution of $\left[O{H}^{⊝}\right]={10}^{-5}M$, then

Product of ionic concentration

$=\left[M{g}^{2+}\right][O{H}^{⊝}{]}^2$

$=\left[0.001\right][{10}^{-5}{]}^2$

$={10}^{-13}$ less than $K_{sp}$ of $Mg(OH{)}_2$ (i.e., $8.9\times {10}^{-12}$)

$Mg(OH{)}_2$ will not precipitate.