Question

Question #c50da

Answer 1

`5.57 * 10^(-8)"M"`

Explanation:

Start by looking at the dissociation equilibrium for barium hydroxide, `"Ba"("OH")_2`

`"Ba"("OH")_ (color(red)(2)(s)) rightleftharpoons "Ba"_ ((aq))^(2+) + color(red)(2)"OH"_ ((aq))^(-)`

Notice that for every `1` mole of barium hydroxide that dissociates, you get `1` mole of barium cations and `color(red)(2)` moles of hydroxide anions.

By definition, the solubility product constant, `K_(sp)`, for this equilibrium looks like this

`K_(sp) = ["Ba"^(2+)] * ["OH"^(-)]^color(red)(2)`

Now, the equilibrium concentration of hydroxide anions in a saturated barium hydroxide solution will always by `color(red)(2)` times that of the barium cations, as given by the aforementioned `1:color(red)(2)` mole ratio that exists between the two ions in solution.

This means that if you take `s` to be the equilibrium concentration of the barium cations, you will have

`["OH"^(-)] = color(red)(2) xx ["Ba"^(2+)] = color(red)(2)s`

If you plug these into the expression for the `K_(sp)`, you will end up with

`K_(sp) = s * (color(red)(2)s)^color(red)(2)`

`K_(sp) = 4s^3`

This will get you

`s = root(3)(K_(sp)/4)`

Plug in your value to find

`s = root(3)((6.92 * 10^(-22))/4) = color(darkgreen)(ul(color(black)(5.57 * 10^(-8)"M")))`

The answer must be rounded to three sig figs.