(1)425ml of a saturated solution of lanthanum iodate, `La(IO_3)_3`, has `2.93 xx 10^-4` mole of `La^(3+)` (a)what is the concentration of `IO_3^-`? (b)calculate the solubility product of lantham iodate?

So, you know that you're dealing with a `"425-mL"` sample of a saturated solution of lanthanum(III) iodate, `"La"("IO"_3)_3`.

You also know that this solution contains `2.93 * 10^(-4)` moles of lanthanum(III) cations, `"La"^(3+)`.

Use this information to find the molarity of the lanthanum(III) cations - do not forget that molarity uses liters of solution!

`color(blue)(c = n/V)`

`["La"^(3+)] = (2.93 * 10^(-4)"moles")/(425 * 10^(-3)"L") = 6.89 * 10^(-4)"M"`

Now, the equilibrium reaction that describes the partial dissociation of lanthanum iodate in aqueous solution looks like this

`"La"("IO"_3)_text(3(s]) rightleftharpoons "La"_text((aq])^(3+) + color(red)(3)"IO"_text(3(aq])^(-)`

Notice that you have a `1:1` mole ratio between lanthanum(III) iodate and the lanthanum(III) cations. This tells you that every mole of the solid that dissolves in solution will produce `1` mole of lanthanum(III) cations.

Likewise, the `1:color(red)(3)` mole ratio that exists between the solid and the iodate anions, `"IO"_3^(-)`, tells you that every mole of the solid that dissolves in solution will produce `color(red)(3)` moles of iodate anions.

So, using this mole ratio, you can say that the molarity of the iodate anions will be `color(red)(3)` times higher than that of the lanthanum(III) cations.

`["IO"_3^(-)] = color(red)(3) xx ["La"^(3+)]`

`["IO"_3^(-)] = color(red)(3) xx 6.89 * 10^(-4)"M" = color(green)(2.07 * 10^(-3)"M"`

The expression of the solubility product constant, `K_(sp)`, for lanthanum(III) iodate will look like this

`K_(sp) = ["La"^(3+)] * ["IO"_3^(-)]^color(red)(3)`

Plug in your values to get

`K_(sp) = 6.89 * 10^(-4) * (2.07 * 10^(-3))^color(red)(3)`

`K_(sp) = color(green)(6.12 * 10^(-12))`

The answers are rounded to three sig figs.