the writing of balanced full, ionic and net ionic equations for reactions in aqueous solution including precipitation, neutralisation and metal displacement reactions, with state symbols

What this dot point is asking

VCAA wants you to write three forms of equation for any reaction in aqueous solution: the full (molecular) equation, the ionic equation with strong electrolytes split into their ions, and the net ionic equation with spectator ions cancelled. Every species must have a state symbol (aq), (s), (l) or (g).

The answer

Which species split, which stay together

In the ionic equation, only strong electrolytes in solution are written as separated ions. Everything else stays intact.

Split into ions (write as separate ions with (aq)):

• Strong acids: $HCl$, $HBr$, $HI$, $HNO_3$, $H_2SO_4$ (first proton), $HClO_4$.
• Strong bases: $NaOH$, $KOH$, other group 1 hydroxides, $Ca(OH)_2$ (where dissolved), $Ba(OH)_2$.
• Soluble ionic compounds in water: most group 1 salts, ammonium salts, nitrates, most chlorides and sulfates.

Stay intact (do not split):

• Solids, with $(s)$. $AgCl(s)$, $CaCO_3(s)$, $Cu(s)$.
• Pure liquids, with $(l)$. $H_2O(l)$.
• Gases, with $(g)$. $CO_2(g)$, $H_2(g)$.
• Weak acids and weak bases. $CH_3COOH(aq)$, $NH_3(aq)$, $HF(aq)$, $H_2CO_3(aq)$.
• Molecular substances dissolved in water. $C_6H_{12}O_6(aq)$, $C_2H_5OH(aq)$.

The three-step procedure

1. Write the full balanced equation with state symbols. Use solubility rules to decide whether a product is $(aq)$ or $(s)$.
2. Expand all strong electrolytes into their constituent ions. Leave everything else as written.
3. Cancel spectator ions (ions that appear identically on both sides). Check both mass and charge balance.

The result is the net ionic equation: the part of the chemistry that actually changed.

Three common reaction types in VCE Unit 2

Precipitation. Two soluble salts are mixed and an insoluble product forms. Swap the partners and check solubility rules. Example: $Pb(NO_3)_2(aq) + 2KI(aq) \to PbI_2(s) + 2KNO_3(aq)$ gives the net ionic equation $Pb^{2+}(aq) + 2I^-(aq) \to PbI_2(s)$.

Acid-base neutralisation. A strong acid plus a strong base in stoichiometric amount gives a salt and water. The net ionic equation collapses to:

This is the same for every strong-acid strong-base neutralisation. The cation and anion are spectators.

For a strong acid plus a weak base, the weak base stays molecular in the net ionic equation:

For an acid plus a carbonate or hydrogen carbonate, $CO_2$ and $H_2O$ are products: $2H^+(aq) + CO_3^{2-}(aq) \to H_2O(l) + CO_2(g)$.

Metal displacement. A more reactive metal displaces a less reactive one from its salt. Net ionic equations show only the metal and the ion that changes:

The anion of the original salt ($SO_4^{2-}$, $NO_3^-$, $Cl^-$) is always a spectator.

State symbols matter

VCE marks state symbols. The convention:

• $(s)$ solid (precipitates, undissolved metals, ionic solids).
• $(l)$ pure liquid (essentially only water in a typical VCE equation).
• $(g)$ gas (the $CO_2$ from a carbonate-acid reaction, the $H_2$ from a reactive metal in acid).
• $(aq)$ aqueous (dissolved in water).

Forgetting state symbols, or putting $(aq)$ on $H_2O$ in a neutralisation, loses marks.

Checking your work

Two checks for every net ionic equation:

1. Mass balance. The same number of each element on each side.
2. Charge balance. The total charge on the left equals the total charge on the right. For $H^+(aq) + OH^-(aq) \to H_2O(l)$: $+1$ plus $-1$ equals $0$ on both sides.

If either check fails, recount coefficients.

Common traps

Splitting weak acids in an ionic equation

$CH_3COOH$ stays as $CH_3COOH(aq)$, not $H^+ + CH_3COO^-$. Only strong acids fully ionise.

Splitting solids

$CaCO_3(s)$ stays intact even when it is reacting. Only $(aq)$ ionic species split.

Forgetting to balance charges before cancelling spectators

$H^+ + 2OH^- \to H_2O$ is wrong: the charge is $-1$ on the left and $0$ on the right.

Cancelling species that have changed in number

If $K^+$ appears as 2 on the left and 2 on the right, both cancel. If it appears as 2 on the left and 1 on the right, only 1 cancels and the leftover stays.

