Question

1. Calculate the molar solubility of LaF; in each of the following (Ksp. LaF3 = 2.0x10-19). a) pure water b) 0.50 M NaF c) 0.25 M La(NO3)3

Answer 1

a)

At equilibrium:
LaF3      <---->     La3+     +         3 F-    

                     s                  3s      

Ksp = [La3+][F-]^3
2*10^-19=(s)*(3s)^3
2*10^-19= 27(s)^4
s = 9.277*10^-6 M
Answer: 9.3*10^-6 M

b)
NaF here is Strong electrolyte
It will dissociate completely to give [F-] = 0.5 M
At equilibrium:
LaF3      <---->     La3+     +         3 F-    
                     s                  0.5 + 3s
Ksp = [La3+][F-]^3
2*10^-19=(s)*(0.5+ 3s)^3
Since Ksp is small, s can be ignored as compared to 0.5
Above expression thus becomes:
2*10^-19=(s)*(0.5)^3
2*10^-19= (s) * 0.125
s = 1.6*10^-18 M
Answer: 1.6*10^-18 M

c)

La(NO3)3 here is Strong electrolyte
It will dissociate completely to give [La3+] = 0.25 M
At equilibrium:
LaF3      <---->     La3+     +         3 F-    
                     0.25 +s            3s      
Ksp = [La3+][F-]^3
2*10^-19=(0.25 + s)*(3s)^3
Since Ksp is small, s can be ignored as compared to 0.25
Above expression thus becomes:
2*10^-19=(0.25)*(3s)^3
2*10^-19= 0.25 * 27(s)^3
s = 3.094*10^-7 M
Answer: 3.1*10^-7 M

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