A solution of NaF is added dropwise to a solution that is 0.0173 M in Ba 2. When the concentration of F- exceeds___M BaF2 will precipiate. Neglect volume changes. For BaF2

0.0099M = [F⁻]

Explanation:

For BaF2, Ksp = 1.7x10⁻⁶

When BaF₂ is in solution, the equilibrium between the solid and the dissociated ions occurs as follows:

BaF₂(s) ⇄ Ba²⁺(aq) + 2F⁻(aq)

Where Ksp = 1.7x10⁻⁶ is defined as:

1.7x10⁻⁶ = [Ba²⁺] [F⁻]²

Where [] are equilibrium concentrations of each ion in solution.

That means you will add F⁻ until its concentration exceeds:

1.7x10⁻⁶ = [0.0173] [F⁻]²

9.827x10⁻⁵ = [F⁻]²

When more F⁻ is added, BaF₂ begins its precipitation.

the answer is 75.0%.