A solution of NaF is added dropwise to a solution that is 0.0144 M in Ba2+. When the concentration of F- exceeds M, BaF2 will precipitate. Neglect volume changes. For BaF2, Ksp = 1.7

When the concentration of F- exceeds 0.0109 M, BaF2 will precipitate.

Explanation:

Ba²⁺(aq) + 2 F⁻(aq) <> BaF₂(s)

When BaF₂ precipitates, the Ksp relation is given by

Ksp = [Ba²⁺] [F⁻]²

[Ba²⁺] = 0.0144 M

[F⁻] = ?

Ksp = (1.7 × 10⁻⁶)

1.7 × 10⁻⁶ = (0.0144) [F⁻]²

[F⁻]² = (1.7 × 10⁻⁶)/0.0144 = 0.0001180555

[F⁻] = √0.0001180555 = 0.01086 M = 0.0109 M

Hope this Helps!!!

c u c l 2 + 2 n a n o 3 → c u ( n o 3 ) 2 + 2 n a c l

  long weird answer but i hope this works

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