• 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)  $\frac{36{{M}^{4}}}{{{x}^{4}}}$

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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