Reactions of acids with metals, carbonates and bases explained: HSC Chemistry Module 6

Syllabus dot point

“Predict and write balanced molecular, ionic and net ionic equations for reactions of acids with active metals, metal carbonates and hydrogencarbonates, and bases (including metal oxides and hydroxides)”

10 min readUpdated 22 Jun 2026

Quick answer

Acids react with active metals to give salt and hydrogen, with carbonates and hydrogencarbonates to give salt, water and $CO_2$ , and with bases (metal oxides, hydroxides, ammonia) to give a salt and water; to write the net ionic equation, split aqueous strong electrolytes, cancel spectators, and check that atoms and charge balance.

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What this dot point is asking
The answer
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In plain English

Acids are proton donors looking for somewhere to dump an $H^+$ , and there are only four "somewheres" you need for this dot point. Hand a proton to an active metal and it grabs it to make hydrogen gas, so the metal itself disappears into solution. Hand a proton to a carbonate or a hydrogencarbonate and the leftover carbon dioxide fizzes off as bubbles, which is exactly why an antacid tablet dropped in stomach acid froths. Hand a proton to a base, whether that is a hydroxide, a metal oxide, or ammonia, and it just gets swallowed up to make water (or, for ammonia, an ammonium ion) with no gas at all. Writing the "ionic equation" is nothing more than deleting the spectators: any ion that shows up unchanged on both sides of the arrow, still fully dissolved and doing nothing, gets crossed out so the net ionic equation shows only the atoms that actually change.

What this dot point is asking

NESA wants you to predict the products of acid reactions with active metals, metal carbonates and hydrogencarbonates, and bases (metal oxides and hydroxides), write balanced molecular, full ionic and net ionic equations, identify spectator ions, and describe the observable evidence (bubbling, dissolution, temperature change, gas tests). The chemistry builds on the Arrhenius properties of acids and feeds directly into enthalpy of neutralisation and titration analysis.

The answer

Reaction type 1: acid with an active metal

General form: acid + metal -> salt + hydrogen.

2HCl_{(aq)} + Mg_{(s)} \rightarrow MgCl_{2(aq)} + H_{2(g)}

Net ionic:

2H^+_{(aq)} + Mg_{(s)} \rightarrow Mg^{2+}_{(aq)} + H_{2(g)}

This is a redox reaction. The metal is oxidised; $H^+$ is reduced.

Activity series: Only metals more reactive than hydrogen displace it from dilute acids. K, Na, Ca, Mg, Al, Zn, Fe, Pb (slowly) react; Cu, Ag, Au do not. Lead reacts slowly because of an insoluble lead salt coating.
Observations: Bubbling (hydrogen gas), the metal disappears, the solution may warm.
Gas test: Hydrogen gives a "pop" with a lit splint.

The activity series below ranks the metals most likely to appear in Module 6 acid-reaction questions, from most to least reactive with dilute acids:

Reaction type 2: acid with a metal carbonate

General form: acid + carbonate -> salt + water + carbon dioxide.

2HCl_{(aq)} + Na_2CO_{3(aq)} \rightarrow 2NaCl_{(aq)} + H_2O_{(l)} + CO_{2(g)}

Net ionic (for the soluble carbonate above):

2H^+_{(aq)} + CO_3^{2-}{}_{(aq)} \rightarrow H_2O_{(l)} + CO_{2(g)}

For an insoluble carbonate ( $CaCO_3$ , $MgCO_3$ ), keep the carbonate as a solid in the ionic equation:

2H^+_{(aq)} + CaCO_{3(s)} \rightarrow Ca^{2+}_{(aq)} + H_2O_{(l)} + CO_{2(g)}

Observations. Bubbling (carbon dioxide), the carbonate dissolves.

Gas test. Carbon dioxide turns limewater ( $Ca(OH)_2{}_{(aq)}$ ) milky/cloudy. The chemistry: $CO_2 + Ca(OH)_2 \rightarrow CaCO_3 \downarrow + H_2O$ .

Reaction type 3: acid with a metal hydrogencarbonate

General form: acid + hydrogencarbonate -> salt + water + carbon dioxide.

