Question

In: Chemistry

Calculate the solubility of LaF3 (Ksp = 2 x 10-19) in grams per liter in: a)...

Calculate the solubility of LaF3 (Ksp = 2 x 10-19) in grams per liter in: a) pure water b) 0.010 M KF solution

Expert Solution

Solubility in pure water : Let S (mol/L) be the solubility

LaF₃ La³⁺ + 3F^-

1 mole of LaF₃ produces 1 mole of La³⁺ and 3 moles of F ^-

concentration of [ La³⁺ ] = S and

concentration of [F^-] = 3S

Solubility product constant ,K_sp = [ La³⁺ ] [F^-]³

2x10^-19 = S x (3S)³

27S⁴ = 2x10^-19

S = 9.28x10^-6 mol/L

Therefore the solubility in pure water is 9.28x10^-6 mol/L

Solubility in 0.010M KF : Let S (mol/L) be the solubility

KF K⁺ + F^-

0.010M 0.01 M 0.01 M

So [F^-] = 3S + 0.01

[ La³⁺ ] = S

Solubility product constant ,K_sp = [ La³⁺ ] [F^-]³

2x10^-19 = S x (3S + 0.01)³

2x10^-19 = S x ( 0.01)³ Since solubility is very low, 3S+0.01 can be taken as 0.01

S = 2.0x10^-13 mol/L

Therefore the solubility is 2.0x10^-13 mol/L

queen_honey_blossom answered 9 months ago

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