# The ksp of Ba(IO3)2 at 25 is 6×10^10 . What is the molar solubility of Ba(IO3)2 ?

• The balanced equation is:

Ba(IO3)2 <-> Ba2 + 2IO3

I….._………….0……..0

C…_………..+x……+2x

F…_………….x………2x

Once balanced, input information at equilibrium:

Ksp = [Ba][IO3]^2

6.0×10^-10 = (x)(2x)^2

6.0×10^-10 = 4x^3

x = 0.00053 M = 0.00053 mol/L

This value for x also equals the molar solubility (mol/L) of Ba(IO3)2.

• Ksp = [Ba+2][IO3]^2

If x g-ions of Ba+2 are in solution, so are 2x g-ions of the iodate. Then

4x^3= 6×10^-10 or x^3=150x^-12. Then x=5.2×10^-4, which is the molar solubility of the salt.

• Ba Io3 2

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Ksp = [Ba^2+(aq)][CrO4^2-(aq)] = 2.1*10^-10 [CrO4^2-(aq)] = 2.1*10^-10 / [Ba^2+(aq)] Since BaCrO4 has very low solubility, we can assume that in 0.0015M Na2CrO4(aq), [CrO4^2-(aq)] = 0.0015 mol/litre. So, 1.5*10^-3 = 2.1*10^-10 /[Ba^2+(aq)] then, [Ba^2+(aq)] = 2.1*10^-10 / 1.5*10^-3 = 1.4*10^-7 mol/litre. On dissolving BaCrO4, 1 mole of BaCrO4 produces 1 mole of Ba^2+ ions Therefore, Solubility of BaCrO4 = 1.4*10^-7 mol/litre Since BaCrO4 has Mr = 253 Solubility = 253 * 1.4*10^-7 g/litre = 3.54*10^-5 g/litre.