Why the group reagents work, and naming a salt

Precipitate or not?

A salt $\text{MA}$ dissolves until its ions reach a limit. At the limit, $[\text{M}^+][\text{A}^-] = K_{sp}$, the solubility product. Mix two solutions and work out the ionic product $Q$ from the concentrations you have.

• If $Q > K_{sp}$, a precipitate forms.
• If $Q < K_{sp}$, nothing forms.

A group reagent works by controlling one ion. In group II and group IV the hydrogen sulphide supplies sulphide ions, and the acid or base decides how many there are.

Why the sulphide group splits in two

Hydrogen sulphide is a weak acid in two steps. In water saturated with the gas, $[\text{H}_2\text{S}] = 0.10\ \text{mol/L}$. The two steps together give

with $K_{a1}K_{a2} = 1.1 \times 10^{-26}$. The $[\text{H}^+]$ is squared. So a small change of pH changes the sulphide concentration enormously.

• In dilute hydrochloric acid at pH 0.5, $[\text{S}^{2-}]$ is about $1 \times 10^{-26}\ \text{mol/L}$. Only a sulphide with an extremely small $K_{sp}$, such as copper sulphide ($K_{sp}$ about $6 \times 10^{-37}$), precipitates.
• With ammonium hydroxide at pH 9, $[\text{S}^{2-}]$ is about $1 \times 10^{-9}\ \text{mol/L}$, which is $10^{17}$ times more. Now zinc, nickel, cobalt and manganese sulphides precipitate too.

Move $p$ from 0.5 to 9. The blue line passes the dashed lines one after another. Copper sulphide is first. Zinc sulphide is next, near pH 3. Manganese sulphide comes last. That is why group IV needs a basic solution.

Values of $K_{sp}$ for sulphides differ between data tables by several powers of ten. The argument does not change: the gaps between the sulphides are huge.

Why ammonium chloride is added for group III

Ammonium hydroxide is a weak base. Its ionisation is

Add ammonium chloride. The ammonium ion is a common ion. It pushes the equilibrium back and lowers $[\text{OH}^-]$, so that

The hydroxides of iron(III) and aluminium have such tiny $K_{sp}$ values that they still precipitate at this low $[\text{OH}^-]$. The hydroxide of magnesium has a much larger $K_{sp}$ (about $5.6 \times 10^{-12}$), and it stays in solution. Without ammonium chloride, magnesium hydroxide, and the hydroxides of group IV, would also appear.

Zinc, nickel and cobalt also stay in solution at this stage because they form ammine complexes. The idea of $Q$ and $K_{sp}$ alone does not tell the whole story for them.

Why carbonate in group V

Ammonium carbonate is added to a solution that contains ammonium chloride and ammonium hydroxide. The medium is basic, so there are plenty of carbonate ions, and barium and calcium carbonates have small $K_{sp}$. Magnesium carbonate is much more soluble. Ammonium ions also turn carbonate ions into hydrogencarbonate ions, which lowers $[\text{CO}_3^{2-}]$ further. So magnesium is left for group VI.

Practise with new numbers

Naming a salt from the observations

A salt is a pair: one cation and one anion. Work in this order.

1. Look. Colour, smell, heating. Blue or green points to copper or nickel. Pink points to cobalt.
2. Flame test. Brick red, apple green or bluish green gives a first guess for the cation.
3. Anion tests. Dilute acid, concentrated acid, barium chloride.
4. Cation tests. Group reagents and then the confirmatory test.
5. Name the pair. The salt must agree with every observation. One test never names an ion alone.

Example A. White crystals. The flame is brick red. Dilute hydrochloric acid on the salt gives brisk effervescence, and the odourless gas turns lime water milky. Ammonium oxalate gives a white precipitate. The salt is calcium carbonate.

Example B. Green crystals. Sodium hydroxide gives a green precipitate that does not dissolve in excess. Excess ammonium hydroxide gives a deep blue solution. Dimethylglyoxime gives a rosy red precipitate. Barium chloride gives a white precipitate that stays in dilute hydrochloric acid. The salt is nickel(II) sulphate.

Try it in Python

The program works out the sulphide ion concentration at three values of pH and tells which sulphides can precipitate.

Common mistakes

• Saying the group reagent works on charge alone. It works because $Q$ passes $K_{sp}$ for some ions and not for others.
• Forgetting that pH acts through $[\text{H}^+]^2$. One pH unit changes the sulphide concentration by a factor of 100.
• Adding ammonium hydroxide without ammonium chloride for group III. Group IV and magnesium hydroxides come out too.
• Naming a salt from one test. Check every observation against the pair.

Remember these

• Precipitation: $Q > K_{sp}$.
• $\text{S}^{2-}$ falls steeply as the acidity rises: it is proportional to $1 / [\text{H}^+]^2$.
• Ammonium chloride lowers $[\text{OH}^-]$, so only the hydroxides of iron(III) and aluminium form in group III.
• A salt is a pair that fits all the observations.

References

• OpenStax Chemistry 2e, Chapter 15, sections 15.1 and 15.2 on precipitation and the solubility product.
• Vogel's Qualitative Inorganic Analysis, the theory of the group separations.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. Why does hydrogen sulphide precipitate copper sulphide in dilute acid but not manganese sulphide?

2. What does ammonium chloride do to the concentration of hydroxide ions in ammonium hydroxide?

3. A solution at pH 3 is saturated with hydrogen sulphide. It holds copper(II) ions and manganese(II) ions, each at $0.010\ \text{mol/L}$. Which sulphide precipitates?

4. Why does an analyst say that the ammonium test is done on the original salt?