HCl_{(aq)} + NaHCO_{3(aq)} \rightarrow NaCl_{(aq)} + H_2O_{(l)} + CO_{2(g)}

Net ionic:

H^+_{(aq)} + HCO_3^-{}_{(aq)} \rightarrow H_2O_{(l)} + CO_{2(g)}

This is the chemistry of common antacids (sodium bicarbonate neutralising stomach acid).

Reaction type 4: acid with a base (neutralisation)

General form: acid + base -> salt + water.

With a soluble hydroxide:

HCl_{(aq)} + NaOH_{(aq)} \rightarrow NaCl_{(aq)} + H_2O_{(l)}

Net ionic:

H^+_{(aq)} + OH^-_{(aq)} \rightarrow H_2O_{(l)}

This single net ionic equation describes every strong acid + strong base reaction.

With a metal oxide (a basic oxide):

2HCl_{(aq)} + CuO_{(s)} \rightarrow CuCl_{2(aq)} + H_2O_{(l)}

Net ionic:

2H^+_{(aq)} + CuO_{(s)} \rightarrow Cu^{2+}_{(aq)} + H_2O_{(l)}

With ammonia:

HCl_{(aq)} + NH_{3(aq)} \rightarrow NH_4Cl_{(aq)}

Net ionic:

H^+_{(aq)} + NH_{3(aq)} \rightarrow NH_4^+{}_{(aq)}

Observations. Heat is released. With a coloured oxide ( $CuO$ black), the solid dissolves and the solution takes on the colour of the metal cation ( $Cu^{2+}$ blue).

The four acid reaction types and their gas tests

Acid + active metal $\rightarrow$ salt + $H_2$ . Test: lit splint gives a "pop".
Acid + carbonate $\rightarrow$ salt + $H_2O$ + $CO_2$ . Test: limewater ( $Ca(OH)_{2(aq)}$ ) turns milky/cloudy.
Acid + hydrogencarbonate $\rightarrow$ salt + $H_2O$ + $CO_2$ . Same $CO_2$ test as above.
Acid + base (hydroxide, metal oxide, or ammonia) $\rightarrow$ salt + $H_2O$ (or $NH_4^+$ for ammonia). No gas test needed; the evidence is heat release and, for a coloured oxide, a colour change as the solid dissolves.
Spectator ion rule. Any ion with the same formula and the same coefficient on both sides of the full ionic equation is a spectator and is cancelled to give the net ionic equation.

Writing ionic and net ionic equations

Write a balanced molecular equation with state symbols.
Split every aqueous strong electrolyte into its ions. Strong acids ( $HCl$ , $HNO_3$ , $H_2SO_4$ , $HClO_4$ ), strong bases ( $NaOH$ , $KOH$ , $Ca(OH)_2$ , $Ba(OH)_2$ ), and soluble salts split. Solids, liquids, gases, and weak electrolytes do not split.
Cancel ions that appear unchanged (same species, same coefficient) on both sides. These are the spectators.
Check that the net ionic equation balances for atoms and for charge.

Exam technique for equation questions

When a question says "write the ionic equation", always check whether it wants the FULL ionic equation (every strong electrolyte split, spectators still present) or the NET ionic equation (spectators cancelled). If the command word is just "ionic equation" with no further clarification, give the net ionic equation and state the spectator separately, since that is what most marking guidelines reward. Always carry state symbols through every equation, including the ionic ones: $(aq)$ , $(s)$ , $(l)$ , $(g)$ are worth marks on their own. For a diprotic acid such as $H_2SO_4$ reacting with a monoprotic base or a hydrogencarbonate, check the mole ratio is 1:2, not 1:1, before you balance.

Worked example

Predict the products and write the balanced molecular, full ionic and net ionic equations for the reaction of dilute sulfuric acid with solid magnesium oxide.

Step 1: Reaction type. Acid + metal oxide -> salt + water.

Step 2: Molecular equation.

H_2SO_{4(aq)} + MgO_{(s)} \rightarrow MgSO_{4(aq)} + H_2O_{(l)}

Step 3: Full ionic equation. Split aqueous strong electrolytes:

2H^+_{(aq)} + SO_4^{2-}{}_{(aq)} + MgO_{(s)} \rightarrow Mg^{2+}_{(aq)} + SO_4^{2-}{}_{(aq)} + H_2O_{(l)}

Step 4: Net ionic equation. Cancel $SO_4^{2-}$ :

2H^+_{(aq)} + MgO_{(s)} \rightarrow Mg^{2+}_{(aq)} + H_2O_{(l)}

Step 5: Check. Atoms: 2 H, 1 Mg, 1 O on each side. Charge: $+2$ on each side. Balanced.

