Calculate the solubility of LaF3 in grams per liter in the following solutions. (Ksp = 2. 10-19.)
a) pure water
Ans: .00182g/L
b) 0.037 M KF solution
c) 0.055 M LaCl3 solution
NEED b&c
PLEASE HELP
To calculate the solubility of LaF3 in grams per liter in the given solutions, we will use the concept of the solubility product constant (Ksp). The Ksp expression for the dissociation of LaF3 is given as follows:
LaF3 ⇌ La3+ + 3F-
a) Pure Water Solution:
In pure water, there are no other ions present to react with LaF3. Therefore, we can assume that the concentration of La3+ and F- ions is equal to the solubility of LaF3.
Let's assume the solubility of LaF3 in grams per liter in pure water is x g/L.
From the balanced equation, we know that the molar ratio of LaF3 to La3+ is 1:1. Therefore, the concentration of La3+ ions (in mol/L) is also x mol/L.
Since each LaF3 molecule releases three fluoride ions (F-) upon dissolution, the concentration of F- ions (in mol/L) is 3x mol/L.
Since the Ksp expression for LaF3 is: Ksp = [La3+][F-]³, we can substitute the concentrations:
Ksp = (x)(3x)³
2.10⁻¹⁹= 27x⁴
x⁴ = (2.10⁻¹⁹)/27
x ≈ 2.66 x 10⁻⁶ mol/L
To convert the solubility from mol/L to g/L, we need to multiply it by the molar mass of LaF3, which is 195.903 g/mol:
Solubility = (2.66 x 10⁻⁶ mol/L) x (195.903 g/mol)
Solubility ≈ 0.000519 g/L
Therefore, the solubility of LaF3 in pure water is approximately 0.000519 g/L (or 0.00182 g/L based on the answer given).
b) KF Solution:
In a 0.037 M KF solution, the F- ions from KF will react with La3+ to form LaF3. Assuming complete dissolution, we can calculate the concentration of F- ions (in mol/L) from the KF solution.
Since KF dissociates into K+ and F- ions, the concentration of F- in the solution is 0.037 M.
From the balanced equation, we know that the molar ratio of LaF3 to F- is 1:3. Therefore, the concentration of LaF3 (in mol/L) will be three times the concentration of F-.
Concentration of LaF3 = 3 x 0.037 M = 0.111 M
To convert the concentration from mol/L to g/L, we can use the molar mass of LaF3:
Solubility = (0.111 mol/L) x (195.903 g/mol) ≈ 21.71 g/L
The solubility of LaF3 in a 0.037 M KF solution is approximately 21.71 g/L.
c) LaCl3 Solution:
In a 0.055 M LaCl3 solution, the LaCl3 will dissociate into La3+ and Cl- ions. The Cl- ions do not react with the LaF3, so we only need to consider the concentration of La3+.
Since LaCl3 dissociates into La3+ and 3 Cl- ions, the concentrations of La3+ and Cl- ions are equal to 0.055 M.
Since the molar ratio of LaF3 to La3+ is 1:1, the concentration of LaF3 (in mol/L) will also be 0.055 M.
To convert the concentration from mol/L to g/L, we can use the molar mass of LaF3:
Solubility = (0.055 mol/L) x (195.903 g/mol) ≈ 10.77 g/L
The solubility of LaF3 in a 0.055 M LaCl3 solution is approximately 10.77 g/L.
Note: These values are approximate and may differ slightly depending on the accuracy of the given data and calculations.
To calculate the solubility of LaF3 in grams per liter in solutions, we need to use the given Ksp value and the concentrations of the solutions. Let's calculate the solubility in each solution:
b) 0.037 M KF solution:
Since KF is a soluble salt, we need to calculate the concentration of fluoride ions (F-) in the solution.
KF dissociates as follows: KF → K+ + F-
Since the concentration of KF is given as 0.037 M, the concentration of fluoride ions will also be 0.037 M.
Since LaF3 dissociates as follows: LaF3 ↔ La3+ + 3F-, the concentration of fluoride ions will be equal to 3 times the solubility of LaF3.
Let's assume the solubility of LaF3 in grams per liter in this solution is "x" g/L.
Thus, [F-] = 0.037 M = 3x
Solving for x, we get:
x = (0.037 M) / 3
Using x = (0.037 M) / 3 = 0.0123 M, we can now calculate the solubility in grams per liter:
Molar mass of LaF3 = 195.9 g/mol (La) + 3(18.998 g/mol) (F) = 231.89 g/mol
Solubility of LaF3 in grams per liter = (0.0123 M) * (231.89 g/mol)
= 2.85 g/L (approx.)
Therefore, the solubility of LaF3 in a 0.037 M KF solution is approximately 2.85 g/L.
c) 0.055 M LaCl3 solution:
Similarly, we need to calculate the concentration of La3+ ions in the LaCl3 solution.
LaCl3 dissociates as follows: LaCl3 → La3+ + 3Cl-
Since the concentration of LaCl3 is given as 0.055 M, the concentration of La3+ ions will also be 0.055 M.
Since LaF3 dissociates as follows: LaF3 ↔ La3+ + 3F-, the concentration of La3+ ions will be equal to the solubility of LaF3.
Let's assume the solubility of LaF3 in grams per liter in this solution is "y" g/L.
Thus, [La3+] = 0.055 M = y
Using y = 0.055 M, we can now calculate the solubility in grams per liter:
Solubility of LaF3 in grams per liter = (0.055 M) * (231.89 g/mol)
= 12.75 g/L (approx.)
Therefore, the solubility of LaF3 in a 0.055 M LaCl3 solution is approximately 12.75 g/L.
Let x = solubility LaF3.
...........LaF3 ==> La^3+ + 3F^-
............x.......x.......3x
...........KF ==> K^+ + F^-
initial.0.037......0....0
change...-0.037..0.037...0.037
equil.....0.....0.037...0.037
Ksp LaF3 = (La^3+)(F^-)^3
From the above charts.
(La^3+) = x
(F^-) = 3x from the LaF3 and 0.037 from the KF to make a total of 3x+0.037
Solve for x
c is done the same way except
(La^3+) = x from LaF3 + 0.0055 from LaCl3.
(F^-) = 3x
Solve for x.