# 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)?

Jun 30, 2017

$\left[{\text{IO}}_{3}^{-}\right] = 8.0 \cdot {10}^{- 5}$ $\text{M}$

#### 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, ${\text{IO}}_{3}^{-}$, in a saturated solution of lead(II) iodate.

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

${\text{Pb"("IO"_ 3)_ (2(s)) rightleftharpoons "Pb"_ ((aq))^(2+) + color(red)(2)"IO}}_{3 \left(a q\right)}^{-}$

Notice that every mole of lead(II) iodate that dissociates produces $1$ mole of lead(II) cations and $\textcolor{red}{2}$ moles of iodate anions in solution.

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

$\left[{\text{IO"_3^(-)] = color(red)(2) * ["Pb}}^{2 +}\right]$

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

${K}_{s p} = {\left[{\text{Pb"^(2+)] * ["IO}}_{3}^{-}\right]}^{\textcolor{red}{2}}$

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

${K}_{s p} = s \cdot {\left(\textcolor{red}{2} s\right)}^{\textcolor{red}{2}}$

which is equivalent to

$2.6 \cdot {10}^{- 13} = 4 {s}^{3}$

Rearrange to solve for $s$

$s = \sqrt[3]{\frac{2.6 \cdot {10}^{- 13}}{4}} = 4.02 \cdot {10}^{- 5}$

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

$\left[{\text{Pb}}^{2 +}\right] = 4.02 \cdot {10}^{- 5}$ $\text{M}$

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.