Question

A $Zn$ salt is mixed with $({NH}_4{)}_2$ of molarity of $0.021M$. Then the amount of ${Zn}^{2+}$ that will remain unprecipitated in $12mL$ of the solution is $1.10\times {10}^{-22}g/12mL.$

$K_{{sp}_{ZnS}}=3\times {10}^{-24}$

• A. True
• B. False

Answer 1

The correct option is B False

$\left[{\left({NH}_4\right)}_{2}S\right]=0.021M$
$\left[S^{2-}\right]=0.021M$
$\because$ At equilibrium $\left[{Zn}^{2+}\right]\left[S^{2-}\right]=K_{{sp}_{ZnS}}$
$\therefore$ $\left[{Zn}^{2+}\right]=\frac{K_{{sp}_{ZnS}}}{\left[S^{2-}\right]}=\frac{3\times {10}^{-24}}{0.021}=1.42\times {10}^{-22}M$
$\therefore$ $\left[{zn}^{2+}\right]$ left in solution $=1.42\times {10}^{-22}\times 65g{litre}^{-1}$
$=\frac{1.42\times {10}^{-22}\times 65\times 12}{1000}g/12mL=1.10\times {10}^{-22}g/12mL$