JCECE Engineering JCECE Engineering Solved Paper-2013

  • 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

    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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