A) \[A{{g}_{2}}S{{O}_{4}}\]
B) \[BaS{{O}_{4}}\]
C) \[CaS{{O}_{4}}\]
D) All of these
Correct Answer: B
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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