JEE Main & Advanced Chemistry Equilibrium / साम्यावस्था Sample Paper Topic Test - Ionic Equilibrium

  • question_answer
    Solubility of \[PbC{{l}_{2}}\] in 0.5 M NaCl is  \[{{S}_{1}}\] and that in 0.5 M \[Pb{{(N{{O}_{3}})}_{2}}\] is \[{{S}_{2}}\] then which of 5 the following is correct:

    A) \[\frac{{{S}_{1}}}{{{S}_{2}}}=1\]

    B) \[{{S}_{1}}-{{S}_{2}}>0\]

    C) \[{{S}_{1}}-{{S}_{2}}<0\]

    D) \[\frac{{{S}_{1}}}{{{S}_{2}}}>1\]

    Correct Answer: C

    Solution :

    [c] \[{{S}_{1}}=\frac{{{K}_{sp}}}{{{0.5}^{2}}},\,{{S}_{2}}=\sqrt{\frac{{{K}_{sp}}}{0.5}}\]
    \[\frac{{{S}_{1}}}{{{S}_{2}}}=\frac{{{K}_{sp}}}{{{0.5}^{2}}}\times \sqrt{\frac{0.5}{{{K}_{sp}}}}=\frac{\sqrt{{{K}_{sp}}}}{0.5\times \sqrt{0.5}}=\frac{\sqrt{{{K}_{sp}}}}{0.707\times 0.5}\]
    But  \[\because \]            \[{{K}_{sp}}<<<1\]
    \[\therefore \]      \[\frac{{{S}_{1}}}{{{S}_{2}}}<1\]                    \[\therefore \] \[{{S}_{1}}<{{S}_{2}}\]
    \[\therefore \]      \[{{S}_{1}}-{{S}_{2}}<0\]


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