Unit 8: Qualitative Analysis (Cations)
Using precipitation and complex formation reactions to identify unknown metal cations in solution.
8.11 Tests with Sodium Hydroxide (Fe²⁺, Fe³⁺, Cu²⁺, Ca²⁺, Zn²⁺, Pb²⁺, Al³⁺)
Adding aqueous sodium hydroxide (NaOH), a strong base, to a solution is a key test. The hydroxide ions ($OH^-$) react with many metal cations to form insoluble metal hydroxide precipitates, each with a characteristic color. Some of these hydroxides are amphoteric and will redissolve if excess NaOH is added.
- $Fe^{2+}(aq) + 2OH^-(aq) \rightarrow Fe(OH)_2(s)$: Forms a green precipitate, insoluble in excess.
- $Fe^{3+}(aq) + 3OH^-(aq) \rightarrow Fe(OH)_3(s)$: Forms a red-brown precipitate, insoluble in excess.
- $Cu^{2+}(aq) + 2OH^-(aq) \rightarrow Cu(OH)_2(s)$: Forms a pale blue precipitate, insoluble in excess.
- $Ca^{2+}(aq) + 2OH^-(aq) \rightarrow Ca(OH)_2(s)$: Forms a white precipitate, insoluble in excess.
- $Zn^{2+}, Al^{3+}, Pb^{2+}$: All form white precipitates which are amphoteric and dissolve in excess NaOH to form colorless solutions.
Solved Examples:
-
An unknown solution produces a green precipitate when NaOH is added. The
precipitate does not dissolve in excess. Identify the cation.
Solution: A green hydroxide precipitate is characteristic of the Iron(II) ion ($Fe^{2+}$). -
What does it mean for a hydroxide to be amphoteric?
Solution: It means the hydroxide can react with both acids and strong bases. In this context, it means the precipitate will redissolve in excess sodium hydroxide.
8.12 Tests with Aqueous Ammonia (NH₃)
Aqueous ammonia is a weak base, providing a lower concentration of $OH^-$ ions. It also acts as a ligand, forming soluble complex ions with certain cations. Comparing the results of adding NH₃ with the results from NaOH is a powerful identification tool.
- $Fe^{2+}$ and $Fe^{3+}$: Form the same green and red-brown precipitates, respectively. Both are insoluble in excess NH₃.
- $Al^{3+}$ and $Pb^{2+}$: Form white precipitates ($Al(OH)_3$, $Pb(OH)_2$) which are insoluble in excess NH₃. This is a key difference from the NaOH test.
- $Ca^{2+}$: Forms no precipitate. Aqueous ammonia is too weak a base to provide enough $OH^-$ to precipitate the hydroxide of this reactive metal.
- $Zn^{2+}(aq) + 2OH^-(aq) \rightarrow Zn(OH)_2(s)$: Forms a white precipitate, which dissolves in excess NH₃ to form the colorless tetraamminezinc(II) complex ion, $[Zn(NH_3)_4]^{2+}$.
- $Cu^{2+}(aq) + 2OH^-(aq) \rightarrow Cu(OH)_2(s)$: Forms a pale blue precipitate, which dissolves in excess NH₃ to form the deep blue tetraamminecopper(II) complex ion, $[Cu(NH_3)_4]^{2+}$.
Solved Examples:
-
A solution forms a white precipitate with NaOH that dissolves in excess. A
separate sample forms a white precipitate with NH₃ that also dissolves in
excess. Identify the cation.
Solution: The only cation whose hydroxide precipitate dissolves in both excess NaOH (due to being amphoteric) and excess NH₃ (due to complex ion formation) is the Zinc ion ($Zn^{2+}$). -
How can you distinguish between a solution of aluminum chloride and a
solution of calcium chloride?
Solution: Add aqueous ammonia. The aluminum chloride solution will form a white precipitate ($Al(OH)_3$) that is insoluble in excess. The calcium chloride solution will show no reaction.
8.13 Tests with Hydrochloric Acid (Pb²⁺, Ag⁺)
According to the solubility rules, most chloride salts are soluble. However, there are a few important exceptions. Adding a source of chloride ions, such as dilute hydrochloric acid (HCl), can be used to identify these specific cations.
- $Ag^+(aq) + Cl^-(aq) \rightarrow AgCl(s)$: Forms a white precipitate of silver chloride.
