Unit 9: Qualitative Analysis Part 4
Using complex formation reactions to identify unknown metal cations in solution.
9.30 Using Complex Formation Reactions to Identify Cations (NH₃ vs NaOH)
Qualitative analysis is the branch of chemistry concerned with identifying the substances present in a sample. We can identify many metal cations by observing their reactions with sodium hydroxide (a strong base) and aqueous ammonia (a weak base).
Both reagents provide hydroxide ions ($OH^-$) that form insoluble metal hydroxide precipitates with many cations. However, some of these precipitates react further with excess reagent. Specifically, certain metal hydroxides that are amphoteric will redissolve in excess strong base (NaOH), while others will redissolve in excess aqueous ammonia by forming soluble complex ions. Comparing the results from both tests is a powerful identification tool.
| Cation | Reaction with NaOH(aq) | Reaction with NH₃(aq) |
|---|---|---|
| $Ca^{2+}$ | White precipitate, insoluble in excess. | No precipitate (ammonia is too weak a base). |
| $Al^{3+}$ | White precipitate, soluble in excess (forms colorless solution). | White precipitate, insoluble in excess. |
| $Pb^{2+}$ | White precipitate, soluble in excess (forms colorless solution). | White precipitate, insoluble in excess. |
| $Zn^{2+}$ | White precipitate, soluble in excess (forms colorless solution). | White precipitate, soluble in excess (forms colorless solution). |
| $Fe^{2+}$ | Green precipitate, insoluble in excess. (Turns brown on standing). | 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 (forms deep blue solution). |
The key differentiators are the amphoteric cations ($Al^{3+}$, $Pb^{2+}$, $Zn^{2+}$) which redissolve in excess NaOH, and the complex-forming cations ($Cu^{2+}$, $Zn^{2+}$) which redissolve in excess NH₃.
Solved Examples:
-
An unknown solution forms a white precipitate with a few drops of NaOH. The
precipitate dissolves when excess NaOH is added. What cations could be
present?
Solution: The precipitate is an amphoteric hydroxide. The possible cations are Aluminum ($Al^{3+}$), Lead ($Pb^{2+}$), or Zinc ($Zn^{2+}$). -
To the solution from Q1, a fresh sample is tested with aqueous ammonia. A
white precipitate forms but it does NOT dissolve in excess ammonia. Which cation
is present?
Solution: Of the three possibilities ($Al^{3+}$, $Pb^{2+}$, $Zn^{2+}$), only the hydroxide of Zinc ($Zn(OH)_2$) dissolves in excess ammonia. Since the precipitate is insoluble, the cation must be either Aluminum ($Al^{3+}$) or Lead ($Pb^{2+}$). (Further tests, like with KI, would be needed to distinguish between Al and Pb). -
A solution produces a pale blue precipitate with a few drops of aqueous
ammonia. When more ammonia is added, the precipitate dissolves to form a deep
blue solution. Identify the cation.
Solution: The formation of a blue precipitate that redissolves in excess ammonia to form a deep blue solution is the characteristic test for the Copper(II) ion ($Cu^{2+}$). The final solution contains the $[Cu(NH_3)_4]^{2+}$ complex ion. -
How would you distinguish between a solution containing $Fe^{2+}$ ions and
one containing $Fe^{3+}$ ions?
Solution: Add a few drops of NaOH or NH₃ to each solution. The solution with $Fe^{2+}$ will form a green precipitate. The solution with $Fe^{3+}$ will form a red-brown precipitate. -
A solution gives no precipitate when aqueous ammonia is added. What cation
is likely present?
Solution: Aqueous ammonia is a weak base and is not strong enough to precipitate the hydroxides of very reactive metals. The cation is likely from Group 1 or Group 2, such as Calcium ($Ca^{2+}$) or Sodium ($Na^+$). -
Write the ionic equation for the formation of the precipitate when NaOH is
added to a solution of iron(III) chloride.
Solution: $Fe^{3+}(aq) + 3OH^-(aq) \rightarrow Fe(OH)_3(s)$ -
Write the ionic equation for the formation of the soluble complex ion when
excess ammonia is added to a solution containing zinc ions.
Solution: $Zn(OH)_2(s) + 4NH_3(aq) \rightarrow [Zn(NH_3)_4]^{2+}(aq) + 2OH^-(aq)$ (Note: some sources show 6 ammonia ligands for zinc, $[Zn(NH_3)_6]^{2+}$, which is also correct). -
You are given two unlabeled colorless solutions, one containing zinc nitrate
and the other aluminum nitrate. Describe a one-step test to identify which is
which.
Solution: Add aqueous ammonia dropwise and then in excess to a sample of each solution. Both will form a white precipitate initially. The solution that is zinc nitrate will see its precipitate redissolve in excess ammonia to form a colorless solution. The precipitate in the aluminum nitrate solution will remain insoluble. -
Why does the green precipitate of $Fe(OH)_2$ turn brown when left to stand
in air?
Solution: The iron(II) ions in the precipitate are oxidized by the oxygen in the air to iron(III) ions. The green $Fe(OH)_2$ turns into the red-brown $Fe(OH)_3$. -
A solution forms a white precipitate with NaOH that is insoluble in excess.
What does this tell you?
Solution: This tells you the cation is not amphoteric. A likely candidate from the common list would be Magnesium ($Mg^{2+}$), although Calcium ($Ca^{2+}$) also behaves this way. It rules out Al³⁺, Pb²⁺, and Zn²⁺.