Common traps

Splitting insoluble solids. $CaCO_3$ , $MgO$ , $CuO$ are all solids in the relevant equations. Do not split them into ions, even when they appear in an ionic equation.

Forgetting the diprotic acid stoichiometry. $H_2SO_4$ provides two protons, so reactions with monoprotic bases or hydrogencarbonates need a 2:1 ratio.

Writing $H^+$ instead of $H_3O^+$ in ionic equations. Both are accepted at HSC. Be consistent within an answer.

Missing the gas test: Markers expect students to know the limewater test for $CO_2$ and the pop test for $H_2$ .
Saying copper reacts with dilute HCl: Copper sits below hydrogen in the activity series. It does not react with dilute non-oxidising acids. (It does react with concentrated $HNO_3$ via a different, oxidative mechanism, but this is beyond HSC.)
Ignoring state symbols: Markers deduct for missing $(aq)$ , $(s)$ , $(l)$ or $(g)$ on ionic equations.

Examples in context

Example 1. Limestone neutralisation of acid mine drainage at Captains Flat. Captains Flat in southern NSW houses a legacy lead and zinc mine site discharging acidic water rich in sulfate and dissolved metals. NSW EPA contractors built passive treatment beds packed with crushed limestone ( $CaCO_3$ ). The dominant chemistry is $CaCO_{3(s)} + 2H^+_{(aq)} \rightarrow Ca^{2+}_{(aq)} + H_2O + CO_{2(g)}$ , neutralising the proton load and releasing carbon dioxide visible as bubbling at the bed surface. The net ionic equation is the same one HSC students write for limestone plus hydrochloric acid. Officers monitor downstream pH; once it rises above 6 the heavy metals precipitate as hydroxides and are filtered out of the discharge.

Example 2. BHP Newcastle pickle line acid attack on mild steel. Steel from the Newcastle steelworks is descaled in a pickle line by dipping in 15 percent sulfuric acid. The chemistry is $Fe_{(s)} + H_2SO_4{}_{(aq)} \rightarrow FeSO_4{}_{(aq)} + H_{2(g)}$ , the textbook acid-plus-active-metal reaction. The net ionic equation reduces to $Fe + 2H^+ \rightarrow Fe^{2+} + H_2$ . Plant operators must vent the hydrogen quickly because the line runs at 80 degrees C, well above the lower flammability limit of $H_2$ in air. Spent acid is recovered as ferrous sulfate, a feedstock for water treatment chemicals. The HSC reaction-type framework explains why iron pickling proceeds at all while copper (below H in the activity series) does not.

Try this

Q1. Write balanced molecular and net ionic equations for the reaction between sulfuric acid and sodium hydrogen carbonate. [3 marks]

Cue. Molecular: $H_2SO_4 + 2NaHCO_3 \rightarrow Na_2SO_4 + 2H_2O + 2CO_2$ . Net ionic: $H^+ + HCO_3^- \rightarrow H_2O + CO_2$ .

Q2. A 2.50 g sample of pure calcium carbonate is added to excess 1.00 mol L $^{-1}$ HCl. Calculate the volume of $CO_2$ produced at 25 degrees C and 100 kPa (molar volume 24.79 L mol $^{-1}$ ). [3 marks]

Cue. $n(CaCO_3) = 2.50 / 100.09 = 0.0250$ mol; $n(CO_2) = 0.0250$ mol; $V = 0.0250 \times 24.79 = 0.620$ L.

Q3. Predict the products and write the net ionic equation for each of: (a) magnesium ribbon plus dilute nitric acid; (b) zinc oxide plus hydrochloric acid; (c) aqueous ammonia plus hydrochloric acid. [2+2+2 marks]

Cue. (a) $Mg + 2H^+ \rightarrow Mg^{2+} + H_2$ . (b) $ZnO + 2H^+ \rightarrow Zn^{2+} + H_2O$ . (c) $NH_3 + H^+ \rightarrow NH_4^+$ .

