Question

Suppose a solution contains 0.23 M Pb^2+ and 0.49 M AI^3+. Calculate the pH range that would allow AI(OH)3 to precipitate but not Pb(OH)2. The Ksp values for AI(OH)3 and Pb(OH)2 can be found here.

Answer 1

Solution: Lets calculate concentration of OH- for both Al(OH)3 and Pb(OH)2

Ksp for Al(OH)3 = 1.9E-33

We know ksp expression for Al(OH)3 is

Ksp = [Al3+] [OH-]³

1.9E-33 = 0.49 * [OH-]³

[OH-]= 1.57E-11

pOH = -log [OH-]

= -log ( 1.57E-11)

=10.80

pH = 14 – pOH

= 14-10.80= 3.20

Lets calculate pH for Pb(OH)2

Ksp = [Pb2+ ] [ OH-]²

2.8 E-16 = 0.23 * [OH-]²

[OH-]= 3.49E-8

pOH = -log (3.49E-8)

= 7.46

pH = 14 – 7.46 = 6.54

we see that pH of Al(OH)3 is less than Pb(OH)2

so order is

(min)3.20 < pH < 6.54 (max)