• # question_answer On adding$0.1\,\,M$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

Solution :

Solubility of$BaS{{O}_{4}},\,\,(x)=\sqrt{{{K}_{sp}}}=\sqrt{{{10}^{-11}}}$                 $=3.16\times {{10}^{-6}}mol\,\,{{L}^{-1}}$ 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$BaS{{O}_{4}}$, hence it will precipitate first.

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