It is of interest to decide if an analytical separation of the metal ions can be

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 0.109 M in Fe2+ and 0.105 M in Ni2+. Ksp 3.50x10-11 3Ksp 6.60x10 To analyze this problem, answer the following questions. (1) What carbonate concentration is needed to precipitate 99.9% of the metal that forms the least soluble carbonate? (2 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? (3) What is the upper limit on the carbonate ion concentration if the more soluble compound is not to precipitate? (4) If the [CO32-] is at this upper limit, what percentage of the metal that forms the least soluble carbonate remains in solution?

Explanation / Answer

1) FeCO3 is least soluble carbonate because Ksp of FeCO3 is lower than Ksp of NiCO3 and concentrations are almost equal

concentration of Fe2+ remains after 99.9% removal = 0.000109M

Ksp = [Fe2+][CO32-]= 3.50×10-11M2

Substituting remaining concentration of Fe2+

0.000109M × [ CO32-] = 3.50×10-11M2

     [CO32-] = 3.211×10-7M

Therefore,

[CO32-] needed = 3.21×10-7M

2) More soluble metal carbonate is NiCO3

Ksp = [Ni2+][CO32-] = 6.60×10-9M2

substituting, [CO32-] obtained from 1)

[Ni2+] × 3.21×10-7M × 6.60×10-9M2

   [Ni2+] = 6.60×10-9M2/3.21×10-7M = 0.021M

Therefore,

No, all the metal ions that forms more soluble carbonate will not remains in solution.

3) Ksp = [Ni2+][CO32-] = 6.60×10-9M2

substituting the given concentration value of Ni2+

0.105M×[CO32-] = 6.60×10-9M2

   [CO32-] = 6.60×10-9M2/0.105M

[CO32-] = 6.29×10-8M

Therefore,

Concentration limit of CO32- to avoid NiCO3 precipitation = 6.29×10-8M

4)

Ksp = [Fe2+] [CO32-] = 3.50×10-11M2

Substituting [CO32-] obtained in 3)

[Fe2+] × 6.29×10-8M = 3.50×10-11M2

[Fe2+] = 3.50×10-11M2/6.29×10-8M = 0.000556M

% of Fe2+ remains in solution = (0.000556M/0.109M)×100= 0.510%

  

  

  

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