Questions related to precipitation of Fe3+ and Al3+ by Ca(OH)2

Dec 2018
I have a problem to resolve for my upcoming exam.

A liquid containing 1mol/L of cobalt (Co2+) as cobalt sulfate is contaminated with iron and aluminum and should be purified by removing these impurities by adding slaked lime (Ca(OH)2). The concentration of individual impurities in the treated solution is not to exceed 10^-3 mol/L.

1. What pH would you recommend for this operation?
2. Following the lime addition, what would be the composition of the solution at the chosen pH?
3. How much lime needs to be added per L of solution? will lime dissolve fully?
4. How much gypsum per kg of cobalt in the feed will be produced in each case?

Log Ksp for the reactions:
Fe3+ + 3OH- = Fe(OH)3 38.8
Al3+ + 3OH- = Al(OH)3 33.5
Co2+ + 2OH- = Co(OH)2 14.9
Ca2+ + 2OH- = Ca(OH)2 5.26

Many thanks!

My attempt:

a. from:
Co2+ + 2OH- = Co(OH)2 14.9
K=[Co2+][OH-]^2/[Co(OH2)] = 1 * [OH-]^2 = 10^(-14.9)
so [OH-] = 10^(-7.45)
so pH= 6.55

Al3+ + 3OH- = Al(OH)3 33.5
K=[Al3+][OH-]^3/[Al(OH3)] = 10^(-3) * [OH-]^3 = 10^(-33.5)
so [OH-] = 10^(-10.2)
so pH= 3.8

No need to check Fe since it will be lower than that for Al.

therefore 3.8<pH<6.55
Choose pH=5

b) pH=5, so [OH-]=10^(-9)
for Co2+, no precipitation, therefore Co2+ = 1M
for Al3+, [Al3+]= Ksp/([OH-]^3)=10^(-33.5)/(10^(-9)^3)= 10^(-6.5)
for Fe3+, [Fe3+]= Ksp/([OH-]^3)=10^(-38.8)/(10^(-9)^3)= 10^(-11.8)

I am stuck at c)
Last edited: