Question

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

Answer 1

Answer:

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!!!

Answer 2

A solution of NaF is added to a solution that is 0.0144 M in Ba²⁺. When the concentration of F- exceeds 0.011 M, BaF₂ will precipitate.

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 × 10⁻⁶.

Let's consider the solution of BaF₂.

BaF₂(s) ⇄ Ba²⁺(aq) + 2 F⁻(aq)

We can use the solubility product constant (Ksp) to find the equilibrium concentration of F⁻ when [Ba²⁺] is 0.0144 M.

$Ksp = 1.7 \times 10^{-6} = [Ba^{2+} ][F^{-} ]^{2} = (0.0144)[F^{-} ]^{2} \\[F^{-} ] = 0.011 M$

When [F⁻] exceeds 0.011 M, BaF₂ will precipitate.

A solution of NaF is added to a solution that is 0.0144 M in Ba²⁺. When the concentration of F- exceeds 0.011 M, BaF₂ will precipitate.

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