Question

Suppose a solution contains 0.28 M Pb2 and 0.46 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

Pb(OH)2 <=> Pb2+ + 2 OH-

[Pb2+] = 0.28 M

Ksp = [Pb2+][OH-]^2

= 0.28 x [OH-]^2 = 1.43 x 10^(-20)

[OH-] = (1.43 x 10^(-20)/0.28)^(1/2) = 2.260 x 10^(-10) M

pOH = -log[OH-] = -log(2.260 x 10^(-10)) = 9.65

pH = 14 - pOH = 4.35 (for Pb(OH)2 precipitation)

Al(OH)3 <=> Al3+ + 3 OH-

[Al3+] = 0.46 M

Ksp = [Al3+][OH-]^3

= 0.46 x [OH-]^3 = 4.6 x 10^(-33)

[OH-] = (4.6 x 10^(-33)/0.46)^(1/3) = 2.154 x 10^(-11) M

pOH = -log[OH-] = -log(2.154 x 10^(-11)) = 10.67

pH = 14 - pOH = 3.33 (for Al(OH)3 precipitation)

Thus for Al(OH)3 to precipitate without precipitating Pb(OH)2:

3.33 < pH < 4.35