The concentration of "IO"_3^(-)IO−3 ions in a pure, saturated solution of "Ba"("IO"_3)_2Ba(IO3)2 is 1.06xx10^(-3)"M"1.06×10−3M. What is the K_(sp)Ksp for "Ba"("IO"_3)_2Ba(IO3)2?
1 Answer
Explanation:
The problem tells you that a saturated solution of barium iodate,
This means that when barium iodate is dissolved in water, the molar concentration of the dissociated iodate anions will be equal to
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))^(-)Ba(NO3)2(s)⇌Ba2+(aq)+2IO−3(aq)
Notice that every mole of barium iodate that dissociates produces
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^(-)][Ba2+]=12⋅[IO−3]
["Ba"^(2+)] = 1/color(red)(2) * 1.06 * 10^(-3)"M" = 5.30 * 10^(-4)"M"[Ba2+]=12⋅1.06⋅10−3M=5.30⋅10−4M
Now, the solubility product constant,
K_(sp) = ["Ba"^(2+)] * ["IO"_3^(-)]^color(red)(2)Ksp=[Ba2+]⋅[IO−3]2
Plug in your values to find
K_(sp) = 5.30 * 10^(-4)"M" * (1.06 * 10^(-3)"M")^color(red)(2)Ksp=5.30⋅10−4M⋅(1.06⋅10−3M)2
K_(sp) = 5.96 * 10^(-10)"M"^3Ksp=5.96⋅10−10M3
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.
