Question

# Part B) Calculate the solubility of LaF3 in grams per liter in 1.2×10−2 M KF solution....

Part B) Calculate the solubility of LaF3 in grams per liter in 1.2×10−2 M KF solution.

Part C) Calculate the solubility of LaF3 in grams per liter in 5.0×10−2 M LaCl3 solution.

#### Homework Answers

Answer #1

PART B : -

LiF3 (s) <-> La3+(aq) + 3 F-(aq)

So the ionic molarities in saturated solution satisfy
Ksp = [La3+] * [F-]3

Ksp = solubility product, -------it has to be given , i think it is given in part A

In a 1.2 * 10-2 = 0.012 KF solution there are already 0.012M fluoride ions. Thus,

[La3+] = x
[F-]= 0.012 + 3*x
Since x is small compared to 0.012, you can approximate:

Ksp = [La3+] *[F⁻]3
=>Ksp = x∙(0.012)3= 1.728*10-6 x
=> x = Ksp / 1.728*10-6 mol / L    (substitute the value of Ksp and calcualte x )

PART C :-

In a 0.05M LaCl3 solution there are already 0.05M lanthanum ions. Thus,
[La3+] = 0.05 + x
[F-]= 3x
Since x is small compared to 0.05, you can approximate:

=> Ksp = [La3+]*[F-]3

By substituting the value of Ksp we can find the value of the solubility.

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