Calculate the solubility of LaF3 in grams per liter in the following solutions. The Ksp of LaF3 is 2.01 e-19.
(a) pure water
(b) 0.051 M KF solution
(c) 0.065 M LaCl3 solution
LaF3 ==> La^+3 + 3F^-
Ksp = (La^+3)(F^-)^3
Set up an ICE chart, substitute into Ksp and solve for solubility. Post your work if you get stuck.
c)
1.79x10^-5
To calculate the solubility of LaF3 in grams per liter in the given solutions, we need to consider the equilibrium constant (Ksp) and the molar concentrations of the relevant ions in the solutions.
(a) Pure water: In pure water, there are no other ions present except for those derived from the dissociation of LaF3. The dissociation equation for LaF3 is:
LaF3(s) ⇌ La3+(aq) + 3F-(aq)
Let's assume the solubility of LaF3 as "x". Since the stoichiometric coefficient of LaF3 is 1, the concentration of La3+ ions in the solution will also be "x". The concentration of F- ions will be 3x (due to the stoichiometric ratio).
Using the Ksp expression, we can write:
Ksp = [La3+][F-]^3
2.01 × 10^(-19) = x * (3x)^3
2.01 × 10^(-19) = 27x^4
Solving this equation will give you the value of "x", which represents the solubility of LaF3 in pure water in moles per liter. To convert moles to grams, you need to know the molar mass of LaF3 (which is 195.9 g/mol) and multiply it by "x".
(b) 0.051 M KF solution: When LaF3 is added to a KF solution, the F- ions from KF will interact with La3+ ions, forming LaF6^3- complex ions. The LaF3 will dissolve by breaking apart because of the stronger interaction between La3+ and F- ions compared to La3+ and Cl- ions.
Since each La3+ ion combines with six F- ions, the concentration of La3+ ions will be the same as the concentration of F- ions originating from KF. Therefore, the concentration of La3+ ions will be 0.051 M.
Using the Ksp expression, we can write:
Ksp = [La3+][F-]^3
2.01 × 10^(-19) = (0.051)^1 * (0.051)^3
2.01 × 10^(-19) = 0.051 * 0.051^3
Solving this equation will give you the concentration of LaF3 in moles per liter. To convert moles to grams, multiply this concentration by the molar mass of LaF3 (195.9 g/mol).
(c) 0.065 M LaCl3 solution: In this case, we have a LaCl3 solution, which already contains La3+ ions. The chloride ions from LaCl3 will interact with the La3+ ions, reducing the concentration of La3+ in the solution. Therefore, the La3+ concentration will be less than 0.065 M.
Similar to case (b), you can use the Ksp expression to solve for the concentration of LaF3. The concentration of La3+ ions will be less than 0.065 M due to the formation of the LaCl-complex ions. Multiply this concentration by the molar mass of LaF3 to get the solubility in grams per liter.
Please note that the calculated values represent the maximum solubility of LaF3 under the given conditions, assuming all the F- ions in the solution are derived from LaF3 dissociation.