How to Write State Symbols in Precipitation Equations

Two clear solutions are mixed, and a white solid appears. A student writes \(\ce{AgCl(aq)}\) in the equation because the reaction happened in water. Can that state symbol be correct? No. If silver chloride forms a solid precipitate, it must be written \(\ce{AgCl(s)}\), even while the solid is suspended in water.

To write state symbols in a precipitation equation, first convert the chemical names into correct formulas. Then use solubility rules to identify the insoluble product, label it \((s)\), label the dissolved substances \((aq)\), and balance the equation.

The important skill is recognising what is actually present in the mixture, rather than giving everything the same state symbol because it happens to be in one beaker.

01What do the state symbols actually tell you?

Imagine stirring salt into a glass of water, then adding a little fine chalk powder.

The dissolved salt spreads throughout the water. The chalk remains as tiny solid particles, even if they’re suspended and take a while to settle.

Both substances are inside a liquid mixture, but they aren’t in the same physical state.

The chalk comparison helps distinguish dissolved material from suspended solid particles. It has one important limitation: a precipitation reaction produces new solid particles from ions already dissolved in water. We aren’t simply adding an existing powder.

Why doesn’t \((aq)\) mean the same thing as \((l)\)?

Suppose a test tube contains sodium nitrate dissolved in water.

The water is \(\ce{H2O(l)}\). The dissolved sodium nitrate is represented as \(\ce{NaNO3(aq)}\).

If you wrote \(\ce{NaNO3(l)}\), you would be describing liquid sodium nitrate itself, not sodium nitrate dissolved in water.

This is one of the easiest state-symbol mistakes to avoid.

What is inside an aqueous ionic solution?

Aqueous sodium chloride isn’t made up mainly of intact \(\ce{NaCl}\) pairs floating through water.

Instead, dissolved sodium chloride is present as separated, hydrated ions:

\[
\ce{NaCl(aq) -> Na+(aq) + Cl-(aq)}
\]

Water molecules surround the ions and help keep them dispersed.

When an insoluble ionic compound forms, some ions become part of a solid crystal lattice. The other dissolved ions remain in the water.

A state symbol therefore describes the state of the particular substance being written, not the overall appearance of the beaker.

02How an experiment reveals the correct state symbols

Consider a simple precipitation investigation using silver nitrate and sodium chloride solutions.

The question is whether mixing the two solutions produces a new solid or simply leaves all the ions dissolved.

The original arrangement

Imagine this apparatus:

1. A clean test tube contains aqueous sodium chloride, \(\ce{NaCl(aq)}\).
2. A dropper contains aqueous silver nitrate, \(\ce{AgNO3(aq)}\).
3. Silver nitrate is added to the sodium chloride and the mixture is gently swirled.
4. The mixture is poured through filter paper inside a funnel.
5. A conical flask underneath collects the liquid passing through the filter.

Before mixing, both solutions are clear.

If every ion remained dissolved, what should happen when the mixture is filtered?

The dissolved ions would pass through with the water. There would be no newly formed solid to collect.

But ordinary classroom concentrations of these solutions are expected to produce a white precipitate. The white solid can be retained by the filter paper.

The observation is that a white solid forms.

The chemical identity of that solid is an inference, made using the known reactants and their solubilities.

Why the first observation isn’t enough

A white precipitate doesn’t automatically prove that silver chloride has formed. Many different ionic compounds can produce white solids.

There could also be contamination in the equipment.

A useful revision is to repeat the procedure with a control mixture. Replace silver nitrate with sodium nitrate, keeping the sodium chloride solution, mixing method, and approximate concentrations the same.

This changes the metal ion from silver to sodium while retaining nitrate.

These are illustrative expected observations for ordinary aqueous concentrations, not measurements from an experiment performed for this article.

The control challenges the idea that mixing any two salt solutions must produce a precipitate. Repeating both mixtures with clean apparatus also helps check for contamination.

The observations support the precipitation model, but neither cloudiness nor filtration alone establishes the product’s exact chemical formula.

From the observation to the equation

Silver nitrate contains \(\mathrm{Ag}^{+}\) and \(\ce{NO3-}\) ions. Sodium chloride contains \(\mathrm{Na}^{+}\) and \(\mathrm{Cl}^{-}\) ions.

