JEE Main & Advanced Sample Paper JEE Main Sample Paper-5

  • question_answer
    \[{{A}_{2}}{{B}_{3}}\]is sparingly soluble salt of molecular weight M and solubility x g/L. The ratio of the molar concentration of \[[{{B}^{2-}}]\]to the solubility product of the solution will be

    A)  \[\frac{7{{M}^{4}}}{54{{x}^{4}}}\]                           

    B)  \[\frac{3}{28\,}\frac{{{M}^{4}}}{{{x}^{4}}}\]

    C)  \[\frac{{{M}^{4}}}{36{{x}^{4}}}\]                             

    D)  \[\frac{36{{M}^{4}}}{{{x}^{4}}}\]

    Correct Answer: C

    Solution :

     Idea This problem can be solved by using the concept of solubility product. Students are advised to follow the trend to solve the question. Write ionisation reaction Determine \[{{K}_{sp}}\] and \[{{B}^{2-}}\] Finally calculate ratio of\[\frac{{{B}^{2-}}}{{{K}_{sp}}}\] Given \[x=\frac{s}{M}=1,s=\frac{x}{M}\] lonisation reaction \[{{A}_{2}}{{B}_{3}}\underset{2s}{\mathop{2{{A}^{3+}}}}\,+\underset{3s}{\mathop{3B{{B}^{2-}}}}\,\] \[{{K}_{sp}}={{(2s)}^{2}}{{(3s)}^{3}}=4{{s}^{2}}\times 27{{s}^{3}}=108{{s}^{5}}\] \[\frac{[{{B}^{-2}}]}{{{K}_{sp}}}=\frac{3s}{108{{s}^{5}}}=\frac{1}{36{{s}^{4}}}\]\[=\frac{1}{36\times {{\left( \frac{x}{M} \right)}^{4}}}=\frac{{{M}^{4}}}{36{{\times }^{4}}}\] TEST Edge Problems including dissociation constant formation of precipitate are also asked frequently so students are advised to understand relation between \[{{K}_{sp}}\] and formation of precipitate.

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