How to Write Net Ionic Equations for Acid-Carbonate Reactions

A piece of marble fizzes when you add hydrochloric acid. The bubbles tell you a gas is escaping, but they don’t tell you which ions belong in the net ionic equation.

Here’s the first decision. Should you write the carbonate as \(\ce{CO3^2-(aq)}\), or keep it as \(\ce{CaCO3(s)}\)?

Because marble is solid calcium carbonate, you must keep it together. For calcium carbonate reacting with dilute hydrochloric acid, the net ionic equation is:

\[ \begin{aligned} &\ce{CaCO3(s) + 2H+(aq)}\\ &\ce{-> Ca^2+(aq) + CO2(g)}\\ &\quad\ce{+ H2O(l)} \end{aligned} \]

To write a net ionic equation for an acid-carbonate reaction:

1. Write the balanced molecular equation, including state symbols.
2. Split strong acids and soluble ionic compounds into their aqueous ions.
3. Keep solids, liquids, gases, and weak acids together.
4. Cancel spectator ions that appear unchanged on both sides.
5. Check that atoms and total electrical charge balance.

That third step is where many otherwise correct answers go wrong. The carbonate’s physical state, and the acid’s strength, determine which particles you should actually write.

01What does the acid-carbonate experiment actually show?

Imagine a conical flask containing small marble chips. A dropping funnel feeds dilute hydrochloric acid into the flask, and a separate delivery tube carries any escaping gas into a test tube containing clear limewater.

Follow the apparatus from left to right:

• The marble chips sit at the bottom of the reaction flask.
• Hydrochloric acid enters through the dropping funnel and contacts their surfaces.
• Gas produced in the flask travels through the delivery tube.
• The gas bubbles through limewater, which is a dilute solution of calcium hydroxide.

The limewater test tube remains open to the atmosphere, so gas can escape rather than building up pressure.

Before adding the acid, predict what you would see if the carbonate really produces carbon dioxide. You should expect bubbles in the reaction flask and a change in the limewater.

Observation is not the same as particle-level evidence

The first design could simply involve watching acid fizz on marble. Its weakness is that bubbles alone don’t identify the gas.

Passing the gas through limewater provides an additional test.

Carbon dioxide reacts with limewater:

\[ \begin{aligned} &\ce{CO2(g) + Ca(OH)2(aq)}\\ &\ce{-> CaCO3(s) + H2O(l)} \end{aligned} \]

The cloudiness comes from tiny particles of solid calcium carbonate forming in the limewater.

Notice the important limitation: even this improved experiment doesn’t directly show that calcium ions are dissolved in the original reaction flask, or that chloride ions have remained unchanged.

Those are conclusions we make using chemical formulas, solubility, and conservation of atoms and charge.

A small calculation from the gas measurement

Suppose the limewater apparatus is replaced by a gas syringe.

An illustrative experiment collects 49.6 mL of carbon dioxide from calcium carbonate reacting with excess hydrochloric acid.

Assume:

• The temperature is \(25^\circ\mathrm{C}\) and pressure is \(100\ \mathrm{kPa}\).
• The molar gas volume under these conditions is \(24.8\ \mathrm{L\,mol^{-1}}\).
• All carbon dioxide is collected as dry gas, with no leaks or dissolution losses.
• Calcium carbonate is the only substance producing carbon dioxide, and it reacts completely.

The carbonate reaction produces one mole of carbon dioxide for every mole of calcium carbonate consumed.

Convert the volume:

\[ 49.6\ \mathrm{mL}=0.0496\ \mathrm{L} \]

Using \(n=V/V_m\), where \(n\) is the amount of gas in moles, \(V\) is its volume, and \(V_m\) is molar gas volume:

\[ \begin{aligned} n(\ce{CO2}) &=\frac{0.0496}{24.8}\\ &=0.00200\ \mathrm{mol} \end{aligned} \]

Therefore:

\[ n(\ce{CaCO3})=0.00200\ \mathrm{mol} \]

With the molar mass of calcium carbonate, \(M=100.1\ \mathrm{g\,mol^{-1}}\):

\[ \begin{aligned} m&=nM\\ &=(0.00200)(100.1)\\ &=0.200\ \mathrm{g} \end{aligned} \]

The gas collected corresponds to approximately \(0.200\ \mathrm{g}\) of calcium carbonate reacting.

