11th Class Chemistry Equilibrium / साम्यावस्था

  • question_answer 1)
      \[{{K}_{{{a}_{1}}}},{{K}_{{{a}_{2}}}}\] and \[{{K}_{{{a}_{3}}}}\] are the respective ionisation constants for the following reactions: \[{{H}_{2}}S\underset{{}}{\leftrightarrows}{{H}^{+}}+H{{S}^{-}}\] \[H{{S}^{-}}\underset{{}}{\leftrightarrows}{{H}^{+}}+{{S}^{2-}}\] \[{{H}_{2}}S\underset{{}}{\leftrightarrows}2{{H}^{+}}+{{S}^{2-}}\] The correct relationship between \[{{K}_{{{a}_{1}}}},{{K}_{{{a}_{2}}}}\] and \[{{K}_{{{a}_{3}}}}\] is: (a) \[{{K}_{{{a}_{3}}}}={{K}_{{{a}_{1}}}}\times {{K}_{{{a}_{2}}}}\]               (b) \[{{K}_{{{a}_{3}}}}={{K}_{{{a}_{1}}}}+{{K}_{{{a}_{2}}}}\] (c) \[{{K}_{{{a}_{3}}}}={{K}_{{{a}_{1}}}}-{{K}_{{{a}_{2}}}}\]            (d) \[{{K}_{{{a}_{3}}}}={{K}_{{{a}_{1}}}}/{{K}_{{{a}_{2}}}}\]

    Answer:

      (a) \[{{H}_{2}}S\underset{{}}{\leftrightarrows}{{H}^{+}}+H{{S}^{-}}\]                \[{{K}_{{{a}_{1}}}}=\frac{[{{H}^{+}}][H{{S}^{-}}]}{[{{H}_{2}}S]}\] \[H{{S}^{-}}\underset{{}}{\leftrightarrows}{{H}^{+}}+{{S}^{2-}}\]                             \[{{K}_{{{a}_{2}}}}=\frac{[{{H}^{+}}][{{S}^{2-}}]}{[H{{S}^{-}}]}\] \[{{H}_{2}}S\underset{{}}{\leftrightarrows}2{{H}^{+}}+{{S}^{2-}}\]                          \[{{K}_{{{a}_{3}}}}=\frac{[{{H}^{+}}][{{S}^{2-}}]}{[{{H}_{2}}S]}\]\[{{K}_{{{a}_{3}}}}={{K}_{{{a}_{1}}}}\times {{K}_{{{a}_{2}}}}\]               


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