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 0.108 M in Fe²⁺ and 0.101 M in Ca²⁺.

FeCO3	Ksp = 3.50×10-11
CaCO3	Ksp = 3.80×10-9

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?  M

(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? _____yesno

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

(4) If the [CO₃^2-] is at this upper limit, what percentage of the metal that forms the least soluble carbonate remains in solution?

Answer 1

Q1.

The least soluble if FeCO3, due to lower Ksp value, so, this will precipitate first,

for 99.9% --> 0.108*(1-0.999) = 0.000108 M of Fe+2 left

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

(3.5*10^-11) = (0.000108)([CO3-2])

[CO3-2] = ((3.5*10^-11))/(0.0001089) = 3.2139*10^-7 M

Q2.

NO, some of Ca+2 is expected to precipitate, since there is some CO3-2 present

Q3.

upper limit:

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

3.8*10^-9 = (0.101)*(CO3-2)

[CO3-2] = (3.8*10^-9)/0.101 = 3.762*10^-8 M is the min level required for no precipitation

Q4.

if CO3-2 achieves --> 3.2139*10^-7

find % precipitateion

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

3.8*10^-9 = [Ca+2]*(3.2139*10^-7)

[Ca+2] = (3.8*10^-9) / (3.2139*10^-7) = 0.01182

% = Ca left / Ca initial * 100% = (0.01182) / (0.101) * 100 = 1.70 % has preciptiated