Selective Precipitation

Selective Precipitation of the Barium Magnesium Group

ABSTRACT
The purpose of this lab was to identify the cations from the barium magnesium group present in an assigned unknown solution through selective precipitation. This was a qualitative lab where the theory of the common ion effect was used in several steps. Once the experiment was completed, it was determined that unknown solution #4 contained Ba2+, Sr2+, and Mg2+ ions.

INTRODUCTION
Selective precipitation is a part of chemistry that allows for the qualitative analysis of ions present in a solution. A large industry that employs this type of chemistry is the mining industry. In the mining industry metals are not typically pure when first collected. Through the use of selective precipitation and the knowledge of Ksp values, certain metal ions can be precipitated out.

EXPERIMENTAL
The Selective Precipitation of the Barium Magnesium Group experiment instructions were followed directly from the handout printed from the course website.

RESULTS AND REVIEW

In the first step of the experiment, the ionization of NH4OH (a weak base) was decreased by NH4Cl. NH4Cl is strongly ionized in solution, meaning the [NH4+] increases in solution. According to Le-Chatlier’s principle, the formation of NH4OH is favored as there was a higher concentration of NH4+ to combine with the [OH-]. This caused the [OH-] to be considerably reduced, and the NH4OH to become an even weaker base. The pH was decreased as the [OH-] was decreased. In solutions with more basic pH there would be a shift towards the products and a decrease in the solubility of certain precipitates.

When (NH4)2CO3 was added to the solution there was a reaction with the ions present in the unknown solution to form in this particular unknown solution: BaCO3 and SrCO3. If Ca2+ was present then CaCO3 would have formed in the white precipitate as well.

(NH4)2CO3(aq) + Ba2+(aq)  BaCO3(s) + NH4+(aq) Ksp BaCO3(s) = 2.58 x 10-9
(NH4)2CO3(aq) + Sr2+(aq)  SrCO3(s) + NH4+(aq) Ksp SrCO3(s) = 5.6 x 10-10
(NH4)2CO3(aq) + Ca2+(aq)  CaCO3(s) + NH4+(aq) Ksp CaCO3(s) = 6.0 x 10-9
(NH4)2CO3(aq) + Mg2+(aq)  MgCO3(s) + NH4+(aq) Ksp MgCO3(s) = 6.82 x 10-6

MgCO3 would not have formed in this reaction as it’s Ksp value was much larger than the other three, it would have been more soluble and not formed a precipitate. The supernatant (S1) was decanted off with the possibility of it containing Mg2+ and NH4+.

Once the first precipitate was dissolved into its ions once more, K2CrO4 was added to the solution. A yellow precipitate formed, indicating the presence of Ba2+. This is known through the Ksp values of the possible precipitates. The Ksp vales of SrCrO4and CaCrO4 are much larger than the Ksp value of BaCrO4. The yellow precipitate in this reaction contains BaCrO4(s) (P2).

K2CrO4(aq) + Ba2+  BaCrO4(s) + K+ Ksp BaCrO4(s) = 1.17 x 10-10
K2CrO4(aq) + Sr2+  SrCrO4(s) + K+ Ksp SrCrO4(s) = 2.2 x 10-5
K2CrO4(aq) + Ca2+  CaCrO4(s) + K+ Ksp CaCrO4(s) = 7.1 x 10-4

The second supernatant (S2) was decanted off into another test tube to test for Sr2+ and Ca2+.

The yellow precipitate(P2) was dissolved in HCl, the high [H+] combined with the CrO4 leaving the Ba2+ in solution. To confirm the Ba2+ ion was indeed in solution, (NH4)2SO4 was added to the solution, a white precipitate formed, confirming the Ba2+ion. The Ksp is also small, indicating that BaSO4is not very soluble and should form a precipitate.

(NH4)2SO4(aq) + Ba2+(aq)  BaSO4(s) + NH4+(aq) Ksp BaSO4(s) = 1.08 x 10-10

The second supernatant (S2) was evaporated until its volume was approximately 1mL. 1mL of triethanolamine was added to the solution in addition to (NH4)2SO4. A white precipitate formed indicating the presence of Sr­2+. The supernatant (S3) was decanted and saved for later to determine if the unknown contained Ca2+. The Ksp value for SrSO4 is smaller than CaSO4 indicating that the precipitate formed contained Sr2+.

(NH4)2SO4(aq) + Sr2+(aq)  SrSO4(s) + NH4+(aq) Ksp SrSO4(s) = 3.44 x 10-7
(NH4)2SO4(aq) + Ca2+(aq)  CaSO4(s) + NH4+(aq) Ksp CaSO4(s) = 4.93 x 10-5

The third supernatant (S3) was made slightly basic with the addition of NH4OH. (NH4)2C2O4 was added to the solution, with no precipitate formed. If there was Ca2+ ions present, there would have been a precipitate.

(NH4)2C2O4(aq) + Ca2+(aq)  CaC2O4(s) + 2NH4+ (aq) Ksp Ca2C2O4(s) = 2.32 x 10-9

If Ca2+ was present it would have formed a precipitate as the Ksp value for Ca2C2O4 is small.

The first supernatant (S1), which had the possibility of containing Mg2+ and NH4+, had 1 drop of NH4OH and 5 drops of Na2HPO4 and placed in hot water for several minutes. A white precipitate formed, indicating the formation of Mg2+.

Na2HPO4(aq) + Mg2+(aq)  MgNH4PO4(s) + Na+(aq) Ksp MgNH4PO4(s) = 3.0 x 10-13

A precipitate did form, the Ksp value of MgNH4PO4(s) confirms that a precipitate would form.

The MgNH4PO4(s) precipitate was dissolved in HCl
MgNH4PO4(s) + HCl(aq)  Mg2+(aq) + NH4Cl(aq) + H3PO4(aq)

5 drops of magnesium reagent were added to the solution as was 6 drops of NaOH, until the solution was basic.

NaOH(aq) + Mg2+(aq)  Mg(OH)2(aq) + Na+(aq) Ksp Mg(OH)2 = 5.61 x 10-12

A blue precipitate was formed that confirmed the presence of the Mg2+ ion.

To test for the presence of NH4+, 15 drops of unknown solution #4 was added to a clean test tube with 8 drops of NaOH added. Litmus paper was dipped into the boiling water and held over the test tube. The litmus paper did not turn blue as would indicate the presence of NH4+. There was no NH4+ present in our uknown.

CONCLUSION
Through selective precipitation and knowledge of the product Ksp values, it was determined that unknown solution #4 contained the Ba2+, Sr2+, and Mg2+ ions.