The concentration of `"IO"_3^(-)` ions in a pure, saturated solution of `"Ba"("IO"_3)_2` is `1.06xx10^(-3)"M"`. What is the `K_(sp)` for `"Ba"("IO"_3)_2`?

The problem tells you that a saturated solution of barium iodate, `"Ba"("IO"_3)_2`, has a concentration of iodate anions, `"IO"_3^(-)`, equal to `1.06 * 10^(-3)"M"`.

This means that when barium iodate is dissolved in water, the molar concentration of the dissociated iodate anions will be equal to `1.06 * 10^(-3)"M"`.

The dissociation equilibrium for barium iodate in aqueous solution looks like this

`"Ba"("NO"_ 3)_ (color(red)(2)(s)) rightleftharpoons "Ba"_ ((aq))^(2+) + color(red)(2)"IO"_ (3(aq))^(-)`

Notice that every mole of barium iodate that dissociates produces `1` mole of barium cations, `"Ba"^(2+)`, and `color(red)(2)` moles of iodate anions.

This means that in a saturated barium iodate solution, the concentration of iodate anions will be twice as high as the concentration of barium cations.

This means that the latter will be equal to

`["Ba"^(2+)] = 1/color(red)(2) * ["IO"_3^(-)]`

`["Ba"^(2+)] = 1/color(red)(2) * 1.06 * 10^(-3)"M" = 5.30 * 10^(-4)"M"`

Now, the solubility product constant, `K_(sp)`, for this dissociation equilibrium is defined as

`K_(sp) = ["Ba"^(2+)] * ["IO"_3^(-)]^color(red)(2)`

Plug in your values to find

`K_(sp) = 5.30 * 10^(-4)"M" * (1.06 * 10^(-3)"M")^color(red)(2)`

`K_(sp) = 5.96 * 10^(-10)"M"^3`

Solubility product constants are usually given without added units, which means that you answer is

`K_(sp) = color(green)(|bar(ul(color(white)(a/a)color(black)(5.96 * 10^(-10))color(white)(a/a)|)))`

The answer is rounded to three sig figs.