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Chhattisgarh PMT Chhattisgarh PMT Solved Paper-2004

  • question_answer
    At \[{{20}^{o}}C\]the \[A{{g}^{+}}\] ion concentration in a saturated solution of \[A{{g}_{2}}Cr{{O}_{4}}\] is \[1.5\times {{10}^{-4}}mol/L\]. At \[{{20}^{o}}C\]the solubility product of \[A{{g}_{2}}Cr{{O}_{4}}\] would be

    A)  \[3.3750\times {{10}^{-12}}~\]

    B)  \[1.6875\times {{10}^{-10}}\]

    C) \[1.6875\times {{10}^{-12}}\]  

    D)  \[1.6875\times {{10}^{-11}}\]

    Correct Answer: C

    Solution :

    Given, \[2[A{{g}^{+}}]=1.5\times {{10}^{-4}}mol/L\] \[[A{{g}^{+}}]=0.75\times {{10}^{-4}}\] \[[A{{g}^{+}}]=[CrO_{4}^{2-}]\]                 \[=0.75\times {{10}^{-4}}mol/L\] \[A{{g}_{2}}Cr{{O}_{4}}\rightleftharpoons 2A{{g}^{+}}+CrO_{4}^{2-}\]                 \[{{K}_{sp}}={{[A{{g}^{+}}]}^{2}}.[CrO_{4}^{2-}]\]                 \[{{K}_{sp}}={{(1.5\times {{10}^{-4}})}^{2}}\times (0.75\times {{10}^{-4}})\]                 \[{{K}_{sp}}=2.25\times {{10}^{-8}}\times 0.75\times {{10}^{-4}}\]                 \[{{K}_{sp}}=1.6875\times {{10}^{-12}}\]


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