The Ksp for LaF3 is 2.0 x 10-19. What is the solubility of LaF, in water in moles per liter (Molar solubility)?

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Chapter1: Chemical Foundations
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Can you explain the problem please? And where did the X4 came from?

The Ksp for LaF3 is 2.0 x 10-19. What is the solubility of LaF in water in moles per liter
(Molar solubility)?
Solution
Again, set up an equilibrium table according to the solubility equilibrium:
La 3+ (aq) + 3F- (aq)
LaF3 (s)
Initial (M)
Eq (M)
0
0
X
3x
Ksp =[La3+] [F-13
=> Ksp = (x)(3x)³
=> 2 × 10-19 = 27x4
2.0 × 10-19
=
x4
(7.4 × 10-21) 1/4 = x
X
27
x = 9.0 × 10-6 M = [La³+] This is the Molar solubility of La 3+
Thus, by 1:1 formula stoichiometry => 9.0 x 106 M = [LaF3]
Transcribed Image Text:The Ksp for LaF3 is 2.0 x 10-19. What is the solubility of LaF in water in moles per liter (Molar solubility)? Solution Again, set up an equilibrium table according to the solubility equilibrium: La 3+ (aq) + 3F- (aq) LaF3 (s) Initial (M) Eq (M) 0 0 X 3x Ksp =[La3+] [F-13 => Ksp = (x)(3x)³ => 2 × 10-19 = 27x4 2.0 × 10-19 = x4 (7.4 × 10-21) 1/4 = x X 27 x = 9.0 × 10-6 M = [La³+] This is the Molar solubility of La 3+ Thus, by 1:1 formula stoichiometry => 9.0 x 106 M = [LaF3]
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