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Ksp= 2×10−19 for LaF3 Calculate the solubility of LaF3 in grams per liter in a solution that is 0.011 M in KF

Sagot :

Solubility of LaF3: "S"

First, we need to write the reaction of solubility:

LaF3 (s) <=> La+3 + 3 F-

Initial - 0 +0.011 (F-, from KF)

Change +S +3.S

Equilibrium S 0.011 + 3.S

Now, we know:

Ksp = [La³⁺]x[F⁻]³ = 2×10^−19 = S x (0.011+3.S)³

We work with: 2×10^−19 = S x (0.011+3.S)³, and we clear the "S"

=> S = 1.5x10^-13 mol/L x (195.9 g/mol) = 2.93x10^-11 g/L

The molar mass of LaF3 = 195.9 g/mol

Answer: S = 2.93x10^-11 g/L

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