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.

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.

To calculate the solubility of barium iodate in the resulting solution, you need to apply the principles of stoichiometry and use the given concentrations of barium chloride and potassium iodate. Here are the steps to determine the solubility:

Step 1: Write the balanced equation for the reaction between barium chloride (BaCl2) and potassium iodate (KIO3) to form barium iodate (Ba(IO3)2).
BaCl2 + KIO3 → Ba(IO3)2 + KCl

Step 2: Determine the limiting reactant. The limiting reactant is the one that is completely consumed in the reaction and determines the maximum amount of product that can be formed. To find the limiting reactant, you need to compare the number of moles of each reactant.

Moles of BaCl2 = volume (L) × concentration (M)
= 0.100 L × 0.014 M
= 0.0014 moles

Moles of KIO3 = volume (L) × concentration (M)
= 0.100 L × 0.03 M
= 0.003 moles

According to the balanced equation, 1 mole of BaCl2 reacts with 1 mole of KIO3 to produce 1 mole of Ba(IO3)2. So, the ratio of BaCl2 to KIO3 is 1:1.

Since the moles of BaCl2 and KIO3 are in a 1:1 ratio, we can see that the reaction has a balanced stoichiometry. Therefore, the stoichiometry does not dictate that either reactant would run out or be in excess. Both BaCl2 and KIO3 are used up in a 1:1 stoichiometric ratio.

Step 3: Since both reactants are completely consumed and used in a 1:1 stoichiometric ratio, the concentration of barium iodate (Ba(IO3)2) formed will be equal to the lowest initial concentration of either reactant. Therefore, the solubility of barium iodate in the solution is 0.014 M.

The solubility of barium iodate in the resulting solution is 0.014 M.