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JEE Main & Advanced Chemistry Equilibrium / साम्यावस्था Question Bank Critical Thinking

  • question_answer
    The molar solubility \[(mol\ {{L}^{-1}})\] of a sparingly soluble salt \[M{{X}_{4}}\] is \['s'\]. The corresponding solubility product is \[{{K}_{sp}}\]. \['s'\]is given in terms of \[_{{{K}_{sp}}}\]by the relation            [AIEEE 2004]

    A)                 \[s={{(256{{K}_{sp}})}^{1/5}}\] 

    B)                 \[s={{(128{{K}_{sp}})}^{1/4}}\]

    C)                 \[s={{({{K}_{sp}}/128)}^{1/4}}\]               

    D)                 \[s={{({{K}_{sp}}/256)}^{1/5}}\]

    Correct Answer: D

    Solution :

               \[M{{X}_{4}}\to \underset{s}{\mathop{M}}\,+\underset{4s}{\mathop{4X}}\,\] ; \[{{K}_{sp}}={{(4s)}^{4}}s\]; \[{{K}_{sp}}=256{{s}^{5}}\]                                 \[s={{\left( \frac{{{K}_{sp}}}{256} \right)}^{1/5}}\].


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