- $Pb^{2+}(aq) + 2Cl^-(aq) \rightarrow PbCl_2(s)$: Forms a white precipitate of lead(II) chloride. An important distinguishing feature is that lead(II) chloride is soluble in hot water, while silver chloride is not.
This test is particularly useful for distinguishing between the cations that form white, amphoteric hydroxides. Of $Al^{3+}$, $Zn^{2+}$, and $Pb^{2+}$, only $Pb^{2+}$ will form a precipitate with HCl.
Solved Examples:
-
A solution forms a white precipitate with NaOH that dissolves in excess. A
separate sample of the solution also forms a white precipitate when HCl is
added. Identify the cation.
Solution: Forming a white amphoteric hydroxide narrows it down to Al³⁺, Zn²⁺, or Pb²⁺. Forming a precipitate with HCl confirms the presence of the Lead(II) ion ($Pb^{2+}$). -
How could you distinguish between a white precipitate of AgCl and
PbCl₂?
Solution: Heat the mixture. The PbCl₂ precipitate will dissolve in hot water, while the AgCl precipitate will remain insoluble.
8.14 Confirmatory Tests for Cations
While precipitation tests are very useful, sometimes a specific, secondary test is needed to confirm the identity of an ion.
- Calcium ($Ca^{2+}$): The most reliable confirmatory test is the flame test. A sample of the compound on a clean wire loop will impart a brick-red color to a Bunsen flame.
- Lead ($Pb^{2+}$): Add a solution containing iodide ions (e.g., potassium iodide, KI). Lead(II) ions will form a brilliant bright yellow precipitate of lead(II) iodide ($PbI_2$).
- Iron(III) ($Fe^{3+}$): Add a solution of potassium thiocyanate (KSCN). Iron(III) ions will form a soluble complex ion that gives a deep blood-red color. This test is extremely sensitive.
- Ammonium ($NH_4^+$): Add sodium hydroxide solution and gently warm the
mixture. The ammonium ion will be converted to ammonia gas ($NH_3$), which has a
characteristic sharp smell and will turn damp red litmus paper blue.
$NH_4^+(aq) + OH^-(aq) \xrightarrow{\Delta} NH_3(g) + H_2O(l)$
Solved Examples:
-
After identifying a cation as either Al³⁺ or Pb²⁺ using NaOH and NH₃, what
further test would you perform?
Solution: Add a few drops of potassium iodide (KI) solution. If the cation is Pb²⁺, a bright yellow precipitate of PbI₂ will form. If it is Al³⁺, no reaction will occur. -
A student warms a solution with NaOH and holds a piece of damp red litmus
paper over the test tube. The paper turns blue. What ion is
present?
Solution: The production of a gas that turns damp red litmus blue is the definitive test for the Ammonium ion ($NH_4^+$). The gas is ammonia ($NH_3$).
8.15 Summary of Cation Precipitation Tests
This table summarizes the key tests for identifying common cations using NaOH and NH₃.
| Cation | Reaction with NaOH(aq) | Reaction with NH₃(aq) |
|---|---|---|
| $Ca^{2+}$ | White precipitate, insoluble in excess. | No precipitate. |
| $Al^{3+}$ | White precipitate, soluble in excess. | White precipitate, insoluble in excess. |
| $Pb^{2+}$ | White precipitate, soluble in excess. | White precipitate, insoluble in excess. |
| $Zn^{2+}$ | White precipitate, soluble in excess. | White precipitate, soluble in excess. |
| $Fe^{2+}$ | Green precipitate, insoluble in excess. | Green precipitate, insoluble in excess. |
| $Fe^{3+}$ | Red-brown precipitate, insoluble in excess. | Red-brown precipitate, insoluble in excess. |
| $Cu^{2+}$ | Blue precipitate, insoluble in excess. | Blue precipitate, soluble in excess (deep blue solution). |
Solved Examples:
-
An unknown solution is blue. What cation is likely present and what would
you expect to see upon adding excess aqueous ammonia?
Solution: The cation is likely Copper(II) ($Cu^{2+}$). Upon adding excess ammonia, the initial blue precipitate would redissolve to form a deep blue solution. -
Describe the expected observations when testing a solution of aluminum
chloride.
Solution: With NaOH, a white precipitate would form and then redissolve in excess. With NH₃, a white precipitate would form and remain insoluble in excess.