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)#?

1 Answer
Jun 30, 2017

Answer:

#["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.