Question

Question #4045b

Answer 1

The equilibrium concentration of `S^(2-)` is `5.5 * 10^(-7)"M"`.

All you really have to do to solve this problem is use an ICE table for the equilibrium reaction given. The initial concentrations of `H^(+)` and `S^(2-)` will be zero.

`" "HS_((aq))^(-) rightleftharpoons H_((aq))^(+) + S_((aq))^(2-)`
I....1.00................0.............0
C....(-x)................(+x).........(+x)
E...1.00-x.............x..............x

By definition, the equilibrium constant for this reaction will be

`K_c = ([H^(+)] * [S^(2-)])/([HS""^(-)]) = (x * x)/(1.00 - x) = x^2/(1.00 - x)`

Since the equilibrium constant is so small, you can approximate (1.00 - x) with 1.00. This will get you

`K_c = x^2/(1.00) = 3 * 10^(-13) => x = 5.5 * 10^(-7)`

Since `x` represents the equilibrium concentration of both `H^(+)` and `S^(2-)`, the answer will be

`[S^(2-)] = color(green)(5.5 * 10^(-7)"M")`