Question

In: Chemistry

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

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

Solutions

Expert Solution

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


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