Question

Zn salt is mixed with $(N{H}_4{)}_{2}S$ of $0.021M$. What amount of $Z{n}^{2+}$ will remain unprecipitated in $12$ mL of the solution?
$(K_{sp}$ of $ZnS=4.51\times {10}^{-24})$

• A. $1.677\times {10}^{-22}g$
• B. $1.677\times {10}^{-12}g$
• C. $1.677\times {10}^{-15}g$
• D. $1.677\times {10}^{-2}g$

Answer 1

The correct option is A $1.677\times {10}^{-22}g$

$\left[\right(N{H}_4{)}_{2}S]=0.021M$

$\therefore \left[S^{2-}\right]=0.021M$

$\therefore$ At equilibrium, $\left[Z{n}^{2+}\right]\left[S^{2-}\right]=K_{sp}$ of ZnS

$\therefore \left[Z{n}^{2+}\right]=\frac{4.51\times {10}^{-24}}{0.021}=2.15\times {10}^{-22}\times 65g/litre$

$=\frac{2.15\times {10}^{-22}\times 65\times 12}{1000}g/12ml$

$=1.677\times {10}^{-22}g$ per $12mL$