Question

At 298 K, the equilibrium constant for the following reaction is 1.00×10-7:H2S(aq) + H2O H3O+(aq) + HS-(aq)The equilibrium constant for a second reaction is 1.00×10-19:HS-(aq) + H2O H3O+(aq) + S2-(aq)Use this information to determine the equilibrium constant for the reaction:H2S(aq) + 2H2O 2H3O+(aq) + S2-(aq)K =

Answer 1

Step-by-step explanation:

As the first reaction equation is as follows.

`H_(2)S(aq) \rightleftharpoons H^(+)(aq) + HS^(-)(aq)` ..... (1)

So, expression for equilibrium constant will be as follows.

`K'_(c) = ([H^(+)][HS^(-)])/([H_(2)S])` =
`1.00 x 10^(-7)`

The second reaction equation is as follows.

`HS^(-)(aq) \rightaleftharpoons H^(+)(aq) + S^(2-)(aq)` ...... (2)

So, expression for equilibrium constant will be as follows.

`K`

Hence, the net reaction equation will be (1) + (2) as follows.

`H_(2)S(aq) \rightleftharpoons 2H^(+)(aq) + S^(2-)(aq)`

`K_(c)` =

=
`[H^(+)][HS^(-)]{[H_(2)S]` x
`([H^(+)][S^(2-)])/([HS^(-)])`

=
`K'_(c) * K`

=
`1.00 * 10^(-7) * 1.00 * 10^(-19)`

=
`1.00 * 10^(-26)`

Thus, we can conclude that the equilibrium constant for the reaction:
`H_(2)S(aq) \rightleftharpoons 2H^(+)(aq) + S^(2-)(aq)` is K =
`1.00 * 10^(-26)`.