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#?

The problem tells you that a saturated solution of barium iodate, #"Ba"("IO"_3)_2#, has a concentration of iodate anions, #"IO"_3^(-)#, equal to #1.06 * 10^(-3)"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 * 10^(-3)"M"#.

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

#"Ba"("NO"_ 3)_ (color(red)(2)(s)) rightleftharpoons "Ba"_ ((aq))^(2+) + color(red)(2)"IO"_ (3(aq))^(-)#

Notice that every mole of barium iodate that dissociates produces #1# mole of barium cations, #"Ba"^(2+)#, and #color(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

#["Ba"^(2+)] = 1/color(red)(2) * ["IO"_3^(-)]#

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

Now, the solubility product constant, #K_(sp)#, for this dissociation equilibrium is defined as

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

Plug in your values to find

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

#K_(sp) = 5.96 * 10^(-10)"M"^3#

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

#K_(sp) = color(green)(|bar(ul(color(white)(a/a)color(black)(5.96 * 10^(-10))color(white)(a/a)|)))#

The answer is rounded to three sig figs.