Answer:
The Ba(IO₃)₂ will precipitate out in the solution.
Step-by-step 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.