In a real experiment, some carbon dioxide can dissolve in the liquid or escape before collection. The calculation therefore depends on the stated assumptions; a measured gas volume is not automatically the total amount produced.

02Which particles belong in the ionic equation?

Now move from what the experiment shows to what’s happening among the particles.

Calcium carbonate is an ionic solid. It contains calcium ions, \(\mathrm{Ca}^{2+}\), and carbonate ions, \(\mathrm{CO_{3}}^{2-}\), arranged in a crystal lattice.

When hydrochloric acid reaches the marble surface, hydrogen ions react with carbonate. Carbon dioxide and water form, while calcium ions can enter the surrounding solution.

Chloride ions remain dissolved without taking part in this chemical change.

This gives us the main rule for constructing ionic equations: write substances according to the particles present in the stated physical conditions, not simply according to whether their formulas contain ions.

Use this table when deciding what to separate.

For example, sodium carbonate dissolved in water is represented as:

\[ \ce{Na2CO3(aq) -> 2Na+(aq) + CO3^2-(aq)} \]

But a marble chip remains:

\[ \ce{CaCO3(s)} \]

You don’t write it as separate aqueous ions on the reactant side.

One further detail: \(\ce{H+(aq)}\) is the standard shorthand for a hydrated proton in water. It does not mean bare hydrogen ions float around independently of water molecules.

03The contrasting case: What if the carbonate is already dissolved?

Now imagine two beakers.

The first contains solid calcium carbonate chips. The second contains a clear solution of sodium carbonate.

You add excess dilute hydrochloric acid to both and let both reactions go to completion.

Would the two net ionic equations be identical?

No. Both reactions produce carbon dioxide, but the starting carbonate is represented differently.

For dissolved sodium carbonate:

\[ \begin{aligned} &\ce{Na2CO3(aq) + 2HCl(aq)}\\ &\ce{-> 2NaCl(aq) + CO2(g)}\\ &\quad\ce{+ H2O(l)} \end{aligned} \]

Sodium carbonate has already separated into sodium and carbonate ions.

Hydrochloric acid supplies hydrogen and chloride ions. Sodium and chloride appear unchanged on both sides, so both are spectators.

That leaves:

\[ \begin{aligned} &\ce{CO3^2-(aq) + 2H+(aq)}\\ &\ce{-> CO2(g) + H2O(l)} \end{aligned} \]

Compare the two cases:

The underlying acid-carbonate chemistry is the same: two hydrogen ions react per carbonate unit to produce one carbon dioxide molecule and one water molecule.

But a net ionic equation describes the actual starting and finishing species. That is why state symbols change the equation.

If constructing the balanced molecular equation is still difficult, the separate guide to predicting acid-carbonate products and balancing equations develops that earlier step.

04A second trap: An aqueous acid isn’t always completely ionised

Hydrochloric acid is a strong acid. In dilute aqueous solution, it is effectively completely ionised.

Ethanoic acid, the acid found in vinegar, is different.

Its formula is \(\ce{CH3COOH}\). It is a weak acid, meaning only a fraction of its molecules are ionised at any moment.

The ionisation equilibrium is:

\[ \begin{aligned} &\ce{CH3COOH(aq)}\\ &\ce{<=> H+(aq) + CH3COO-(aq)} \end{aligned} \]

Here’s the tempting mistake: seeing \((aq)\) and immediately splitting ethanoic acid into hydrogen and ethanoate ions.

That would incorrectly treat the weak acid as completely ionised.

In a standard complete ionic equation, we keep weak acids in their molecular form. This is a representation of their predominant form in solution, not a claim that absolutely no ions exist.

