How do you calculate the concentration of iodate ions in a saturated solution of lead (II) iodate, Pb(IO_3)_2Pb(IO3)2? The K_(sp) = 2.6 xx 10^(-13)Ksp=2.6×10−13?

1 Answer
Stefan V.
Jun 30, 2017

["IO"_3^(-)] = 8.0 * 10^(-5)[IO−3]=8.0⋅10−5 "M"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^(-)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))^(-)Pb(IO3)2(s)⇌Pb2+(aq)+2IO−3(aq)

Notice that every mole of lead(II) iodate that dissociates produces 11 mole of lead(II) cations and color(red)(2)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+)][IO−3]=2⋅[Pb2+]

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

K_(sp) = ["Pb"^(2+)] * ["IO"_3^(-)]^color(red)(2)Ksp=[Pb2+]⋅[IO−3]2

If you take ss 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)Ksp=s⋅(2s)2

which is equivalent to

2.6 * 10^(-13) = 4s^32.6⋅10−13=4s3

Rearrange to solve for ss

s = root(3)( (2.6 * 10^(-13))/4) = 4.02 * 10^(-5)s=3√2.6⋅10−134=4.02⋅10−5

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

["Pb"^(2+)] = 4.02 * 10^(-5)[Pb2+]=4.02⋅10−5 "M"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.

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