RAJASTHAN PMT Rajasthan - PMT Solved Paper-2006

  • question_answer
    The solubility of\[S{{b}_{2}}{{S}_{3}}\], in water is \[1.0\times {{10}^{-5}}\] \[mol/L\] at\[298\,\,K\]. 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 :

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


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