Question

If $K_{sp}$ of $ZnS$ is $1.1\times {10}^{-21}$, the minimum volume of water is required for the dissolution of 0.097 g ZnS is: (Zn = 65, S = 32)

• A. $3.16\times {10}^{5}litre$
• B. $3.16\times {10}^{7}litre$
• C. $3.16\times {10}^{9}litre$
• D. $3.16\times {10}^{3}litre$

Answer 1

The correct option is B $3.16\times {10}^{7}litre$

Given

$K_{sp}=1.1\ast {10}^{-21}$

Mass of ZnS = 0.097g

Solution

$ZnS=Z{n}^{2+}+S^{2-}$

Therefore

$\left[Z{n}^{2+}\right]=\left[S^{2-}\right]=s$

Hence

$K_{sp}=s\ast s=s^2$

$s=(1.1\ast {10}^{-21}{)}^{\frac{1}{2}}$

$s=3.31\ast {10}^{-11}M$

Molecular mass of ZnS = 65+32=97

Solubility in g/L =$3.31\ast {10}^{-11}\ast 97$

$=321.71\ast {10}^{-11}g/L$

Hence

3.22*10^{-9}g of ZnS will dissolve in 1 L water

0.097g of ZnS will dissolve in$\frac{0.097}{3.22\ast {10}^{-9}}$ L of water

$=0.03164\ast 109=3.16\ast {10}^7$ L of water

The correct option is B