what is the molar solubility of barium fluoride in 0.15M NaF at 25 degrees C?

Ah, the magical world of solubility! Well, let me tell you a joke before we dive into the numbers. Why was the math book sad? Because it had too many problems! Now, let's tackle your question.

The molar solubility of barium fluoride in 0.15M NaF at 25 degrees C can be calculated using the solubility product constant (Ksp) equation. The Ksp for barium fluoride (BaF2) is 2.0 x 10^-6.

To calculate the molar solubility, we need to consider the balanced chemical equation between BaF2 and NaF:

BaF2 (s) ⇌ Ba2+ (aq) + 2F- (aq)

Since the concentration of NaF is 0.15M, we assume that the concentration of F- ions is the same. Let's call the molar solubility of BaF2 as "s". The concentration of Ba2+ ions will be "s" as well.

Using the Ksp equation, we have:
Ksp = [Ba2+][F-]^2
2.0 x 10^-6 = (s)(s)^2
2.0 x 10^-6 = s^3

By solving this equation, we find that the molar solubility of barium fluoride in 0.15M NaF is approximately 1.26 x 10^-2 M.

So, there you have it! The molar solubility of barium fluoride is like a hidden treasure in a sea of NaF solution – approximately 1.26 x 10^-2 M.

To determine the molar solubility of barium fluoride (BaF2) in a solution of 0.15M sodium fluoride (NaF) at 25 degrees Celsius, we need to make use of the solubility product constant (Ksp) for barium fluoride.

The solubility product constant is the equilibrium expression (Ksp) for the dissolution of a sparingly soluble salt into its ions in solution. The general equation for the dissolution of barium fluoride is:

BaF2(s) ↔ Ba2+(aq) + 2F-(aq)

The equilibrium expression is written as:

Ksp = [Ba2+][F-]^2

The given solution contains 0.15M NaF, which provides a concentration of fluoride ions in solution. Since sodium fluoride (NaF) is a soluble salt, we can consider it to be completely dissociated, giving a concentration of 0.15M for fluoride ions ([F-]).

Now, we need to use the Ksp expression and the known concentration of fluoride ions to calculate the molar solubility of BaF2.

Since the molar solubility of BaF2 is defined as "x", we can substitute the relevant values into the Ksp expression:

Ksp = (x)(0.15)^2

Given that Ksp for BaF2 is approximately 1.7 x 10^-6 at 25 degrees Celsius, we can solve for the molar solubility (x).

1.7 x 10^-6 = (x)(0.15)^2

Solving this equation will give us the molar solubility of BaF2 in the 0.15M NaF solution at 25 degrees Celsius.

This is almost correct, except it should be (s)(2s+.15)^2=Ksp

6.9*10-3

BaF2 ==> Ba^+2 + 2F^-

S = solubility
..S......S........2S

NaF ==> Na^+ + F^-
0.15....0.15....0.15

Ksp = (Ba^+2)(F^-)^2 =
(S)(S+0.15)^2 = Ksp. Solve for S.