Writing the same equation for every neutralisation

Strong acid plus strong base does collapse to $H^+ + OH^- \to H_2O$, but a strong acid plus a weak base (or vice versa) does not. The weak species stays molecular.

Wrong state symbols on the precipitate

$AgCl$, $PbI_2$ and $BaSO_4$ are insoluble; mark them $(s)$. $NaNO_3$, $KCl$ and most other group 1 salts are $(aq)$.

In one sentence

To write a net ionic equation: write the balanced full equation with state symbols, split only strong electrolytes in solution into their ions, cancel spectator ions, and double-check both mass and charge balance.

Examples in context

Example 1. Hard-water scale in Latrobe Valley cooling towers. Engineers at AGL Loy Yang dose cooling-water circuits with anti-scaling additives because the water contains dissolved $\text{Ca}^{2+}$ and $\text{HCO}_3^-$ from limestone aquifers. On heating, the net ionic equation is $\text{Ca}^{2+} + 2 \text{HCO}_3^- \to \text{CaCO}_3 \downarrow + \text{H}_2 \text{O} + \text{CO}_2 \uparrow$. The spectator $\text{Na}^+$ from any added base is omitted. Scale of about $5 \, \text{kg}$ per megalitre forms at $80^{\circ}\text{C}$. Treatment with $\text{Na}_3 \text{PO}_4$ shifts the precipitation to $3 \text{Ca}^{2+} + 2 \text{PO}_4^{3-} \to \text{Ca}_3 (\text{PO}_4)_2 \downarrow$, which forms a softer sludge that the blowdown stream can flush before it bakes onto turbine tubes.

Example 2. Lead-acid battery recycling at the Hume Hub plant. The Hume Hub recycling plant in Sydney processes $5 \, \text{million}$ used lead-acid batteries annually. The first step neutralises the spent sulfuric-acid electrolyte with sodium carbonate slurry: net ionic equation $2 \text{H}^+ + \text{CO}_3^{2-} \to \text{H}_2 \text{O} + \text{CO}_2 \uparrow$. The spectator $\text{Na}^+$ and $\text{SO}_4^{2-}$ remain in solution and are crystallised separately as sodium sulfate decahydrate (Glauber's salt) for sale. Lead pastes (mostly $\text{PbSO}_4$ and $\text{PbO}_2$) are desulfurised similarly: $\text{PbSO}_4 (\text{s}) + \text{CO}_3^{2-} \to \text{PbCO}_3 (\text{s}) + \text{SO}_4^{2-}$. The $\text{PbCO}_3$ is then smelted to recover lead metal, achieving over $95\%$ recovery efficiency.

Try this

Q1. Write the net ionic equation when aqueous silver nitrate is mixed with aqueous sodium chloride. State the spectator ions. [2 marks]

• Cue. $\text{Ag}^+ + \text{Cl}^- \to \text{AgCl} \downarrow$. Spectators: $\text{Na}^+$ and $\text{NO}_3^-$.

Q2. $25.0 \, \text{mL}$ of $0.100 \, \text{mol/L}$ $\text{Pb(NO}_3)_2$ is mixed with $25.0 \, \text{mL}$ of $0.300 \, \text{mol/L}$ $\text{KI}$. (a) Write the net ionic equation. (b) Calculate the mass of precipitate. [4 marks]

• Cue. (a) $\text{Pb}^{2+} + 2 \text{I}^- \to \text{PbI}_2 \downarrow$. (b) $n(\text{Pb}^{2+}) = 2.50 \times 10^{-3}$, $n(\text{I}^-) = 7.50 \times 10^{-3}$; need $2:1$ so $\text{Pb}^{2+}$ limits. $n(\text{PbI}_2) = 2.50 \times 10^{-3}$; mass $= 2.50 \times 10^{-3} \times 461.0 = 1.15 \, \text{g}$.

Q3. Consider the reaction of dilute $\text{HCl}$ with solid $\text{CaCO}_3$. (a) Write the full balanced equation. (b) Write the net ionic equation. (c) Explain why $\text{CaCO}_3$ is not split into ions. [2+2+2 marks]

• Cue. (a) $\text{CaCO}_3 + 2 \text{HCl} \to \text{CaCl}_2 + \text{H}_2 \text{O} + \text{CO}_2$. (b) $\text{CaCO}_3 (\text{s}) + 2 \text{H}^+ \to \text{Ca}^{2+} + \text{H}_2 \text{O} + \text{CO}_2$. (c) Solids are not dissociated; only $(aq)$ strong electrolytes are split.