05How to decide whether an ion is really a spectator

For a straightforward reaction between a soluble carbonate and a strong acid, the metal ion from the carbonate and the negative ion from the acid usually cancel.

But don’t memorise that as an unconditional rule.

An ion is a spectator only if it appears unchanged, in the same state and amount, on both sides of the complete ionic equation.

For example, if the acid’s negative ion combines with a metal ion to form an insoluble product, that ion has participated in another chemical change. You cannot cancel it.

This means you must consider all the stated products and their solubilities, especially if precipitation accompanies the acid-carbonate reaction.

Before finalising an answer, use three checks:

• Species check: Did you keep solids and weak acids together, while splitting strong aqueous electrolytes?
• Spectator check: Did you cancel only ions that appear identically on both sides?
• Conservation check: Do both the atom counts and total electrical charges balance?

The charge check is particularly useful. A net ionic equation can have different charges on its two sides, but their totals must be equal.

06Questions and solutions

Question 1

Solid magnesium carbonate is added to excess dilute nitric acid.

Write the balanced net ionic equation, identify the spectator ion, and predict one visible observation.

Solution 1

The net ionic equation is:

\[ \begin{aligned} &\ce{MgCO3(s) + 2H+(aq)}\\ &\ce{-> Mg^2+(aq) + CO2(g)}\\ &\quad\ce{+ H2O(l)} \end{aligned} \]

The spectator ion is \(\ce{NO3-(aq)}\), and the expected visible observation is effervescence as gas forms.

First write the balanced molecular equation:

\[ \begin{aligned} &\ce{MgCO3(s) + 2HNO3(aq)}\\ &\ce{-> Mg(NO3)2(aq)}\\ &\quad\ce{+ CO2(g) + H2O(l)} \end{aligned} \]

Nitric acid is strong, and magnesium nitrate is soluble.

Separating both into their aqueous ions produces two nitrate ions on each side. They cancel.

Magnesium carbonate remains written as a solid on the left, but its magnesium appears as \(\ce{Mg^2+(aq)}\) on the right.

The tempting mistake is to split solid magnesium carbonate into aqueous ions before the reaction. That would incorrectly describe the starting substance as already dissolved.

The gas is predicted to be carbon dioxide from the reaction chemistry; bubbles alone would not identify it experimentally.

Question 2

An unknown clear solution is prepared by dissolving a mixture containing only sodium carbonate and sodium hydrogencarbonate.

The solution is reacted completely with excess dilute nitric acid.

The following measurements are illustrative and idealised:

Assume that all produced carbon dioxide is accounted for, all carbonate species are converted into carbon dioxide, and no other substances consume acid.

Determine the original amount, in moles, of each compound.

Write both net ionic equations and explain why measuring carbon dioxide alone would not determine the mixture’s composition.

Solution 2

The original solution contained \(0.0040\ \mathrm{mol}\) of sodium carbonate and \(0.0040\ \mathrm{mol}\) of sodium hydrogencarbonate.

The key is that both species produce one mole of carbon dioxide per mole of starting species, but they consume different amounts of hydrogen ions.

For carbonate:

\[ \begin{aligned} &\ce{CO3^2-(aq) + 2H+(aq)}\\ &\ce{-> CO2(g) + H2O(l)} \end{aligned} \]

For hydrogencarbonate:

\[ \begin{aligned} &\ce{HCO3-(aq) + H+(aq)}\\ &\ce{-> CO2(g) + H2O(l)} \end{aligned} \]

Let \(x\) be the initial amount of sodium carbonate, in moles, and \(y\) the initial amount of sodium hydrogencarbonate, in moles.

Each produces one mole of carbon dioxide per mole of compound, so:

\[ x+y=0.0080\ \mathrm{mol} \]

Carbonate consumes two hydrogen ions per formula unit. Hydrogencarbonate consumes one.