Silver chloride is poorly soluble in water, so it forms the precipitate:

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

Sodium and nitrate ions remain dissolved.

The complete balanced equation is:

\[
\begin{aligned}
&\ce{AgNO3(aq) + NaCl(aq)}\\
&\quad\ce{-> AgCl(s) + NaNO3(aq)}
\end{aligned}
\]

Notice that the products have different state symbols.

• \(\ce{AgCl(s)}\) is the insoluble solid formed during mixing.
• \(\ce{NaNO3(aq)}\) represents soluble sodium nitrate remaining in the aqueous mixture.

The white solid is directly observable. The presence of individual dissolved sodium and nitrate ions is inferred from the reactants and the chemical model.

03How to go from chemical names to state symbols

Most state-symbol questions become easier when you separate three jobs: writing the formulas, predicting solubility, and balancing the equation.

Step 1: Translate the names into formulas

Start with the charges of the ions.

For example:

The charges must balance to zero in a neutral ionic compound.

For magnesium nitrate, one magnesium ion has a charge of \(2+\). Each nitrate ion has a charge of \(1-\), so two nitrate ions are needed.

That’s why the formula is \(\ce{Mg(NO3)2}\), not \(\ce{MgNO3}\).

Parentheses show that the entire nitrate ion is repeated.

Step 2: Identify which substances are dissolved

If a question says aqueous silver nitrate, the reactant is \(\ce{AgNO3(aq)}\).

If it says solid silver nitrate, the reactant starts as \(\ce{AgNO3(s)}\).

Don’t use \((aq)\) simply because a substance is soluble. A dry crystal is still a solid until it dissolves.

For products, solubility rules help you decide whether the newly formed compound remains dissolved or precipitates.

The most useful patterns for these questions are:

These are useful approximations, not statements that a solid has exactly zero solubility. Temperature and concentration can affect whether precipitation occurs. Use any specific solubility information supplied in an exam question.

Step 3: Assign the products’ state symbols before balancing

Suppose a soluble calcium salt reacts with sodium carbonate.

The calcium ion and carbonate ion can form poorly soluble calcium carbonate.

That product is written \(\ce{CaCO3(s)}\).

Sodium salts remain soluble, so the accompanying sodium compound receives \((aq)\).

Once you’ve chosen the correct formulas and state symbols, balance the equation using coefficients. Don’t change a subscript to make the atom counts work.

04What a precipitate’s mass can tell you

A state-symbol decision can also be checked against a quantitative prediction.

If you know the reactant concentrations and volumes, you can calculate the amount of precipitate expected from the balanced equation.

This provides another connection between an inferred chemical formula and something measurable.

05A contrasting case: when no precipitate forms

What if you mixed sodium nitrate solution with potassium chloride solution instead?

Both are soluble salts. Their ions remain dissolved, and the other possible combinations, sodium chloride and potassium nitrate, are also soluble.

So, under ordinary dilute conditions, no precipitate forms.

It would be wrong to write:

\[
\ce{NaCl(s) + KNO3(s)}
\]

as products of the mixing process.

Both substances are soluble in water, and there is no evidence that either forms a solid on mixing. There is no net ionic precipitation equation for this mixture.

But what if crystals appear after evaporation?

Suppose the sodium nitrate and potassium chloride mixture remains clear for several minutes.

You then heat it, allowing water to evaporate, and eventually see white crystals.

Does that prove a precipitation reaction occurred when the solutions were first mixed?

No. The water was removed, so dissolved salts could crystallise later.

The timing and method matter:

A common mistake is to look at the final solid and work backwards, assigning \((s)\) to substances at an earlier time.

State symbols belong to the particular stage of the process described by the equation.

This also explains why writing a precipitation equation is different from simply predicting that a salt can eventually crystallise.

For the broader method of identifying products and removing spectator ions, see Precipitation Reactions and Net Ionic Equations for HSC Chemistry.

06Questions and solutions

Assume ordinary aqueous conditions at room temperature unless otherwise specified. Use the usual solubility rules, and neglect minor dissolution of precipitates where a question asks for complete precipitation.

