Question

Suppose a solution contains 0.21 M Pb2+ and 0.44 M Al3+. Calculate the pH range that would allow Al(OH)3 to precipitate but not Pb(OH)2. The ksp values are Al(OH)3 = 4.6 x 10^-33

Pb(OH)2= 1.43 x 10^-22

Answer 1

First Al(OH)3:

Al(OH)3(s) <--> Al3+(aq) + 3 OH-(aq)

Ksp = [Al3+][OH-]^3 = 4.6X10^-33

Pb(OH)2(s) <--> Pb2+(aq) + 2 OH-(aq)

Ksp = [Pb2+][OH-]^2 = 1.43 X10^-20

Now, for each ion, calculate the [OH-] at which that ion will begin to precipitate:

Al(OH)3: 4.6X10^-33 = (0.44)[OH-]^3

[OH-] = 2.2X10^-11 M

Pb(OH)2: 1.43X10^-20 = 0.21 [OH-]^2

[OH-] = 2.6X10^-10 M

From this, you see that Al(OH)3 will begin to precipitate when [OH-] = 2.2X10^-11 M, but Pb(OH)2 will not precipitate until [OH-] = 2.6X10^-10 M.

So, of [OH-] = 2.2X10^-11, pOH = 10.66 and pH = 14 - pOH = 3.34

and then [OH-] = 2.6X10^-10, pOH = 9.58 and pH = 4.41

So, the pH range is 3.34 to 4.41