A solution contains 2.0 × 10^–3 M Ba^2+ and 1.0 × 10^–2 M Sc^3+. A source of F^– is slowly added to the solution. Will BaF2 (Ksp = 1.84 × 10^–7) or ScF3 (Ksp = 5.81 × 10^–24) precipitate first?

I believe that ScF3 would precipitate first but I don't know how to prove this mathematically.

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

I.......solid.....0........0
C........-x.......x........2x
E.......solid.....x........2x

Ksp = (Ba^2+)(F^-)^2
You know (Ba^2+) = 2E-3M
Solve for F^-

..........ScF3 ==> Sc^2+ + 3F^-
I.........solid.....0.......0
C..........-x.......x.......3x
E..........solid....x.......3x

Ksp = (Sc^3+)(F^-)^3
You know (Sc^3+) = 1E-2
Solve for F^-.

Since you are adding F^- drop by drop, the one that ppts first will be the one with the smallest F^- in your calculations. That will be the one that reaches Ksp first.

Well, math is not really my strong suit. I prefer to balance equations with a unicycle. But I'll try to help you out!

To determine which compound will precipitate first, we need to compare the solubility product constants (Ksp) of BaF2 and ScF3. The compound with the smaller Ksp value is less soluble and more likely to precipitate first.

Let's start with BaF2. Its Ksp is 1.84 × 10^–7, while the concentration of Ba^2+ is 2.0 × 10^–3 M and the concentration of F^– is unknown (as we're slowly adding it).

For ScF3, its Ksp is a whopping 5.81 × 10^–24. The concentration of Sc^3+ is 1.0 × 10^–2 M, and the concentration of F^– is again unknown.

Comparing the two Ksp values, we can see that ScF3 has a much smaller Ksp, which means it's way less soluble than BaF2. So, you're absolutely right! ScF3 will likely reach its solubility limit and precipitate first.

But hey, every now and then surprises happen in chemistry, just like those unexpected funny chemical reactions! So keep an eye out for any unexpected clownfish squirting water at you when conducting experiments!

To determine which compound will precipitate first, we can compare the solubility product constants (Ksp) of BaF2 and ScF3 with the ion product (Q) of the solution. The compound with the higher value of Q will be more likely to precipitate.

Let's calculate the value of Q for BaF2 and ScF3 in the given solution.

For BaF2:
Ba2+ + 2F- → BaF2

Since the concentration of Ba2+ is 2.0 × 10^–3 M and the concentration of F- is initially 0, the initial value of Q for BaF2 is 0.

For ScF3:
Sc3+ + 3F- → ScF3

Since the concentration of Sc3+ is 1.0 × 10^–2 M and the concentration of F- is initially 0, the initial value of Q for ScF3 is also 0.

Since both Q values are zero initially, neither compound will precipitate at the start. However, as F- ions are slowly added, the concentration of F- increases while the concentrations of Sc3+ and Ba2+ remain the same.

To determine when the first compound will precipitate, we need to calculate the value of Q at the point when one of the compounds starts to precipitate. The compound with a larger Ksp value will reach Q = Ksp first and start to precipitate.

Let's calculate the Q values for BaF2 and ScF3 when one of them first starts to precipitate:

For BaF2:
The concentration of F- will be 2x, where x is the amount of F- ions added. The concentration of Ba2+ remains constant at 2.0 × 10^–3 M.

So, Q for BaF2 at this point is:
Q = [Ba2+][F-]^2 = (2.0 × 10^–3)(2x)^2 = 8x^2 × 10^–3

For ScF3:
The concentration of F- will be 3x, where x is the amount of F- ions added. The concentration of Sc3+ remains constant at 1.0 × 10^–2 M.

So, Q for ScF3 at this point is:
Q = [Sc3+][F-]^3 = (1.0 × 10^–2)(3x)^3 = 27x^3 × 10^–2

To determine when one of the compounds will start to precipitate, we need to compare the Q values with their respective Ksp values. Whichever compound reaches its Ksp value first will precipitate.

Comparing the values, we have:
Q for BaF2 = 8x^2 × 10^–3
Q for ScF3 = 27x^3 × 10^–2
Ksp for BaF2 = 1.84 × 10^–7
Ksp for ScF3 = 5.81 × 10^–24

We can now set up the following inequality:
8x^2 × 10^–3 < 1.84 × 10^–7
27x^3 × 10^–2 < 5.81 × 10^–24

Simplifying the inequalities:
8x^2 < 1.84 × 10^–4
27x^3 < 5.81 × 10^–22

To solve these inequalities, we can divide both sides by the given values:
x^2 < 2.3 × 10^–5
x^3 < 2.15 × 10^–23

Taking the square root of the first inequality:
x < sqrt(2.3 × 10^–5)

Taking the cube root of the second inequality:
x < cuberoot(2.15 × 10^–23)

Calculating the values:
x < 0.0048
x < 0.013

Based on these calculations, when the concentration of F- ions reaches approximately 0.0048M (or 0.013M), BaF2 will have a larger Q value than ScF3, and BaF2 will begin to precipitate before ScF3.

Therefore, BaF2 will precipitate first in the given solution.

To determine which compound will precipitate first, we need to compare the solubility products (Ksp) of BaF2 and ScF3. We can use the concept of the common ion effect to make this comparison.

The common ion effect states that the solubility of a compound decreases when a common ion is present in the solution. In this case, both BaF2 and ScF3 contain the fluoride ion (F^-) as a common ion. Since F^- is being slowly added to the solution, the concentration of F^- will increase gradually.

To find the point where precipitation occurs, we need to compare the concentrations of F^- with the concentrations of Ba^2+ and Sc^3+ ions. The compound with the lowest solubility product (Ksp) compared to the ion concentrations will precipitate first.

Let's calculate the concentration of F^- when both BaF2 and ScF3 are just about to precipitate:

For BaF2: BaF2(s) ⇌ Ba^2+(aq) + 2F^-(aq)
The concentration of F^- (at equilibrium) is 2x, where x is the concentration of Ba^2+.
The solubility product (Ksp) expression for BaF2 is: Ksp = [Ba^2+][F^-]^2

For ScF3: ScF3(s) ⇌ Sc^3+(aq) + 3F^-(aq)
The concentration of F^- (at equilibrium) is 3x, where x is the concentration of Sc^3+.
The solubility product (Ksp) expression for ScF3 is: Ksp = [Sc^3+][F^-]^3

Now, let's substitute the given values into the expressions:

For BaF2: Ksp = (2x)(2x)^2 = 4x^3
For ScF3: Ksp = (1x)(3x)^3 = 27x^4

Since we want to compare the values of Ksp, we do not need to know the exact concentrations of Ba^2+ and Sc^3+.
We can compare the powers of x directly.

In this case, because the powers of x are 3 for BaF2 (4x^3) and 4 for ScF3 (27x^4), ScF3 will have a lower solubility product (Ksp) than BaF2.

Therefore, ScF3 will precipitate first when the concentration of F^- reaches a level where the solubility product (Ksp) of ScF3 is exceeded.