Question

It is of interest to decide if an analytical separation of the metal ions can be effected by selective precipitation of carbonates from a solution that is 9.84×10-2 M in Fe2+ and 9.84×10-2 M in Ni2+. NiCO3 Ksp = 6.60×10-9 FeCO3 Ksp = 3.50×10-11 To analyze this problem, answer the following questions.

(a) What carbonate concentration is needed to precipitate 99.9% of the metal that forms the least soluble carbonate? M

(b) When 99.9% of the least soluble carbonate has precipitated, will all of the metal that forms the more soluble carbonate still remain in solution? (yes/no)

(c) What is the upper limit on the carbonate ion concentration if the more soluble compound is not to precipitate? M

(d) If the [CO32-] is at this upper limit, what percentage of the metal that forms the least soluble carbonate remains in solution? %

Answer 1

ans)

Leas tsoluble = FeCO3, sicne it has lower Ksp

Q1

find CO3-2 required to precipitate hte least soluble

Ksp = [Fe+2][CO3-2]

99.99% of Fe+2 = (100-99.99) = 0.01 % left

0.01/100*(9.84*10^-2) = 0.00000984 M of Fe+2 left in solution

3.5*10^-11 = (0.00000984)([CO3-2]

[CO3-2] = (3.5*10^-11)/(0.00000984) = 0.000003542 M = 3.55*10^-6 M

Q2

find the other metal at this point

Ksp = [Ni+2][CO3-2]

(6.6*10^-9) = [Ni2+] ( 3.55*10^-6)

[Ni2+] = (6.6*10^-9) / (3.55*10^-6) = 0.00185

therefore, not all the Nickel has precipitated

Q3

upper limit so the soluble compound will not precipitate

Ksp =[Ni2+][CO3-2]

(6.6*10^-9) / (0.0984) = [CO3-2]

[CO3-] = 6.71*10^-8 M then Ni2+ wont form preciptiate

Q4

if CO3-2 is at this value, find metal of leaast soluble

Ksp = [Fe+2][CO3-2]

3.5*10^-11 = [Fe+2] (  6.71*10^-8)

[Fe+2] = (3.5*10^-11) / (  6.71*10^-8) = 0.000522 M

% remains = (0.000522)/(9.84*10^-2)*100 =0.530 %