Therefore:

\[ 2x+y=0.0120\ \mathrm{mol} \]

Subtracting the first equation from the second:

\[ \begin{aligned} x&=0.0120-0.0080\\ &=0.0040\ \mathrm{mol} \end{aligned} \]

Then:

\[ \begin{aligned} y&=0.0080-0.0040\\ &=0.0040\ \mathrm{mol} \end{aligned} \]

We can verify the hydrogen ion consumption:

\[ \begin{aligned} n(\mathrm{H}^{+})&=2x+y\\ &=2(0.0040)+0.0040\\ &=0.0120\ \mathrm{mol} \end{aligned} \]

This agrees with the measurement.

A tempting approach is to divide the hydrogen ion consumption by two and assume the sample is entirely carbonate. That ignores the hydrogencarbonate, which also consumes acid.

The carbon dioxide measurement alone provides only \(x+y\). The acid consumption provides the second independent relationship needed to distinguish the two starting species.

Question 3

Two identical samples each contain \(0.00500\ \mathrm{mol}\) of solid calcium carbonate.

One reacts with excess hydrochloric acid. The other reacts with excess aqueous ethanoic acid.

Both reactions are allowed to finish completely, and all carbon dioxide produced is collected under identical temperature and pressure conditions.

A student observes equal final gas volumes and concludes:

Both net ionic equations must be identical. Both acids provide hydrogen ions, so the acid’s negative ion is always a spectator.

Evaluate this conclusion.

Write both net ionic equations, calculate the amount of gas produced by each sample, and explain whether identical final gas volumes demonstrate identical acid strengths or reaction rates.

Solution 3

The student is incorrect about the net ionic equations. Both samples produce \(0.00500\ \mathrm{mol}\) of carbon dioxide, but the equations differ because hydrochloric acid is strong and ethanoic acid is weak.

For hydrochloric acid:

\[ \begin{aligned} &\ce{CaCO3(s) + 2H+(aq)}\\ &\ce{-> Ca^2+(aq) + CO2(g)}\\ &\quad\ce{+ H2O(l)} \end{aligned} \]

Hydrochloric acid ionises effectively completely. Chloride ions remain aqueous on both sides and cancel.

For ethanoic acid:

\[ \begin{aligned} &\ce{CaCO3(s)}\\ &\quad\ce{+ 2CH3COOH(aq)}\\ &\ce{-> Ca^2+(aq)}\\ &\quad\ce{+ 2CH3COO-(aq)}\\ &\quad\ce{+ CO2(g) + H2O(l)} \end{aligned} \]

Ethanoic acid is retained as molecules on the left. Ethanoate ions appear on the right after the acid has reacted, so they cannot be cancelled.

For either completed reaction:

\[ n(\ce{CO2})=n(\ce{CaCO3}) \]

Therefore:

\[ n(\ce{CO2})=0.00500\ \mathrm{mol} \]

Each sample produces the same amount of carbon dioxide. At identical temperature and pressure, equal amounts of gas occupy equal volumes under the ideal gas model.

But this does not mean the two acids have the same strength or that the reactions proceed at the same rate.

Acid strength concerns the extent of ionisation in water. Reaction rate depends on conditions such as acid concentration, temperature, and exposed carbonate surface area.

The stated measurements describe the final amount of gas, not the rate at which it was produced.

The deeper mistake is assuming that the same final gas yield proves the same reacting species throughout the process. An endpoint measurement cannot establish that.

Question 4

Solid calcium carbonate is reacted with dilute sulfuric acid.

For this question, assume sulfuric acid supplies two available hydrogen ions and one sulfate ion per formula unit, and that calcium sulfate is sufficiently insoluble to precipitate under the stated conditions.

A student writes the usual calcium carbonate-hydrochloric acid net ionic equation and argues that sulfate must be a spectator because it comes from the acid.

Determine the correct net ionic equation.

Explain why sulfate cannot be cancelled, what solid material might remain after acid addition, and whether the net ionic equation alone guarantees complete consumption of the calcium carbonate within a fixed time.

