Question

A solution is 0.020 M in each Ca2+ and Cd2+. Adjusting the pH of the solution to which of the following values would achieve the best separation by precipitation of the hydroxides?

solubility product constants Best pH =
(answer options 8.0-13.0; increasing in .5 increments)

What would be the resulting concentrations of the ions if the solution were adjusted to this pH?

[Ca2+] = M

[Cd2+] = M

Answer 1

Ca(OH)₂ : 5.5 x 10^-6 = [Ca²⁺] [OH^-]²
5.5 x 10^-6 = [0.02] [OH-]²
[OH-] = 1.66 x 10^-2
pOH = 1.78
pH = 12.2
not until you get more basic than 12.2, will Ca(OH)₂ begin to ppt
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Mg(OH)₂ : 6.52 x 10^-15 = [Mg²⁺] [OH^-]²
6.52 x 10^-15 = [0.02] [OH^-]²
[OH-] = 5.71 x 10^-7
pOH = 6.24
pH = 7.76
not until you get more basic than pH = 7.76 will Mg(OH)₂ begin to ppt
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Your Answer : @ pH = 12.0 , because the [OH-] = 10^-2, and

Mg(OH)₂ : 6.52 x 10^-15 = [Mg²⁺] [OH^-]²
6.52 x 10^-15= [Mg²⁺] [10^-2]²
answer: [Mg²⁺] = 6.52 x 10^-11 M

@ PH 12 : 6.52 x 10^-11 M / 0.0200M = all of the Mg⁺² has ppt'd except for 0.0000003% of Mg , while the Ca⁺² is still 100% in solution:
answer: [Ca²⁺] is 0.02M,

& that's a separation.