Topic 1 · page 10 of 10

Test-tube reactions

Every observation is evidence for a species in the ionic equation. A colour fades because an ion is used up, a precipitate appears because an insoluble solid forms, and fizzing means a gas.

14 min read1.12(i)1.12(ii)1.12(iii)

Read first: Writing equations

Describing what you see 1.12(i)1.12(ii)1.12(iii)

An observation is something you can see, hear, feel or smell. The name of a product is not an observation.

Say this	Not this
fizzing, or bubbles of gas	hydrogen is produced
the blue colour fades	copper ions are used up
a white precipitate forms	silver chloride forms
the solid dissolves, giving a colourless solution	the metal reacts
the test tube gets warm	the reaction is exothermic

Displacement reactions 1.12(i)

Metals

A more reactive metal displaces a less reactive metal from a solution of its compound. The more reactive metal goes into solution as ions; the less reactive metal comes out as a solid.

\ce{Zn(s) + CuSO4(aq) -> ZnSO4(aq) + Cu(s)}

\ce{Zn(s) + Cu^2+(aq) -> Zn^2+(aq) + Cu(s)}

Observation	Species in the ionic equation
blue colour of the solution fades	Cu²⁺(aq) used up
a brown or pink-brown solid coats the zinc	Cu(s) formed
the zinc slowly gets smaller	Zn(s) used up
the mixture gets warm	the reaction releases energy

The sulfate ions are spectators: they appear in the full equation on both sides but not in the ionic equation.

Halogens

A more reactive halogen displaces a less reactive halide ion from solution. Reactivity decreases down the group: chlorine > bromine > iodine.

\ce{Cl2(aq) + 2KBr(aq) -> 2KCl(aq) + Br2(aq)}

\ce{Cl2(aq) + 2Br^-(aq) -> 2Cl^-(aq) + Br2(aq)}

Mixture	Observation	Ionic equation
chlorine water + potassium bromide	colourless to orange (or yellow-orange)	$\ce{Cl2 + 2Br^- -> 2Cl^- + Br2}$
chlorine water + potassium iodide	colourless to brown	$\ce{Cl2 + 2I^- -> 2Cl^- + I2}$
bromine water + potassium iodide	orange to brown	$\ce{Br2 + 2I^- -> 2Br^- + I2}$
bromine water + potassium chloride	no change	no reaction

Typical reactions of acids 1.12(ii)

All acids in water contain H⁺(aq). That is why one ionic equation covers each type of reaction, whichever acid you use.

Acid + … → products	Ionic equation (example)	Observations
metal → salt + hydrogen	$\ce{Mg(s) + 2H+(aq) -> Mg^2+(aq) + H2(g)}$	fizzing; the metal dissolves; the gas gives a squeaky pop with a lighted splint
carbonate solution → salt + water + carbon dioxide	$\ce{CO3^2-(aq) + 2H+(aq) -> H2O(l) + CO2(g)}$	fizzing; the gas turns limewater milky
solid carbonate → salt + water + carbon dioxide	$\ce{CaCO3(s) + 2H+(aq) -> Ca^2+(aq) + H2O(l) + CO2(g)}$	fizzing; the solid dissolves
hydrogencarbonate → salt + water + carbon dioxide	$\ce{HCO3^-(aq) + H+(aq) -> H2O(l) + CO2(g)}$	fizzing
metal oxide → salt + water	$\ce{CuO(s) + 2H+(aq) -> Cu^2+(aq) + H2O(l)}$	black solid dissolves on warming; the solution turns blue
alkali → salt + water	$\ce{H+(aq) + OH^-(aq) -> H2O(l)}$	no visible change; the temperature rises

Precipitation reactions 1.12(iii)

When two solutions are mixed and an insoluble combination of ions is possible, that combination comes out of solution as a solid, called a precipitate. The ionic equation shows only the two ions that come together:

\ce{Ag+(aq) + Cl^-(aq) -> AgCl(s)}

Test for	Add	Precipitate	Ionic equation
chloride, Cl⁻	AgNO₃(aq) and dilute HNO₃	white, AgCl	$\ce{Ag+(aq) + Cl^-(aq) -> AgCl(s)}$
bromide, Br⁻	AgNO₃(aq) and dilute HNO₃	cream, AgBr	$\ce{Ag+(aq) + Br^-(aq) -> AgBr(s)}$
iodide, I⁻	AgNO₃(aq) and dilute HNO₃	yellow, AgI	$\ce{Ag+(aq) + I^-(aq) -> AgI(s)}$
sulfate, SO₄²⁻	BaCl₂(aq) and dilute HCl	white, BaSO₄	$\ce{Ba^2+(aq) + SO4^2-(aq) -> BaSO4(s)}$
copper(II), Cu²⁺	NaOH(aq)	blue, Cu(OH)₂	$\ce{Cu^2+(aq) + 2OH^-(aq) -> Cu(OH)2(s)}$
iron(II), Fe²⁺	NaOH(aq)	green, Fe(OH)₂	$\ce{Fe^2+(aq) + 2OH^-(aq) -> Fe(OH)2(s)}$
iron(III), Fe³⁺	NaOH(aq)	orange-brown, Fe(OH)₃	$\ce{Fe^3+(aq) + 3OH^-(aq) -> Fe(OH)3(s)}$

The dilute acid added with silver nitrate or barium chloride removes carbonate ions, which would otherwise give a misleading white precipitate.

Which ions make precipitates?

You can predict a precipitate if you know which compounds are soluble in water.

Usually soluble	Exceptions (insoluble)
all sodium, potassium and ammonium compounds	none
all nitrates	none
most chlorides, bromides and iodides	silver, lead(II)
most sulfates	barium, lead(II); calcium sulfate is only slightly soluble

Usually insoluble	Exceptions (soluble)
most carbonates	sodium, potassium, ammonium
most hydroxides	sodium, potassium, ammonium; calcium hydroxide is slightly soluble

Check yourself

Magnesium ribbon is added to copper(II) sulfate solution. Give two observations and the ionic equation.
Sodium carbonate solution is added to dilute nitric acid. Give one observation and the ionic equation.
Silver nitrate solution is added to potassium iodide solution. Give the observation, the full equation and the ionic equation, with state symbols.
Bromine water is added to potassium chloride solution. What happens? Explain.
Copper(II) oxide is warmed with dilute sulfuric acid. Give two observations and the ionic equation.
Iron(III) chloride solution is mixed with sodium hydroxide solution. Give the observation and the ionic equation.

Answers cover these until you have tried the questions

The blue colour fades; a brown solid forms (also: the mixture warms, the magnesium gets smaller). $\ce{Mg(s) + Cu^2+(aq) -> Mg^2+(aq) + Cu(s)}$
Fizzing (the gas turns limewater milky). $\ce{CO3^2-(aq) + 2H+(aq) -> H2O(l) + CO2(g)}$
A yellow precipitate. Full: $\ce{AgNO3(aq) + KI(aq) -> AgI(s) + KNO3(aq)}$ . Ionic: $\ce{Ag+(aq) + I^-(aq) -> AgI(s)}$
No change. Bromine is less reactive than chlorine, so it cannot displace chloride ions.
The black solid dissolves; the solution turns blue. $\ce{CuO(s) + 2H+(aq) -> Cu^2+(aq) + H2O(l)}$
An orange-brown precipitate. $\ce{Fe^3+(aq) + 3OH^-(aq) -> Fe(OH)3(s)}$