Solution 4

Sulfate is not a spectator under the stated conditions because it becomes part of solid calcium sulfate. The correct net ionic equation is:

\[ \begin{aligned} &\ce{CaCO3(s) + 2H+(aq)}\\ &\quad\ce{+ SO4^2-(aq)}\\ &\ce{-> CaSO4(s) + CO2(g)}\\ &\quad\ce{+ H2O(l)} \end{aligned} \]

Start with the balanced molecular equation:

\[ \begin{aligned} &\ce{CaCO3(s) + H2SO4(aq)}\\ &\ce{-> CaSO4(s) + CO2(g)}\\ &\quad\ce{+ H2O(l)} \end{aligned} \]

Under the question’s assumptions, calcium sulfate is a solid product.

That changes the ionic bookkeeping. The sulfate ion begins dissolved but finishes incorporated into a solid. It does not appear unchanged on both sides.

We can check the net ionic equation.

On the left, the charge is:

\[ 2(+1)+(-2)=0 \]

Every product is electrically neutral, so the right-hand charge is also zero.

The atoms balance: one calcium, one carbon, one sulfur, seven oxygen, and two hydrogen atoms occur on each side.

What about the solid remaining in the flask?

Some of it may be calcium sulfate produced by the reaction, rather than unreacted calcium carbonate. A calcium sulfate layer can also restrict fresh acid from reaching the underlying carbonate.

The equation tells us the stoichiometric chemical change. It doesn’t tell us how rapidly acid reaches the surface or whether a solid coating prevents further reaction over a particular time.

Therefore, without information about reaction conditions, time, and extent, we cannot conclude that all the calcium carbonate disappears.

The student made two linked errors: assuming the acid’s negative ion must be a spectator, and using a balanced equation to infer complete reaction without considering the physical conditions.

Question 5

Two experiments use exactly the same initial amounts of reagents:

• \(0.0100\ \mathrm{mol}\) of dissolved sodium carbonate
• \(0.0100\ \mathrm{mol}\) of dissolved calcium chloride
• \(0.0200\ \mathrm{mol}\) of dilute nitric acid

The experiments differ only in the order of mixing.

Experiment A

First, sodium carbonate solution is mixed with calcium chloride solution. Then nitric acid is added.

Experiment B

First, sodium carbonate solution is reacted completely with nitric acid, and the carbon dioxide is allowed to escape. Then calcium chloride solution is added.

Assume calcium carbonate precipitates essentially completely when calcium and carbonate ions meet before acid is introduced. Assume subsequent acid reaction is complete, carbon dioxide escapes, and all sodium, calcium nitrate, and chloride salts remain dissolved in the final solution.

A student makes two claims:

1. If both experiments have the same overall net ionic equation, they must show the same observations during mixing.

2. If calcium ions disappear when we simplify the overall equation, they cannot have participated in either experiment.

For each experiment, determine whether a precipitate forms, write the net ionic equations for the chemical changes, and calculate the total amount of carbon dioxide produced.

Then derive the overall net ionic equation and evaluate both claims.

Finally, explain whether the overall net ionic equation and final dissolved-ion composition are sufficient to reconstruct the order of mixing.

Solution 5

Both experiments produce \(0.0100\ \mathrm{mol}\) of carbon dioxide and have the same overall net ionic equation. However, Experiment A forms a calcium carbonate precipitate that subsequently dissolves, while Experiment B does not form that precipitate.

Both student claims are false.

The crucial distinction is between the overall chemical change and the intermediate changes that occur along the way.

Experiment A: Precipitation followed by acid reaction

When sodium carbonate and calcium chloride solutions mix, calcium and carbonate ions combine.

The first net ionic equation is:

\[ \begin{aligned} &\ce{Ca^2+(aq) + CO3^2-(aq)}\\ &\ce{-> CaCO3(s)} \end{aligned} \]

The amount of calcium carbonate precipitated is:

\[ n(\ce{CaCO3})=0.0100\ \mathrm{mol} \]

Now add \(0.0200\ \mathrm{mol}\) of nitric acid.

Nitric acid is strong, so it supplies \(0.0200\ \mathrm{mol}\) of hydrogen ions.

