Respuesta :
Explanation:
As the first reaction equation is as follows.
[tex]H_{2}S(aq) \rightleftharpoons H^{+}(aq) + HS^{-}(aq)[/tex] ..... (1)
So, expression for equilibrium constant will be as follows.
[tex]K'_{c} = \frac{[H^{+}][HS^{-}]}{[H_{2}S]}[/tex] = [tex]1.00 x 10^{-7}[/tex]
The second reaction equation is as follows.
[tex]HS^{-}(aq) \rightaleftharpoons H^{+}(aq) + S^{2-}(aq)[/tex] ...... (2)
So, expression for equilibrium constant will be as follows.
[tex]K"_{c} = \frac{[H^{+}][S^{2-}]}{[HS^{-}]} = [tex]1.00 \times 10^{-19}[/tex]
Hence, the net reaction equation will be (1) + (2) as follows.
[tex]H_{2}S(aq) \rightleftharpoons 2H^{+}(aq) + S^{2-}(aq)[/tex]
[tex]K_{c}[/tex] = [tex]\frac{[H^{+}]^{2}[S^{2-}]}{[H_{2}S][/tex]
= [tex][H^{+}][HS^{-}]{[H_{2}S][/tex] x [tex]\frac{[H^{+}][S^{2-}]}{[HS^{-}]}[/tex]
= [tex]K'_{c} \times K"_{c}[/tex]
= [tex]1.00 \times 10^{-7} \times 1.00 \times 10^{-19}[/tex]
= [tex]1.00 \times 10^{-26}[/tex]
Thus, we can conclude that the equilibrium constant for the reaction: [tex]H_{2}S(aq) \rightleftharpoons 2H^{+}(aq) + S^{2-}(aq)[/tex] is K = [tex]1.00 \times 10^{-26}[/tex].
