JEE Main & Advanced AIEEE Solved Paper-2004

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

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

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

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

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

    Correct Answer: D

    Solution :

    For the solute, \[{{A}_{x}}{{B}_{y}}xA+yB\] \[{{K}_{ap}}={{x}^{x}}{{y}^{y}}{{(s)}^{x+y}}\] \[M{{X}_{4}}{{M}^{4+}}+4{{X}^{-}}\] \[x=1,y=4\] \[\therefore \]\[{{K}_{sp}}={{(4)}^{4}}{{(1)}^{1}}{{(s)}^{5}}=256{{s}^{5}}\] \[\therefore \]  \[s={{\left( \frac{{{K}_{sp}}}{256} \right)}^{1/5}}\]


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