# The concentration of "IO"_3^(-) ions in a pure, saturated solution of "Ba"("IO"_3)_2 is 1.06xx10^(-3)"M". What is the K_(sp) for "Ba"("IO"_3)_2?

Aug 2, 2016

${K}_{s p} = 5.96 \cdot {10}^{- 10}$

#### Explanation:

The problem tells you that a saturated solution of barium iodate, "Ba"("IO"_3)_2, has a concentration of iodate anions, ${\text{IO}}_{3}^{-}$, equal to $1.06 \cdot {10}^{- 3} \text{M}$.

This means that when barium iodate is dissolved in water, the molar concentration of the dissociated iodate anions will be equal to $1.06 \cdot {10}^{- 3} \text{M}$.

The dissociation equilibrium for barium iodate in aqueous solution looks like this

${\text{Ba"("NO"_ 3)_ (color(red)(2)(s)) rightleftharpoons "Ba"_ ((aq))^(2+) + color(red)(2)"IO}}_{3 \left(a q\right)}^{-}$

Notice that every mole of barium iodate that dissociates produces $1$ mole of barium cations, ${\text{Ba}}^{2 +}$, and $\textcolor{red}{2}$ moles of iodate anions.

This means that in a saturated barium iodate solution, the concentration of iodate anions will be twice as high as the concentration of barium cations.

This means that the latter will be equal to

$\left[{\text{Ba"^(2+)] = 1/color(red)(2) * ["IO}}_{3}^{-}\right]$

["Ba"^(2+)] = 1/color(red)(2) * 1.06 * 10^(-3)"M" = 5.30 * 10^(-4)"M"

Now, the solubility product constant, ${K}_{s p}$, for this dissociation equilibrium is defined as

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

Plug in your values to find

K_(sp) = 5.30 * 10^(-4)"M" * (1.06 * 10^(-3)"M")^color(red)(2)

${K}_{s p} = 5.96 \cdot {10}^{- 10} {\text{M}}^{3}$

Solubility product constants are usually given without added units, which means that you answer is

${K}_{s p} = \textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{5.96 \cdot {10}^{- 10}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

The answer is rounded to three sig figs.