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5.0 mL of 0.050 M Ba(NO3)2 M and 100.0 mL of 0.10 M KIO3 are added together in a 250 mL erlenmeyer flask. In this problem, ignore ionic strength effects and neglect the possible presence of any complex ions in solution. a. Will Ba(IO3)2 precipitate out of solution? Support your answer with appropriate calculations. b. What are the concentrations of Ba2+ and 103 in the solution? Report your answer using 2 significant figures.

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Answer:

The Ba(IO₃)₂ will precipitate out in the solution.

Explanation:

The solubility of Ba(IO₃)₂ is:

Ba(IO₃)₂(s) → Ba²⁺(aq) + 2 IO₃⁻(aq)

Where ksp, is defined as:

ksp =  [Ba²⁺] [IO₃⁻]² = 1.5x10⁻⁹

If the multiplication of  [Ba²⁺] [IO₃⁻] is higher than ksp, the reaction will produce Ba(IO₃)₂, that means the solid will precipitate.

In solution, concentration of Ba²⁺ and  IO₃⁻ are:

[Ba²⁺] = 0.050M × (5mL / 105mL) = 0.00238M

[IO₃⁻] = 0.10M × (100mL / 105mL) = 0.0952M

Replacing in ksp formula:

[0.00238] [0.0952]² = 2.16x10⁻⁵. As the result is higher than ksp, The Ba(IO₃)₂ will precipitate out in the solution.

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