Solubility Problem - Involves Addition of Another Ion

[vertciel]

Homework Statement

How many moles of solid sodium fluoride should be added to 1.0 L of a saturated solution of barium fluoride, BaF₂, at 25^OC to raise the fluoride concentration to 3.0 x 10^-2 mol/L? The solubility product constant for BaF₂ is K_sp = 1.7 x 10^-6 at 25^OC.

The Attempt at a Solution

BaF₂ ---> Ba²⁺ + 2F^-

Initial Concentrations: 0, 0
Change: + x, + 2x

Therefore:

1.7 x 10^-6 = x(2x)²

x = 0.00752 M.

So: [F^-] = 0.01504 mol/2 L.

I thought to have the desired fluoride concentration, only 1.5 x 10^-2 mol would need to be added. However, the correct answer is

Spoiler

2.6 x 10^{-2 mol}

.

Thank you for your help!

 

[Borek]

What happens when you add F^- to the solution?

 

[Blueshift5]

I would like to provide a response to this solubility problem. Firstly, it is important to note that the solubility product constant (Ksp) for a compound is a measure of its solubility in water, and is typically given in units of mol/L. In this problem, we are given the Ksp for barium fluoride (BaF2) at 25OC, which is 1.7 x 10-6 mol/L.

To solve this problem, we can use the equation for Ksp: Ksp = [Ba2+][F-]2. Since we are trying to raise the fluoride concentration to 3.0 x 10-2 mol/L, we can set up the following equation:

Ksp = (3.0 x 10-2 mol/L)2 = [Ba2+][F-]2

Solving for [Ba2+], we get [Ba2+] = Ksp/[F-]2 = (1.7 x 10-6 mol/L)/[3.0 x 10-2 mol/L]2 = 0.0019 mol/L.

Since we are starting with a saturated solution of barium fluoride (BaF2), we know that the concentration of [Ba2+] is already at its maximum solubility. Therefore, to increase the fluoride concentration to 3.0 x 10-2 mol/L, we need to add more fluoride ions (F-).

The amount of solid sodium fluoride (NaF) needed can be calculated by using the equation: [F-] = [NaF] = moles of NaF/volume of solution.

Therefore, [NaF] = (3.0 x 10-2 mol/L) - (0.0019 mol/L) = 0.0281 mol/L.

Since we are working with a 1.0 L solution, the moles of NaF needed can be calculated as:

moles of NaF = (0.0281 mol/L)(1.0 L) = 0.0281 mol.

Therefore, 0.0281 mol of solid sodium fluoride should be added to 1.0 L of the saturated solution of barium fluoride to achieve a fluoride concentration of 3.0 x 10-2 mol/L.

In summary, the key to solving this problem is to recognize that the solubility product