WB JEE Medical WB JEE Medical Solved Paper-2006

  • question_answer
    The solubility of \[S{{b}_{2}}{{S}_{3}}\]in water is \[1.0\times {{10}^{-5}}\text{mol/L}\]at 298K. What will be its solubility product?

    A)  \[108\times {{10}^{-25}}\]          

    B)  \[1.0\times {{10}^{-25}}\]

    C)  \[144\times {{10}^{-25}}\]          

    D)  \[126\times {{10}^{-24}}\]

    Correct Answer: A

    Solution :

                     \[\underset{\text{s}\,\text{mol/L}}{\mathop{S{{b}_{2}}{{S}_{3}}}}\,\underset{2s}{\mathop{2S{{b}^{3+}}}}\,\underset{3s}{\mathop{3{{S}^{2-}}}}\,\] Solubility product \[({{K}_{sp}})=[S{{b}^{3+}}]{{[{{S}^{2-}}]}^{3}}\] \[={{(2s)}^{2}}{{(3s)}^{2}}=108{{s}^{5}}\]              \[=108\times {{(1.0\times {{10}^{-5}})}^{5}}=108\times {{10}^{-25}}\]


You need to login to perform this action.
You will be redirected in 3 sec spinner