A solution of NaF is added dropwise to a solution that is 0.0122 M in Ba²⁺. When the concentration of F^- exceeds ________ M, BaF₂ will precipitate. Neglect volume changes. For BaF, Ksp = 1.7 × 10^-⁶. A)7.0 × 10^-5 B)1.2 × 10^-2 C)2.1 × 10^-⁸ D)3.0 × 10^-³ E)1.4 × 10^-4

Question

Quizplus · Accepted Answer

First, we can write the balanced chemical equation for the precipitation reaction:

BaS1U1P12+S1S1P0 + x NaFS1U1P12 -> BaFS1U1B12S1U1B0 + x NaS1S1P0

We can use the expression for the solubility product to find when BaF starts to precipitate:

Ksp = [BaFS1U1B12S1U1B0][NaS1S1P0]/[NaFS1U1P12]

Since we're adding NaF dropwise, we can assume that the concentration of NaF does not change significantly as we add more drops. Therefore, we can use the initial concentration of BaS1U1P12+S1S1P0 to find the initial concentration of NaFS1U1P12:

0.0122 M BaS1U1P12+S1S1P0 = x NaFS1U1P12
x = 0.0122 MNaFS1U1P12

Substituting into our expression for Ksp:

1.7 × 10S1U1P1-S1S1P0S1U1P16S1S1P0 = [BaFS1U1B12S1U1B0][NaS1S1P0]/(0.0122 MNaFS1U1P12)

Solving for [NaS1S1P0], we get:

[NaS1S1P0] = 1.2 × 10S1U1P1-2S1S1P0 M

Therefore, the concentration of FS1U1P1-S1S1P0 must exceed 1.2 × 10S1U1P1-2S1S1P0 M for BaFS1U1B12S1U1B0 to start precipitating.

A Solution of NaF Is Added Dropwise to a Solution