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JEE Main & Advanced Sample Paper JEE Main Sample Paper-12

  • question_answer
    pH of a saturated solution of silver salt of monobasic acid HA is found to be 9. Find the \[{{\text{K}}_{\text{sp}}}\] of sparingly soluble salt Ag A(s). Given \[{{\text{K}}_{a}}(HA)={{10}^{-10}}\]

    A)  \[1.1\times {{10}^{-9}}\]                             

    B)  \[1.1\times {{10}^{-10}}\]

    C)  \[1.1\times {{10}^{-11}}\]                           

    D)  \[{{10}^{-12}}\]

    Correct Answer: C

    Solution :

     \[\underset{x}{\mathop{Ag(A)}}\,\underset{x-y}{\mathop{A{{g}^{+}}(aq)}}\,{{A}^{-}}(aq)\] \[\underset{x-y}{\mathop{Ag(A)}}\,+{{H}_{2}}O(\ell )HA\underset{y}{\mathop{(aq)}}\,+O{{H}^{-}}\underset{y}{\mathop{(aq)}}\,\] \[{{K}_{sp}}=x(x-y)\] \[{{K}_{h}}=\frac{{{K}_{w}}}{{{K}_{a}}}=\frac{{{y}^{2}}}{(x-y)};\frac{{{10}^{-14}}}{{{10}^{-10}}}=\frac{{{({{10}^{-5}})}^{2}}}{(x-y)}\] \[x-y={{10}^{-6}}\] \[x={{10}^{-5}}+{{10}^{-6}}\] \[x=1.1\times {{10}^{-5}}\] \[{{K}_{sp}}=1.1\times {{10}^{-5}}\times {{10}^{-6}}\] \[{{K}_{sp}}=1.1\times {{10}^{11}}\]


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