Exam-style questions

Questions in the style of NESA exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

2021 HSC-style4 marks

Dilute hydrochloric acid is added to solid calcium carbonate. Write the balanced molecular equation, the full ionic equation, and the net ionic equation for the reaction. Identify the spectator ion(s).

Show worked answer →

A 4 mark answer needs all three equations and the spectator identification.

Molecular equation.

2HCl_{(aq)} + CaCO_{3(s)} \rightarrow CaCl_{2(aq)} + H_2O_{(l)} + CO_{2(g)}

Full ionic equation. Split soluble strong electrolytes into ions. $CaCO_3$ is a solid (insoluble) so it is not split. $H_2O$ and $CO_2$ are molecular so they are not split.

2H^+_{(aq)} + 2Cl^-_{(aq)} + CaCO_{3(s)} \rightarrow Ca^{2+}_{(aq)} + 2Cl^-_{(aq)} + H_2O_{(l)} + CO_{2(g)}

Net ionic equation. Cancel ions that appear unchanged on both sides ( $2Cl^-$ ).

2H^+_{(aq)} + CaCO_{3(s)} \rightarrow Ca^{2+}_{(aq)} + H_2O_{(l)} + CO_{2(g)}

Spectator ion. $Cl^-_{(aq)}$ . It appears on both sides unchanged and takes no part in the reaction.

Markers reward (1) a correctly balanced molecular equation with state symbols, (2) keeping the solid carbonate intact in the ionic equation, (3) a correctly cancelled net ionic equation, (4) explicit identification of the spectator.

2019 HSC-style3 marks

A small piece of zinc is dropped into dilute sulfuric acid. Describe the observations and write the balanced ionic equation. Explain how the identity of the gas produced could be confirmed.

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Observations. Vigorous bubbling at the zinc surface. The zinc gradually decreases in size and dissolves. The solution may warm noticeably (the reaction is exothermic).

Equation. Sulfuric acid is fully dissociated, so the net ionic equation is:

Zn_{(s)} + 2H^+_{(aq)} \rightarrow Zn^{2+}_{(aq)} + H_{2(g)}

The $SO_4^{2-}$ ion is a spectator and is omitted.

Confirmation of gas. Collect a sample over water in a test tube. Bring a lit splint to the mouth of the tube. A "pop" sound confirms hydrogen gas. (The hydrogen burns explosively in the small volume of air.)

Markers reward (1) two specific observations, (2) a correctly balanced net ionic equation, (3) naming the pop test with the expected result.

Practice questions

Original practice questions graded from foundation to exam level, each with a full worked solution. Try them before revealing the solution.

foundation3 marks

Write the balanced molecular equation, the full ionic equation and the net ionic equation for the reaction of dilute hydrochloric acid with solid zinc, and name the spectator ion.

Show worked solution →

Molecular equation.

2HCl_{(aq)} + Zn_{(s)} \rightarrow ZnCl_{2(aq)} + H_{2(g)}

Full ionic equation. $HCl$ is a strong acid (splits); $ZnCl_2$ is a soluble salt (splits); $Zn$ is a solid and $H_2$ is a gas (neither splits).

2H^+_{(aq)} + 2Cl^-_{(aq)} + Zn_{(s)} \rightarrow Zn^{2+}_{(aq)} + 2Cl^-_{(aq)} + H_{2(g)}

Net ionic equation. Cancel $2Cl^-_{(aq)}$ .

2H^+_{(aq)} + Zn_{(s)} \rightarrow Zn^{2+}_{(aq)} + H_{2(g)}

Spectator ion. $Cl^-_{(aq)}$ .

Marking criteria: 1 mark for the correct balanced molecular equation with states, 1 mark for the correct full ionic equation, 1 mark for the correctly cancelled net ionic equation with the spectator named.

foundation3 marks

State the reagent used to test for each gas, and the positive result observed: (a) hydrogen; (b) carbon dioxide. (c) Explain why bubbling gas through limewater is a chemical test rather than a physical one.

