Question

Suppose a solution contains 0.22 M Pb2 and 0.49 M Al3 . Calculate the pH range that would allow Al(OH)3 to precipitate but not Pb(OH)2. The Ksp values for Al(OH)3 and Pb(OH)2 can be found here.

Answer 1

Ksp of Al(OH)3 = 1.9*10^-33
Ksp of Pb(OH)2 = 2.8*10^-16

Al(OH)3 = AL3+ + 3OH-
                       0.49        3s

Ksp = [Al3+] [OH-]^3
1.9*10^-33 = 0.59 * (3s)^3
1.9*10^-33 = 0.59 * 27*s^3
s = 4.92*10^-12 M

[OH-] = 3s = 1.48*10^-11 M
pOH = -log [OH-]
          = -log (1.48*10^-11)
          = 10.83
pH = 14 - pOH
       = 14 - 10.83
        = 3.17

Pb(OH)2 = Pb2+ + 2OH-
                       0.22        2s

Ksp = [Pb2+] [OH-]^2
2.8*10^-16 = 0.22 * (2s)^2
2.8*10^-16 = 0.22 * 4*s^2
s = 1.78*10^-8 M

[OH-] = 2s = 3.56*10^-8 M
pOH = -log [OH-]
          = -log (3.56*10^-8 )
          = 7.45
pH = 14 - pOH
       = 14 - 7.45
        = 6.55

Clealry pH range = 3.17 to 6.55