Calculate the solubility of LaF3 in grams per liter in the following solutions. The Ksp of LaF3 is 2.01 e-19.
0.013 M KF Solution
This is a problem to illustrate the common ion effect. The common ion, in this case, is F^- from KF and the presence of KF will decrease the solubility of LfF3 in water.
........LaF3 ==> La^3+ + 3F^-
I.......solid.....0.......0
C.......solid.....x.......3x
E.......solid.....x.......3x
and
........KF ==> K^+ + F^-
I.....0.013M....0.....0
C....-0.013...0.013...0.013
E......0......0.013...0.013
Ksp = (La^3+)(F^3-)^3
You know Ksp.
(La^3+) = x
(F^3-) = 3x from LaF3 and 0.013 from KF or 3x+0.013
Solve for x in mols/L
To calculate the solubility of LaF3 in grams per liter in a solution of 0.013 M KF, you need to use the Ksp expression for LaF3, which is given as 2.01 e-19.
The balanced equation for the dissociation of LaF3 is:
LaF3(s) ⇌ La3+(aq) + 3F^-(aq)
From this equation, we can see that one mole of LaF3 dissociates into one mole of La3+ ions and three moles of F- ions.
To use the Ksp expression, we need to determine the concentration of La3+ ions and F- ions in the solution. Since LaF3 dissociates in a 1:1 ratio, the concentration of La3+ ions is the same as the concentration of LaF3 that dissolves.
Let's assume the solubility of LaF3 in grams per liter is S g/L.
The solubility product expression for LaF3 is:
Ksp = [La3+][F-]^3
The concentration of F- ions can be calculated from the concentration of KF, which is given as 0.013 M.
Since KF dissociates completely into K+ and F- ions, the concentration of F- ions is also 0.013 M.
Using the stoichiometry of the dissolution reaction, the concentration of La3+ ions is also 0.013 M.
Substituting these values into the solubility product expression:
2.01 e-19 = (0.013)(S)^3
To solve for S, we rearrange the equation:
S^3 = (2.01 e-19) / (0.013)
S^3 = 1.54 e-17
Taking the cube root of both sides gives:
S ≈ 2.53 e-6
Therefore, the solubility of LaF3 in grams per liter in a 0.013 M KF solution is approximately 2.53 e-6 g/L.
To calculate the solubility of LaF3 in grams per liter in a 0.013 M KF solution, we need to use the concept of the solubility product constant (Ksp).
The solubility product constant (Ksp) is a measure of the equilibrium saturation point of a compound in a solution. It represents the product of the concentrations of the ions formed when the compound dissolves.
The dissociation equation for LaF3 can be written as:
LaF3(s) ⇌ La3+(aq) + 3F-(aq)
The Ksp expression for LaF3 is:
Ksp = [La3+][F-]^3
Given that the Ksp of LaF3 is 2.01 e-19, and assuming that the dissolution of LaF3 in a KF solution is the primary source of La3+ and F- ions:
1. Calculate the equilibrium concentration of F- ions:
Since KF is a strong electrolyte, it will completely dissociate into K+ and F- ions in solution. Therefore, [F-] = [KF].
[F-] = 0.013 M
2. Calculate the equilibrium concentration of La3+ ions (assuming all of the LaF3 dissolved):
From the chemical equation, the concentration of La3+ ions is equal to the concentration of F- ions raised to the power of 1/3.
[La3+] = ([F-])^(1/3)
[La3+] = (0.013 M)^(1/3)
3. Calculate the solubility of LaF3:
The solubility of LaF3 is equal to the concentration of La3+ ions multiplied by the molar mass of LaF3.
Solubility = [La3+] × molar mass of LaF3
To find the molar mass of LaF3:
The molar mass of La is 138.91 g/mol, and the molar mass of F is 18.99 g/mol.
Molar mass of LaF3 = (138.91 g/mol) + 3 × (18.99 g/mol)
4. Substitute the calculated values into the equation to find the solubility of LaF3:
Solubility = ([F-])^(1/3) × molar mass of LaF3