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(a) Hydrogen: Bring a lit splint to the mouth of the collecting tube. A "pop" sound confirms hydrogen gas igniting explosively in the small trapped volume of air.
(b) Carbon dioxide: Bubble the gas through limewater, $Ca(OH)_{2(aq)}$ . The solution turns milky/cloudy.
(c) Chemical test: A new substance (insoluble solid $CaCO_3$ ) forms via a chemical reaction, $CO_2 + Ca(OH)_2 \rightarrow CaCO_3 \downarrow + H_2O$ , rather than the gas simply being observed to change a physical property such as colour or state without any new bond forming.

Marking criteria: 1 mark per correct gas test (reagent + observation), 1 mark for correctly explaining why the limewater test is chemical (new substance formed via reaction) rather than physical.

core5 marks

A 3.20 g sample of solid magnesium oxide is added to excess dilute sulfuric acid and reacts completely. Calculate the mass of magnesium sulfate,

MgSO_4

, produced, to 3 significant figures. (

M(MgO) = 40.30\ \text{g mol}^{-1}

M(MgSO_4) = 120.37\ \text{g mol}^{-1}

Show worked solution →

Step 1: write the balanced equation and identify the mole ratio.

H_2SO_{4(aq)} + MgO_{(s)} \rightarrow MgSO_{4(aq)} + H_2O_{(l)}

The mole ratio of $MgO$ to $MgSO_4$ is 1:1.

Step 2: moles of magnesium oxide.

n(MgO) = \frac{m}{M} = \frac{3.20\ \text{g}}{40.30\ \text{g mol}^{-1}} = 0.07940\ \text{mol}

Step 3: moles of magnesium sulfate (1:1 ratio).

n(MgSO_4) = 0.07940\ \text{mol}

Step 4: mass of magnesium sulfate.

m = n \times M = 0.07940\ \text{mol} \times 120.37\ \text{g mol}^{-1} = 9.557\ \text{g}

Step 5: round to 3 significant figures (matching the 3.20 g data).

m(MgSO_4) = 9.56\ \text{g}

Marking criteria: 1 mark for the correctly balanced equation, 1 mark for correct moles of $MgO$ , 1 mark for correctly applying the 1:1 mole ratio, 1 mark for the mass calculation, 1 mark for the final answer to 3 significant figures with correct units. This is a THEORETICAL yield; excess acid ensures $MgO$ is the limiting reagent, but a real prepared sample may show a lower practical yield due to incomplete reaction or losses on filtration.

core5 marks

The activity-series ladder below ranks metals by their reactivity with dilute acid, with a dashed line marking the hydrogen threshold. Using the diagram, (a) identify which TWO of the following metals would show NO visible reaction with dilute hydrochloric acid: magnesium, copper, zinc, silver; (b) explain, in terms of electron transfer, why the other two DO react.

Show worked solution →

Reading the diagram: The ladder places $K, Na, Ca$ at the top (most vigorous), then $Mg, Al$ , then $Zn, Fe$ , then $Pb$ (slow, coated), a dashed hydrogen threshold, and below it $Cu, Ag, Au$ (no reaction).
(a) No visible reaction: Copper and silver both sit below the dashed hydrogen threshold on the ladder, so neither displaces hydrogen from dilute HCl: no bubbling, no dissolution.
(b) Explanation for magnesium and zinc: Both sit above the hydrogen threshold, meaning both are more reactive than hydrogen and more readily lose electrons than $H_2$ does. Each metal atom is oxidised, losing two electrons to form $M^{2+}_{(aq)}$ , while the two electrons released reduce two $H^+_{(aq)}$ ions to one molecule of $H_{2(g)}$ : $M_{(s)} \rightarrow M^{2+}_{(aq)} + 2e^-$ combined with $2H^+_{(aq)} + 2e^- \rightarrow H_{2(g)}$ . Because copper and silver hold their electrons more strongly than hydrogen (they are less reactive), they cannot transfer electrons to $H^+$ and so no reaction is observed.

Marking criteria: 1 mark for correctly identifying copper as unreactive, 1 mark for correctly identifying silver as unreactive, 1 mark for correctly identifying magnesium and zinc as reactive with reference to the diagram's hydrogen threshold, 2 marks for a correct electron-transfer (oxidation of metal, reduction of $H^+$ ) explanation with a half-equation or clear electron count.

core4 marks

Predict the products and write the net ionic equation for the reaction of dilute nitric acid with solid copper(II) oxide, and describe one observation that would confirm the reaction has occurred.