Question 1

A student mixes aqueous zinc nitrate with aqueous potassium carbonate.

Write the balanced chemical equation, including state symbols. Name the precipitate and explain which part of the result could be observed directly.

Solution 1

Zinc carbonate, \(\ce{ZnCO3(s)}\), is the precipitate, and potassium nitrate remains aqueous.

Zinc nitrate contains \(\mathrm{Zn}^{2+}\) and \(\ce{NO3-}\), giving \(\ce{Zn(NO3)2}\).

Potassium carbonate contains \(\mathrm{K}^{+}\) and \(\mathrm{CO_{3}}^{2-}\), giving \(\ce{K2CO3}\).

The products are zinc carbonate and potassium nitrate.

The balanced equation is:

\[
\begin{aligned}
&\ce{Zn(NO3)2(aq) + K2CO3(aq)}\\
&\quad\ce{-> ZnCO3(s) + 2KNO3(aq)}
\end{aligned}
\]

Zinc carbonate is poorly soluble, while potassium nitrate is soluble.

The student could directly observe a white solid forming. Its identity as zinc carbonate is inferred from the known reactants and solubility rules.

The tempting mistake is to label potassium nitrate \((s)\) simply because a solid is observed somewhere in the mixture. Only the insoluble product receives \((s)\).

Question 2

A student mixes 20.0 mL of 0.0500 mol L\(^{-1}\) barium chloride with 30.0 mL of 0.0200 mol L\(^{-1}\) sodium sulfate.

The white precipitate is filtered out. The clear filtrate is then mixed with another 10.0 mL of 0.0500 mol L\(^{-1}\) sodium sulfate, producing more white solid.

Assume essentially complete precipitation and use \(M(\ce{BaSO4})=233.4\ \mathrm{g\,mol^{-1}}\).

Determine:

• The balanced precipitation equation with state symbols.
• The theoretical mass of solid formed during the first mixing.
• The additional mass of solid formed after the second sulfate addition.
• Why the clear filtrate could still produce another precipitate.

Solution 2

Barium sulfate, \(\ce{BaSO4(s)}\), forms in both stages. The first mixing produces approximately 0.140 g, and the second produces another 0.0934 g because barium ions remain dissolved after the first filtration.

The balanced equation is:

\[
\begin{aligned}
&\ce{BaCl2(aq) + Na2SO4(aq)}\\
&\quad\ce{-> BaSO4(s) + 2NaCl(aq)}
\end{aligned}
\]

First mixing

Convert volumes to litres and use \(n=cV\).

\[
\begin{aligned}
n_{\mathrm{Ba}}&=(0.0500)(0.0200)\\
&=0.00100\ \mathrm{mol}\\
n_{\mathrm{sulfate}}&=(0.0200)(0.0300)\\
&=0.000600\ \mathrm{mol}
\end{aligned}
\]

Barium and sulfate react in a 1:1 mole ratio.

Sulfate is limiting, so 0.000600 mol of barium sulfate forms.

\[
\begin{aligned}
m_1&=(0.000600)(233.4)\\
&=0.14004\ \mathrm{g}\\
&\approx0.140\ \mathrm{g}
\end{aligned}
\]

The symbol \(m_1\) represents the theoretical mass collected from the first mixing.

The amount of barium remaining dissolved is:

\[
\begin{aligned}
n_{\mathrm{Ba,left}}&=0.00100-0.000600\\
&=0.000400\ \mathrm{mol}
\end{aligned}
\]

Second sulfate addition

The added sulfate provides:

\[
\begin{aligned}
n_{\mathrm{added}}&=(0.0500)(0.0100)\\
&=0.000500\ \mathrm{mol}
\end{aligned}
\]

Only 0.000400 mol of barium remains available to react.

Therefore, barium is now limiting, producing another 0.000400 mol of solid.

\[
\begin{aligned}
m_2&=(0.000400)(233.4)\\
&=0.09336\ \mathrm{g}\\
&\approx0.0934\ \mathrm{g}
\end{aligned}
\]

The symbol \(m_2\) represents the additional theoretical precipitate mass.

The critical point is that filtration removes solid barium sulfate, not every dissolved barium ion. The original mixture ran out of sulfate before all the barium could precipitate.

