IGCSE Chemistry 0620: Distinguishing Iron(II), Iron(III) and Copper(II) Ions

An electroplating workshop keeps three reagent bottles of solution for different plating baths: iron(II) sulfate solution, iron(III) chloride solution and copper(II) sulfate solution. After a stock check, the labels are found to be illegible, so a quality-control chemist takes a fresh sample from each bottle.

(a) The chemist adds aqueous sodium hydroxide dropwise, then in excess, to a sample of each solution. State the colour of the precipitate formed with each solution, and state whether each precipitate dissolves in the excess sodium hydroxide. [3]

(b) The chemist then repeats the test using aqueous ammonia instead of sodium hydroxide, adding it dropwise to a fresh sample from each bottle and then in excess. Describe how the result for the copper(II) sulfate solution differs from the results for the other two solutions. [2]

(c) Explain why the excess-ammonia test in (b) gives the chemist a second, independent check that a bottle contains copper(II) sulfate solution, beyond the precipitate colour already seen in (a). [2]

Part (a): Testing all three solutions with sodium hydroxide

Adding aqueous sodium hydroxide dropwise, then in excess, to each solution gives:

Solution	Precipitate colour	Dissolves in excess NaOH?
iron(II) sulfate, $\text{Fe}^{2+}$	green	no
iron(III) chloride, $\text{Fe}^{3+}$	red-brown	no
copper(II) sulfate, $\text{Cu}^{2+}$	light blue	no

Unlike aluminium and zinc hydroxides, none of these three hydroxide precipitates redissolves in excess sodium hydroxide, they simply remain as coloured precipitates, and it is the colour that identifies the ion.

Part (b): Repeating the test with aqueous ammonia

Adding aqueous ammonia dropwise, then in excess, gives a different pattern for copper(II):

Iron(II) sulfate and iron(III) chloride: the green and red-brown precipitates, respectively, remain undissolved in excess ammonia, just as they did with excess sodium hydroxide.
Copper(II) sulfate: the light blue precipitate dissolves in excess aqueous ammonia, forming a deep blue solution.

$\text{Cu(OH)}_2(s) + 4\text{NH}_3(aq) \rightarrow [\text{Cu(NH}_3)_4]^{2+}(aq) + 2\text{OH}^-(aq)$

So the copper(II) sample is the only one of the three where excess ammonia produces a visible change.

Part (c): Why this is a useful second check

The precipitate colours in (a) already identify each ion on their own, so in principle the chemist does not strictly need a second test. However, the excess-ammonia result is valuable as an independent confirmation:

It relies on a completely different property (solubility of the precipitate in excess ammonia) rather than colour, which can be harder to judge with certainty, especially between similar shades.
Only copper(II) hydroxide dissolves in excess ammonia among the three; iron(II) and iron(III) hydroxides do not. So if a precipitate dissolves in excess ammonia to a deep blue solution, this by itself confirms copper(II) ions are present, cross-checking the identification already made from the precipitate colour in (a).

Final answers

(a) $\text{Fe}^{2+}$ : green precipitate; $\text{Fe}^{3+}$ : red-brown precipitate; $\text{Cu}^{2+}$ : light blue precipitate, none dissolves in excess sodium hydroxide.
(b) The copper(II) sulfate precipitate $\boxed{\text{dissolves in excess ammonia to a deep blue solution}}$ ; the iron(II) and iron(III) precipitates remain undissolved.
(c) Only copper(II) hydroxide dissolves in excess ammonia, so this solubility behaviour independently confirms copper(II) ions, separate from the colour evidence in (a).