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 9.04×10-2 M in Mn2+ and 9.07×10-2 M in Ba2+.

BaCO3 Ksp = 8.10×10-9

MnCO3 Ksp = 1.80×10-11

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? _____(yes or no)

(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)

Least soluble carbonate is MnCO3 because Ksp of MnCO3 is lower than Ksp of BaCO3

[Mn2+] after removing 99.9% of [Mn2+] = 0.0000905M

Ksp of MnCO3 = [Mn2+][CO32-] = 1.80×10-11M2

substituting [Mn2+] after 99.9% prcipitation

0.0000905M × [CO32-] = 1.80×10-11M2

[CO32-] = 1.99×10-7M

2)

Ksp of BaCO3 = [Ba2+][CO32-]= 8.10-9M2

substituting carbonate concencentration obtained in 1)

[Ba2+]×1.99×10-7M= 8.10×10-9M2

[Ba2+] = 4.07×10-2M

Therefore,

No, all the ions of metal that form most soluble carbonate will not remain

3)

Ksp of BaCO3 = [Ba2+] [CO32-] = 8.10×10-9M2

substituting given [Ba2+]

0.0907M × [CO32-] = 8.10×10-9M2

[CO32-] = 8.93×10-8M

4)

Ksp of MnCO3 = [Mn2+][CO32-] = 1.80×10-11M2

substituting [CO32-] obtained from 3)

[Mn2+] × 8.93×10-8M = 1.80×10-11M2

[Mn2+] = 2.02×10-4M

percentage of Mn2+ remain in solution = (0.000202M/0.0904)×100= 0.22%

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.