JEE Main & Advanced Chemistry Equilibrium Question Bank Common ion effect, Isohydric solutions, Solubility product, Ionic product of water and salt hydrolysis

  • question_answer The solubility of \[AgCl\] in \[0.2\,\,M\,\,NaCl\] solution \[({{K}_{sp}}\]for \[AgCl=1.20\times {{10}^{-10}})\] is   [MP PET 1996]

    A)                 \[0.2\,\,M\]       

    B)                 \[1.2\times {{10}^{-10}}\,M\]

    C)                 \[0.2\times {{10}^{-10}}\,M\]    

    D)                 \[0.2\times {{10}^{-10}}\,M\]

    Correct Answer: D

    Solution :

             \[\underset{a}{\mathop{AgCl}}\,\]⇌\[\underset{a}{\mathop{A{{g}^{+}}}}\,+\underset{a}{\mathop{C{{l}^{-}}}}\,\]                    \[\underset{0.02}{\mathop{NaCl}}\,\]⇌ \[\underset{0.02\,\,\,\,\,}{\mathop{N{{a}^{+}}}}\,\,\,+\underset{0.02\,\,\,\,\,\,}{\mathop{C{{l}^{-}}}}\,\]                    \[{{K}_{sp}}\,\,AgCl=1.20\times {{10}^{-10}}\]                    \[{{K}_{sp}}\,\,AgCl=[A{{g}^{+}}]\,\,[C{{l}^{-}}]\]\[=a\times [a+0.2]\] \[={{a}^{2}}+0.2a\]                    \[{{a}^{2}}\]is a very small so it is a neglected.                    \[{{K}_{sp}}\,\,AgCl=0.2a\]                    \[1.20\times {{10}^{-10}}=0.2a\]                                 \[a=\frac{1.20\times {{10}^{-10}}}{0.20}=6\times {{10}^{-10}}\]mole


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