Question

How do you calculate the concentration of iodate ions in a saturated solution of lead (II) iodate, `Pb(IO_3)_2`? The `K_(sp) = 2.6 xx 10^(-13)`?

Answer 1

`["IO"_3^(-)] = 8.0 * 10^(-5)` `"M"`

Explanation:

The idea here is that lead(II) iodate is considered insoluble in water, so right from the start, you should expect to find a very low concentration of iodate anions, `"IO"_3^(-)`, in a saturated solution of lead(II) iodate.

The dissociation equilibrium that describes the dissociation of lead(II) iodate looks like this

`"Pb"("IO"_ 3)_ (2(s)) rightleftharpoons "Pb"_ ((aq))^(2+) + color(red)(2)"IO"_ (3(aq))^(-)`

Notice that every mole of lead(II) iodate that dissociates produces `1` mole of lead(II) cations and `color(red)(2)` moles of iodate anions in solution.

This means that, at equilibrium, a saturated solution of lead(II) iodate will have

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

Now, the solubility product constant for this dissociation equilibrium looks like this

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

If you take `s` to be the concentration of lead(II) cations in the solution, i.e. the molar solubility of the salt, you can say that you have

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

which is equivalent to

`2.6 * 10^(-13) = 4s^3`

Rearrange to solve for `s`

`s = root(3)( (2.6 * 10^(-13))/4) = 4.02 * 10^(-5)`

This means that a saturated solution of lead(II) iodate will have

`["Pb"^(2+)] = 4.02 * 10^(-5)` `"M"`

and

`["IO"_3^(-)] = color(red)(2) * 4.02 * 10^(-5)color(white)(.)"M" = color(darkgreen)(ul(color(black)(8.0 * 10^(-5)color(white)(.)"M")))`

I'll leave the answer rounded to two sig figs.