Those hydrogen ions react with the solid calcium carbonate:

\[ \begin{aligned} &\ce{CaCO3(s) + 2H+(aq)}\\ &\ce{-> Ca^2+(aq) + CO2(g)}\\ &\quad\ce{+ H2O(l)} \end{aligned} \]

The acid needed is:

\[ \begin{aligned} n(\mathrm{H}^{+}) &=2n(\ce{CaCO3})\\ &=2(0.0100)\\ &=0.0200\ \mathrm{mol} \end{aligned} \]

Exactly the required amount has been supplied.

Under the stated complete-reaction assumption, the precipitate dissolves, calcium returns to the aqueous phase, and:

\[ n(\ce{CO2})=0.0100\ \mathrm{mol} \]

Experiment A therefore has two observable stages: a solid forms, then the solid reacts with acid as gas is produced.

Experiment B: Acid reaction before calcium is introduced

Here, nitric acid reacts directly with dissolved carbonate ions.

The net ionic equation is:

\[ \begin{aligned} &\ce{CO3^2-(aq) + 2H+(aq)}\\ &\ce{-> CO2(g) + H2O(l)} \end{aligned} \]

The original sodium carbonate provides \(0.0100\ \mathrm{mol}\) of carbonate ions.

The acid provides \(0.0200\ \mathrm{mol}\) of hydrogen ions, exactly enough for complete conversion.

Therefore:

\[ n(\ce{CO2})=0.0100\ \mathrm{mol} \]

After this reaction finishes and carbon dioxide escapes, no carbonate remains under the question’s assumptions.

Adding calcium chloride now introduces calcium and chloride ions, but there is no remaining carbonate available to form calcium carbonate.

So Experiment B produces gas in the first stage but no calcium carbonate precipitate in the second stage.

Why the overall equations are identical

Return to Experiment A.

Its two net ionic steps are:

\[ \begin{aligned} &\ce{Ca^2+(aq) + CO3^2-(aq)}\\ &\ce{-> CaCO3(s)} \end{aligned} \]

and:

\[ \begin{aligned} &\ce{CaCO3(s) + 2H+(aq)}\\ &\ce{-> Ca^2+(aq) + CO2(g)}\\ &\quad\ce{+ H2O(l)} \end{aligned} \]

When we add the two chemical equations, the calcium carbonate produced in the first reaction is consumed in the second.

The aqueous calcium ions also cancel because they are consumed initially and regenerated later.

What remains is:

\[ \begin{aligned} &\ce{CO3^2-(aq) + 2H+(aq)}\\ &\ce{-> CO2(g) + H2O(l)} \end{aligned} \]

That is exactly the net ionic equation for Experiment B.

Both equations describe the same change between the initial and final states, but not the same reaction pathway.

Evaluating the two claims

The first claim is false because identical overall net ionic equations don’t require identical observations during the process.

Experiment A temporarily contains a calcium carbonate precipitate. Experiment B does not.

The second claim is false because calcium ions genuinely participate in Experiment A. They form the precipitate and are released again during acid reaction.

Their cancellation from the combined overall equation does not mean they were chemically inactive throughout.

This is the subtle limitation of a net ionic equation: cancellation removes species with no net change over the chosen sequence. It does not establish that those species never participated in an intermediate reaction.

Finally, the overall net ionic equations and final dissolved-ion compositions are insufficient to reconstruct the order of mixing.

Both experiments finish with the same amounts of dissolved sodium, calcium, chloride, and nitrate ions, along with the same amount of carbon dioxide produced and water formed.

Without observations or measurements taken during mixing, the transient precipitate in Experiment A would not be revealed by the overall net ionic equation.

This is why writing a net ionic equation and explaining an experimental observation are related but different jobs. The equation identifies the net chemical change. To understand the pathway, you must also consider which solids, dissolved ions, and gases can exist at each stage.

That reasoning leads directly into precipitation reactions and net ionic equations, where solubility determines which ions remain dissolved and which become part of a new solid.