Question

2. A solution NaF is add dropwise to a solution that is .0122 M in Ba . When the concentration of F exceeds ______M, BaF2 will precipitate. Neglect volume changes. BaF2 K

Answer 1

To predict when BaF2 will precipitate, the product of the ion concentrations is compared to its Ksp. Without Ksp for BaF2, the exact F- concentration for precipitation cannot be calculated, but exceeding a certain F- concentration will result in BaF2 precipitation.

Step-by-step explanation:

The student's question pertains to the precipitation of BaF2 from a solution containing barium ions (Ba2+) when fluoride ions (F-) from a NaF solution are added. To determine at what concentration of F- the BaF2 will precipitate, the solubility product constant (Ksp) for BaF2 is required. As per the provided examples, when dealing with precipitation reactions, the ion product of the relevant ions is compared to the Ksp to predict whether precipitation will occur.

In the absence of the Ksp value for BaF2 in the question, we use the general approach: if the ion product [Ba2+][F-]2 is greater than the Ksp, then BaF2 will precipitate. Without knowing the exact Ksp value for BaF2, a precise fluoride ion concentration that causes precipitation can't be calculated. However, the principle remains that upon exceeding a certain concentration of F-, BaF2 precipitation is expected.

Answer 2

Step-by-step explanation:

BaF2(s) <------> Ba2+(aq) + 2F-(aq)

Ksp BaF2 = 1.0 x 10^-6.

Ksp BaF2 = [Ba2+(aq)]×[F-(aq)]^2 at equilibrium

When Qsp >Ksp, BaF2 will precipitate

Qsp = [Ba2+(aq)]×[F-(aq)]^2

[Ba2+(aq)]×[F-(aq)]^2 > 1.0 x 10^-6.

0.0122 moldm-3 × [F-(aq)]^2 > 1.0 x 10^-6

[F-(aq)]^2 > 1×10^-6 / 0.0122 mol2dm-6

[F-(aq)]^2 > 81.96 × 10^-6 mol2dm-6

[F- (aq)] > 9.05 × 10^-3 moldm-3

So F- concentration should be more than 9.05 × 10^-3 moldm-3