Unit 8: Qualitative Analysis (Anions)
Using precipitation reactions to identify common negative ions in solution.
8.14 Tests with Silver Nitrate (Halides)
The definitive test for halide ions ($Cl^-, Br^-, I^-$) in solution involves using aqueous silver nitrate ($AgNO_3$). The silver ions ($Ag^+$) react with the halide ions to form insoluble silver halide precipitates, which have distinct colors.
It is crucial to first acidify the solution with dilute nitric acid ($HNO_3$) before adding the silver nitrate. This is to remove any interfering ions, such as carbonate ions ($CO_3^{2-}$), which would also form a precipitate with silver ions ($Ag_2CO_3$). The nitric acid reacts with the carbonate ions to produce carbon dioxide gas, preventing a false positive result.
- Chloride ($Cl^-$): Forms a white precipitate of silver
chloride (AgCl).
$Ag^+(aq) + Cl^-(aq) \rightarrow AgCl(s)$ - Bromide ($Br^-$): Forms a cream precipitate of silver
bromide (AgBr).
$Ag^+(aq) + Br^-(aq) \rightarrow AgBr(s)$ - Iodide ($I^-$): Forms a yellow precipitate of silver
iodide (AgI).
$Ag^+(aq) + I^-(aq) \rightarrow AgI(s)$
Solved Examples:
-
An unknown solution is acidified with nitric acid, and then silver nitrate
solution is added. A cream-colored precipitate forms. Identify the
anion.
Solution: A cream precipitate with acidified silver nitrate is the characteristic test for the Bromide ion ($Br^-$). -
Why is nitric acid used to acidify the solution instead of hydrochloric
acid?
Solution: Hydrochloric acid cannot be used because it contains chloride ions ($Cl^-$). Adding it would introduce the very ion you might be testing for, leading to a false positive result (a white precipitate of AgCl) in any sample.
8.15 Tests with Barium Chloride (Sulfates, Sulfites, Carbonates)
Aqueous barium chloride ($BaCl_2$) is used to test for sulfate ions ($SO_4^{2-}$). Similar to the halide test, the solution must first be acidified with dilute hydrochloric acid (HCl). This is to eliminate interfering carbonate ($CO_3^{2-}$) and sulfite ($SO_3^{2-}$) ions, which would also form white precipitates with barium ions. The acid reacts with them to form gases.
- Sulfate ($SO_4^{2-}$): Forms a dense white precipitate
of barium sulfate ($BaSO_4$), which is insoluble in acids.
$Ba^{2+}(aq) + SO_4^{2-}(aq) \rightarrow BaSO_4(s)$
Distinguishing Sulfites and Carbonates
If a precipitate forms with barium chloride in a neutral solution but dissolves upon adding acid, the ion is either a sulfite or a carbonate.
- Carbonate ($CO_3^{2-}$): Adding acid to the original solution (or the precipitate) will produce effervescence. The gas produced is carbon dioxide ($CO_2$), which can be confirmed by bubbling it through limewater, turning it milky.
- Sulfite ($SO_3^{2-}$): Adding acid will produce sulfur dioxide ($SO_2$) gas, which has a sharp, choking smell and will turn moist acidified potassium dichromate(VI) paper from orange to green.
Solved Examples:
-
An unknown solution is acidified with HCl, and then barium chloride solution
is added. A white precipitate forms. Identify the anion.
Solution: The formation of a white precipitate with acidified barium chloride is the definitive test for the Sulfate ion ($SO_4^{2-}$). -
A solution forms a white precipitate with BaClâ‚‚. When HCl is added, the
precipitate dissolves and a gas is evolved that turns limewater milky. Identify
the anion.
Solution: The anion is the Carbonate ion ($CO_3^{2-}$).
8.16 Summary of Anion Precipitation Tests
| Anion | Test | Observation |
|---|---|---|
| $Cl^-$ | Add $HNO_3$, then $AgNO_3(aq)$ | White precipitate. |
| $Br^-$ | Add $HNO_3$, then $AgNO_3(aq)$ | Cream precipitate. |
| $I^-$ | Add $HNO_3$, then $AgNO_3(aq)$ | Yellow precipitate. |
| $SO_4^{2-}$ | Add $HCl$, then $BaCl_2(aq)$ | White precipitate. |
| $CO_3^{2-}$ | Add $BaCl_2(aq)$ | White precipitate. |
| Then add $HCl$ | Precipitate dissolves, effervescence (gas turns limewater milky). | |
| $SO_3^{2-}$ | Add $BaCl_2(aq)$ | White precipitate. |
| Then add $HCl$ | Precipitate dissolves, gas with sharp smell evolved. |
Solved Examples:
-
Describe a test to distinguish between sodium chloride and sodium iodide
solutions.
Solution: To a sample of each, add a few drops of dilute nitric acid followed by a few drops of silver nitrate solution. The sodium chloride solution will produce a white precipitate. The sodium iodide solution will produce a yellow precipitate. -
Why is it important to perform the test for sulfates in acidic
conditions?
Solution: To prevent false positives from other ions like carbonates and sulfites, which also form white precipitates with barium chloride in neutral conditions but not in acidic conditions. -
A student adds BaClâ‚‚ to an unknown solution and sees no precipitate. Which
anions are likely absent?
Solution: Sulfate ($SO_4^{2-}$), Carbonate ($CO_3^{2-}$), and Sulfite ($SO_3^{2-}$) are likely absent. -
Write the net ionic equation for the test for the sulfate ion.
Solution: $Ba^{2+}(aq) + SO_4^{2-}(aq) \rightarrow BaSO_4(s)$ -
What is the purpose of adding acid in both the halide and sulfate
tests?
Solution: The purpose is to remove interfering carbonate ions (and sulfite ions in the sulfate test) that would otherwise form a precipitate with the testing reagent ($Ag^+$ or $Ba^{2+}$) and give a false positive result. -
A solution gives no precipitate with acidified silver nitrate, but gives a
white precipitate with acidified barium chloride. What is the
anion?
Solution: The anion is the Sulfate ion ($SO_4^{2-}$).