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}},4{{x}^{3}}={{K}_{sp}} $
$ x=\sqrt[3]{\frac{{{10}^{-5}}}{4}}=1\times {{10}^{-2}}mol\,{{L}^{-1}} $
Least solubility is of $BaS0_4$,
hence it will precipitate first.