Question

Suppose a solution contains 0.26 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

The Ksp data:

Al(OH)3 4.6×10–33; Pb(OH)2 1.43×10–20

note that

Al(OH)3 has the "form" of an AB3 salt:

AB3 <--> A+3 + 3B- ; Ksp = [A+3][B-]^3

Get [OH-] required

(4.6*10^-33) = (0.49)(OH-)^3

[OH-] = ((4.6*10^-33) /(0.49))^(1/3) = 2.109*10^-11 M

pOH = -log(2.109*10^-11) = 10.675

pH = !4-pOH = 14-10.675 = 3.325

and Pb(OH)2 has a "form" of an AB2 salt:

AB2 <--> A+2 + 2B- ; Ksp = [A+2][B-]^2

get [OH-]

1.43*10^-20 = (0.26)(OH)^2

[OH-] = ((1.43*10^-20)/(0.26))^(1/2) = 2.345*10^-10

pOH = -log(2.345*10^-10) = 9.62985

ph = 14-9.62985

pH = 4.37015

3.325 < pH < 4.37015

any range bewteween this point will allow Al(OH)3(s) to form, but not Pb(OH)2