Qualitative analysis of cations

String every equilibrium so far into one flowchart: hydrochloric acid brings down silver, lead and mercury(I); sodium hydroxide and ammonia then sort copper, iron, aluminium and zinc for confirmation.

In this lesson

Explain the “separate into groups, then confirm” logic of qualitative analysis, and the difference between group and confirmatory reagents
Justify every step of the scheme with Ksp values, formation constants and amphoterism
Deduce which cations an unknown contains from a set of observations
Recognize the interferences and limits of the scheme

Before spectrometers were everywhere, chemists could name the metals in a rock or a water sample with a rack of test tubes and a dozen reagent bottles. Instruments are far faster today, yet qualitative analysis remains a classic of chemistry teaching — because every step is an equilibrium from the last few lessons: solubility products decide what precipitates, formation constants decide what redissolves, and pH decides which side amphoteric hydroxides take. This lesson strings them into one flowchart.

Separate into groups, then confirm

With many ions present, a single reagent often throws down several precipitates at once, each masking the others. Qualitative analysis therefore works in two stages:

a group reagent sorts the ions by solubility — hydrochloric acid, for example, precipitates only the three group I chlorides;
within each group the ions are separated further, and each is finally confirmed by a specific, eye-catching confirmatory test.

The scheme here handles seven cations: the chloride group ( $\ce{Ag+}$ , $\ce{Pb^2+}$ , $\ce{Hg2^2+}$ ) and $\ce{Cu^2+}$ , $\ce{Fe^3+}$ , $\ce{Al^3+}$ and $\ce{Zn^2+}$ , separated with sodium hydroxide and ammonia.

The seven-cation scheme
Step	Operation	Why it works	Positive result
1	Add 6 M HCl	Tiny Ksp of AgCl, PbCl₂, Hg₂Cl₂	white precipitate (group I)
2–3	Hot water, then K₂CrO₄	PbCl₂ is about 3× more soluble hot	yellow PbCrO₄ → Pb²⁺
4	6 M NH₃ on the residue	K = 3 × 10⁻³ for AgCl + 2NH₃; Hg₂Cl₂ disproportionates	grey-black residue → Hg₂²⁺
5	Acidify the ammonia extract	NH₃ is neutralized and AgCl returns	white AgCl → Ag⁺
6	Add H₂SO₄	removes the ~3 % of Pb²⁺ HCl left	(faint white PbSO₄)
7	Excess NaOH	Al(OH)₃ and Zn(OH)₂ are amphoteric	Fe(OH)₃ and Cu(OH)₂ precipitate
8	Dissolve in HCl, then excess NH₃	Cu forms its ammine in the NH₃/NH₄⁺ buffer	deep-blue solution → Cu²⁺
9	Fe(OH)₃ in HCl, then KSCN	FeSCN²⁺, K₁ = 1.0 × 10³	blood red → Fe³⁺
10–11	Acidify, excess NH₃, then aluminon	Al(OH)₃ precipitates near pH 9–10; Zn stays as its ammine	red lake → Al³⁺
12	Na₂S into the solution	ZnS is insoluble	white ZnS → Zn²⁺

Source: Scheme adapted from Vogel’s Qualitative Inorganic Analysis, 7th ed. (1996); constants from Martell & Smith, Critical Stability Constants.

\ce{Hg2Cl2(s) + 2NH3 -> Hg(l) + HgNH2Cl(s) + NH4+ + Cl-}

Confirming mercury(I): ammonia makes Hg₂Cl₂ disproportionate, and finely divided mercury turns the solid grey-black.

$\ce{Hg}$: mercury(0), black
$\ce{HgNH2Cl}$: mercury(II) amidochloride, white

Worked example

Reading the observations

An unknown gives: 1: white precipitate with HCl. 3: the hot-water extract with K₂CrO₄ gives only a yellow solution, no precipitate. 4: the residue dissolves completely in ammonia. 5: acidifying brings back a white precipitate. 7: excess NaOH gives a red-brown precipitate. 8: dissolved in HCl and treated with excess ammonia, the solution stays colourless and the red-brown solid remains. 9: KSCN turns it blood red. 10–11: a white gelatinous precipitate, red with aluminon. 12: no precipitate with Na₂S. Which ions are present?

Seven cations · every step computed from Ksp and formation constants

Cation qualitative-analysis bench

Unknown #

The chloride group (group I)

Hydroxides and ammine complexes

Test-tube rack (solutions: computed from spectra; solids: documented colours)

Lab notebook

Start with step 1: add 6 M HCl to the unknown.

Which ions do you think the unknown contains?

\ce{Ag+}

silver

\ce{Pb^2+}

lead(II)

\ce{Hg2^2+}

mercury(I)

\ce{Cu^2+}

copper(II)

\ce{Fe^3+}

iron(III)

\ce{Al^3+}

aluminium

\ce{Zn^2+}

zinc

Challenge: pick an unknown number, work through the 12 steps in order, read each observation in the notebook, then tick the ions you think are present and check. Each number is a fixed unknown, so you can compare results with classmates.

Concept check

Why is the group I chloride precipitate treated with hot water before ammonia?

Hot water dissolves AgCl PbCl₂ is about three times more soluble in hot water, separating it from AgCl and Hg₂Cl₂ Hot water makes Hg₂Cl₂ disproportionate Only to wash away excess acid

Try the calculation

1.00 mL of unknown holds 0.10 mmol each of Ag⁺, Pb²⁺ and Hg₂²⁺. What is the least volume (mL) of 6 M HCl that converts all three into chlorides?

Key takeaways

Qualitative analysis sorts ions into groups by solubility, separates them within each group, and confirms each with a specific test.

Group I: HCl precipitates Ag⁺, Pb²⁺, Hg₂²⁺; hot water extracts PbCl₂, and ammonia dissolves AgCl while blackening Hg₂Cl₂.

NaOH uses amphoterism to split Al and Zn from Fe and Cu; an NH₃/NH₄⁺ buffer turns Cu into its deep-blue ammine, precipitates Al(OH)₃ and keeps Zn in solution.

Every conclusion — present or absent — needs a clear positive or negative result behind it.

Data and references

G. Svehla, Vogel’s Qualitative Inorganic Analysis, 7th ed. (Longman, 1996).

A. E. Martell & R. M. Smith, Critical Stability Constants (Plenum, 1974–1989), via D. Harvey, Analytical Chemistry 2.1, Appendices 10–12.

W. M. Haynes (ed.), CRC Handbook of Chemistry and Physics, 97th ed. (2016): solubility of lead(II) chloride.

N. N. Greenwood & A. Earnshaw, Chemistry of the Elements, 2nd ed. (1997): mercury(I) chemistry and precipitate colours.