Show worked solution →

Reaction type. Acid + metal oxide (a base) $\rightarrow$ salt + water.

Molecular equation.

2HNO_{3(aq)} + CuO_{(s)} \rightarrow Cu(NO_3)_{2(aq)} + H_2O_{(l)}

Net ionic equation. $HNO_3$ is a strong acid and $Cu(NO_3)_2$ is a soluble salt, so both split; $CuO$ is a solid and does not split; the $NO_3^-$ ions are spectators and cancel.

2H^+_{(aq)} + CuO_{(s)} \rightarrow Cu^{2+}_{(aq)} + H_2O_{(l)}

Observation. The black solid $CuO$ gradually dissolves and disappears, and the solution turns blue as the hydrated $Cu^{2+}_{(aq)}$ ion forms.

Marking criteria: 1 mark for the correct molecular equation, 1 mark for the correctly balanced net ionic equation with states, 1 mark for identifying the nitrate ion as the cancelled spectator, 1 mark for a specific colour/dissolution observation (not just "a reaction occurs").

exam6 marks

A student is given four unlabelled solids: calcium carbonate, zinc metal, copper metal, and sodium hydroxide. Design a sequence of tests, using only dilute hydrochloric acid and any standard gas tests, that would allow the student to identify all four unlabelled solids. Justify each step using the reaction types and activity series covered in this dot point.

Show worked solution →

This is a 6-mark DESIGN/JUSTIFY question: markers reward a logical decision sequence tied to observable, distinguishable evidence, not just a list of the four reactions.

Band 6 plan.

Add dilute HCl to a small sample of each solid separately and observe.
Sort results into three groups first by whether bubbling occurs, then confirm gas identity, then use the no-reaction group to separate metal from base by a second property (e.g. solubility/appearance) since HCl reacts with both remaining candidates in different visible ways.
Use the activity series explicitly to justify why copper alone shows no reaction with dilute HCl.

Model answer.

Add a few drops of dilute hydrochloric acid to a small sample of each of the four solids in separate test tubes and observe.

Calcium carbonate will fizz vigorously as it reacts, releasing a colourless gas. Confirm the gas is $CO_2$ by bubbling it through limewater; a milky/cloudy result confirms carbonate, via $CaCO_{3(s)} + 2H^+_{(aq)} \rightarrow Ca^{2+}_{(aq)} + H_2O_{(l)} + CO_{2(g)}$ .
Zinc metal will also bubble, but the gas is colourless and does not cloud limewater. Instead, test it with a lit splint: a "pop" confirms hydrogen, consistent with $Zn_{(s)} + 2H^+_{(aq)} \rightarrow Zn^{2+}_{(aq)} + H_{2(g)}$ . Zinc sits above hydrogen in the activity series so it displaces $H_2$ readily.
Copper metal will show no bubbling, no dissolution and no colour change, because copper sits below hydrogen in the activity series and cannot transfer electrons to $H^+_{(aq)}$ in a dilute, non-oxidising acid.
Sodium hydroxide, added to dilute HCl, produces no visible gas or bubbling either, since $HCl_{(aq)} + NaOH_{(aq)} \rightarrow NaCl_{(aq)} + H_2O_{(l)}$ is a neutralisation with no gaseous product; however, unlike copper, the solid sodium hydroxide dissolves completely and the mixture releases a noticeable amount of heat (mildly warm to touch), whereas the copper sample remains a solid metal piece with no temperature change.

Combining these four distinct sets of evidence, bubbling plus limewater test, bubbling plus pop test, no reaction at all, and no gas but dissolution with heat release, uniquely identifies all four solids using only dilute HCl and standard gas tests.

Marker's note: full marks require (1) all four solids correctly matched to a distinct observable outcome, (2) both required confirmatory gas tests named with their positive result, (3) explicit use of the activity series to justify copper's inertness, and (4) a clear distinguishing observation (heat/dissolution, not just "no gas") separating sodium hydroxide from copper, since both give no bubbling.

Practise this

Sources & how we know this

Chemistry Stage 6 Syllabus (2017) — NESA
Chemistry HSC exam papers — NESA