Question

A solution of naf is added dropwise to a solution that already contains 0.0144 m ba2 . When the concentration of f- exceeds _____m, BaF₂ will precipitate. ksp for BaF₂ = 1.7 x 10-6.

Answer 1

When the concentration of fluoride ions (F⁺) exceeds 0.0343 M in a solution containing 0.0144 M of Ba²⁺, barium fluoride (BaF₂) will precipitate, as this concentration will make the ion product surpass the solubility product constant (Ksp) of 1.7 x 10⁻⁶ for BaF₂.

Step-by-step explanation:

To determine when BaF₂ will precipitate, we need to compare the ion product (Q) of Ba²⁺ and F⁺ with the solubility product constant (Ksp) given as 1.7 x 10⁻⁶ for BaF₂. Precipitation will occur when Q exceeds Ksp. The ion product Q is calculated as [Ba²⁺][F⁺]², where [Ba²⁺] is 0.0144 M. For precipitation to occur, [F⁺]² must be greater than Ksp / [Ba²⁺]. Therefore, [F⁺] must exceed √(Ksp / [Ba²⁺]).

After calculating:

Ksp / [Ba²⁺] = 1.7 x 10⁻⁶ / 0.0144 M = 1.1806 x 10⁻⁴
[F⁺] must exceed √(1.1806 x 10⁻⁴) = 0.0343 M

So, when the concentration of F⁺ exceeds 0.0343 M, BaF₂ will precipitate.