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.
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
1 answer