DeaneN245743 DeaneN245743
  • 03-11-2022
  • Chemistry
contestada

Ksp= 2×10−19 for LaF3 Calculate the solubility of LaF3 in grams per liter in a solution that is 0.055 M in LaCl3

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RosaleaN682543 RosaleaN682543
  • 03-11-2022

First, we write our reaction:

We call solubility to "S"

LaF₃(s) → La³⁺(aq) + 3 F⁻(aq)

Initial - 0.055M (LaCl3) 0

Change +S +3.S

Eq. 0.055+S 3.S

Ksp = 2×10^−19 = [La³⁺]x[F⁻]³ = (0.055+S)x(3.S)³

2×10^−19 = (0.055+S)x(3.S)³ => S = 5.12x10^-7 mol/L x (195.9 g/mol) =>

S = 1.00x10^-4 g/L

(195.9 g/mol = The molar mass of LaF3)

Answer: S = 1.00x10^-4 g/L

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