Question

On adding 0.1M solution each of $[A{g}^{+}],[B{a}^{2+}],(C{a}^{2+}]$ in $N{a}_{2}S{O}_4$ solution, species first precipitated is $[K_{sp}BaS{O}_4={10}^{-11},K_{sp}CaS{O}_4={10}^{-6}$ $and{K}_{sp}A{g}_{2}S{O}_4={10}^{-5}]$

• A. $A{g}_{2}S{O}_4$
• B. $BaS{O}_4$
• C. $CaS{O}_4$
• D. All of these

Answer 1

Answer: (B)

Solubility of $BaS{O}_4,(x)=\sqrt{K_{sp}}=\sqrt{{10}^{-11}}$
$=3.16\times {10}^{-6}\text{mol}{\text{L}}^{-1}$
Solubility of $CaS{O}_4,$
$x=\sqrt{K_{sp}}=\sqrt{{10}^{-6}}=1.0\times {10}^{-3}mol{L}^{-1}$
Solubility of $A{g}_{2}S{O}_4,x=\sqrt[3]{\frac{K_{sp}}{4}}$
because for $A{g}_{2}S{O}_4,4x^3=K_{sp}$
$x=\sqrt[3]{\frac{{10}^{-5}}{4}}=1\times {10}^{-2}mol{L}^{-1}$
Least solubility is of $BaS{0}_4$,
hence it will precipitate first.