BaCl2 + 2KIO3 ==> Ba(IO3)2 + 2KCl
millimols BaCl2 = M x mL = 1.4
mmols KIO3 = 3
The is a limiting regent (LR) and solubility product problem rolled into one. First, what is the LR?
1.4 mmols BaCl2 will form 1.4 mmols Ba(IO3)2 if we have all of the KIO3 needed.
3 mmols KIO3 will form 1.5 mmols Ba(IO3) if we have all of the BaCl2 needed. But we don't.
The correct value for mols product formed in LR problems is ALWAYS the smaller value so BaCl2 is the limiting reagent in this problem. All of the BaCl2 will be used and 2.8 mmols KIO3 will be used leaving no BaCl2 and leaving 0.2 millimols KIO3. (KIO3) = mmols/mL = 0.2 mmols/200 mL = 0.001M
Now we do the Ksp part.
..........Ba(IO3)2 ==> Ba^2+ + 2IO3^-
I.........solid........0........0.001
C.........solid........x........x
E.........solid........x........0.001+x
Ksp = (Ba^2+)(IO3^-)^2
Substitute the E line into Ksp expression and solve for x = (Ba^2+) = [Ba(IO3)2] solubility.
Calculate the solubility of barium iodate in a solution that results when 100 ml of 0.014 M barium chloride are mixed with 100 ml of 0.03 M potassium iodate.
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