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 0.107 M in Fe2+and 9.30×10-2 M in Ba2+.

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?  %

Explanation / Answer

1)

FeCO3 is least soluble because Ksp of FeCO3 is lower than Ksp of BaCO3

Initial concentration of Fe2+ = 0.107M

after precipitating 99.9%, [Fe2+] = 0.000107M

Solubility equillibrium of FeCO3

FeCO3(s) <-------> Fe2+(aq) + CO32-(aq)

Ksp = [Fe2+]×[CO32-] = 3.50×10-11

Substituting [Fe2+] calculated above

0.000107M× [CO32-] = 3.50×10-11M2

[CO32-] = 3.27×10-7M

2)

Solubility equillibrium of BaCO3 is

BaCO3(s) <--------> Ba2+(aq) + CO32-(aq)

Ksp = [ Ba2+][CO32-] = 8.10×10-9M2

Substituting the [CO32-] required to precipitate 99.9% Fe2+

   [Ba2+]× 3.27×10-7M = 8.10×10-9M2

[Ba2+] = 8.10×10-9 M2/3.27×10-7M = 2.48×10-2M

Therefore,

No, all the ions of the matal that form more soluble carbinate will not remain

3)   

Ksp = [Ba2+][CO32-] = 8.10×10-9M2

substituting given [Ba2+]

0.0930M × [CO32-] = 8.10×10-9M2

[CO32-] =8.10×10-9M2/0.0930M

[CO32-] = 8.71×10-8M

Therefore,

Upper limit [CO32+] to avoid Ba2+ precipitation is 8.71×10-8M

4)

Ksp = [Fe2+] [CO32-] = 3.50×10-11M2

Substituting upper limit concentration of Ba2+ to avoid precipitation of BaCO3

[Fe2+] × 8.71×10-8M = 3.50×10-11M2

[Fe2+] = 4.02×10-4M

percentage of Fe2+ remains = (0.000402M/0.107M)×100 = 0.38%

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