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JEE Main & Advanced JEE Main Online Paper (Held On 08 April 2018)

  • question_answer
    An aqueous solution contains 0.10 M \[{{\text{H}}_{\text{2}}}\text{S}\] and 0.20 M \[\text{HCl}\]. If the equilibrium constants for the formation of \[\text{H}{{\text{S}}^{-}}\] from \[{{\text{H}}_{\text{2}}}\text{S}\] is \[\text{1}\text{.0}\times \text{1}{{\text{0}}^{-7}}\] and that of \[{{S}^{2-}}\] from  \[\text{H}{{\text{S}}^{\text{-}}}\]ions is \[\text{1}\text{.2}\times \text{1}{{\text{0}}^{-13}}\] then the concentration of \[{{S}^{2-}}\]ions in aqueous solution is:                       [JEE Main Online 08-04-2018]

    A)  \[6\times {{10}^{-21}}\]                

    B)  \[5\times {{10}^{-19}}\]

    C)  \[5\times {{10}^{-8}}\]                  

    D)  \[3\times {{10}^{-20}}\]

    Correct Answer: D

    Solution :

    \[\text{HCl(aq)}\xrightarrow{{}}\underset{\text{0}\text{.2M}}{\mathop{{{\text{H}}^{\text{+}}}\text{(aq)+}}}\,\text{C}{{\text{l}}^{\text{-}}}\text{(aq)}\] \[{{\text{H}}_{2}}S(aq){{H}^{+}}(aq)+H{{S}^{-}}(aq),{{k}_{1}}=1\times {{10}^{-7}}\] \[\text{H}{{S}^{-}}(aq){{H}^{+}}(aq)+{{S}^{-2}}(aq),{{k}_{2}}=1.2\times {{10}^{-13}}\] \[{{\text{H}}_{2}}{{S}^{-}}(aq)2{{H}^{+}}(aq)+{{S}^{-2}}(aq),{{K}_{eq}}={{k}_{1}}\times {{k}_{2}}\] \[\text{0}\text{.1M}\] \[{{\text{K}}_{eq.}}=1.2\times {{10}^{-20}}=\frac{{{[0.2]}^{2}}[{{S}^{-2}}]}{0.1}\] \[\text{ }\!\![\!\!\text{ }{{\text{S}}^{\text{-2}}}\text{ }\!\!]\!\!\text{ =3 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-20}}}\text{M}\]


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