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\[BaS{{O}_{4}},(x)=\sqrt{{{K}_{sp}}}=\sqrt{{{10}^{-11}}}\] \[=3.16\times {{10}^{-6}}\text{mol}{{\text{L}}^{-1}}\] 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 BaS04, hence it will precipitate first.
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