Question

What is the sequence by which the cations below will precipitate out of solution as the pH is raised?

(Ksp of: Cu(OH)2=4.8x10-20, Sn(OH)2=3x10-27, Cr(OH)3=1.6x10-30)

a. Cu(OH)2, Sn(OH)2, Cr(OH)3
b. Sn(OH)2, Cu(OH)2, Cr(OH)3
c. Sn(OH)2, Cr(OH)3, Cu(OH)2
d.Cr(OH)3, Sn(OH)2, Cu(OH)2

Answer 1

To answer this question completely, one needs to know the concentration of the cation. Let's say the cation concentration is 0.1 M in each case. I'll do Cu(OH)2 for you. You do the others.
..................Cu(OH)2 ==> Cu^2+ + 2OH^-
Ksp = (Cu^2+)(OH^-) = 4.8E-20. Solve for (OH^-).
(OH^-) = sqrt [Ksp/(Cu^2+)] = sqrt (4.8E-20/0.1) = sqrt (48E-20) = 6.9E-10 M
Calculate (OH^-) similarly for the other two salts, the sequence them by pH. If it will make it easier, convert the (OH^-) in each case to pH.
Post your work if you get stuck.