A clear filtrate can still contain reactive ions. Treating clear liquid as pure water would give the wrong prediction for the second stage.

Question 3

Two mixtures are prepared from initially clear aqueous solutions.

Flask A: Silver nitrate is mixed with sodium chloride. A white solid appears immediately. When the mixture is filtered, white material stays on the filter paper.

Flask B: Sodium nitrate is mixed with potassium chloride. The mixture stays clear, and immediate filtration leaves no visible solid. However, white crystals appear after the water is evaporated.

A student makes three claims:

1. Silver chloride should be labelled \((aq)\) because Flask A still contains liquid.
2. The crystals in Flask B prove that a precipitation reaction occurred when the solutions were mixed.
3. The white appearance of both solids proves that they have the same chemical composition.

Assess all three claims. Write the balanced precipitation equation for Flask A and explain what chemical equation, if any, is justified for the initial mixing in Flask B.

Solution 3

All three claims are incorrect. Flask A produces solid silver chloride during mixing, while Flask B provides no evidence of precipitation until water is deliberately removed, and the appearance of a white solid does not uniquely identify its composition.

Claim 1

Silver chloride must be written \(\ce{AgCl(s)}\), because the solid is retained by the filter paper.

The balanced equation is:

\[
\begin{aligned}
&\ce{AgNO3(aq) + NaCl(aq)}\\
&\quad\ce{-> AgCl(s) + NaNO3(aq)}
\end{aligned}
\]

Silver chloride is poorly soluble. Sodium nitrate remains dissolved.

The liquid surrounding a precipitate doesn’t determine the precipitate’s state symbol.

Claim 2

Flask B shows no precipitation reaction during the initial mixing.

Sodium nitrate, potassium chloride, sodium chloride, and potassium nitrate are all soluble under the stated conditions.

The ions remain aqueous, so there is no net ionic precipitation equation to write.

Crystals appearing after evaporation result from the later removal of water. That doesn’t establish that any solid existed during the original mixing.

The observation alone also doesn’t identify which salts have crystallised from the mixed solution.

Claim 3

A white solid isn’t chemically distinctive enough to identify a substance.

Both silver chloride and many soluble salts can appear white as solids.

To identify a substance, the student needs further evidence, such as its known reactants, solubility behaviour, or appropriate chemical tests.

The deeper mistake is treating appearance as a complete particle-level explanation. Direct observations support an interpretation, but they don’t automatically reveal every chemical species present.

Question 4

An initially clear aqueous sample contains both silver nitrate and barium nitrate.

A student adds enough sodium chloride solution to precipitate essentially all the silver ions, with some chloride remaining in excess. A white solid forms and is removed by filtration.

Sodium sulfate solution is then added to the clear filtrate. A second white precipitate forms.

Assume ordinary dilute conditions and ignore minor solubility and complex-ion effects.

Identify the two precipitates, and write balanced net ionic equations with state symbols for both stages.

List the main dissolved ions present immediately after the first filtration. Explain why the second precipitate isn’t simply some of the original white solid that passed through the filter.

Solution 4

The first precipitate is silver chloride, \(\ce{AgCl(s)}\), and the second is barium sulfate, \(\ce{BaSO4(s)}\). The second solid forms because barium ions survive the first filtration in aqueous solution.

First stage: silver chloride precipitation

The relevant ions are silver and chloride.

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

Silver chloride is poorly soluble.

Barium chloride, however, is soluble. Adding chloride ions therefore doesn’t remove the barium ions as a solid.

Once the silver chloride has been filtered out, the main dissolved ions are:

• \(\ce{Ba^2+(aq)}\), from the original barium nitrate.
• \(\ce{NO3-(aq)}\), from the original nitrate salts.
• \(\ce{Na+(aq)}\), from the added sodium chloride.
• \(\ce{Cl-(aq)}\), because chloride was added in excess.

These ions are not directly visible, but their presence follows from the known starting materials, addition, and solubility behaviour.

Second stage: barium sulfate precipitation

Adding sodium sulfate supplies sulfate ions.

Barium sulfate is poorly soluble, so the remaining barium ions form another solid:

\[
\ce{Ba^2+(aq) + SO4^2-(aq) -> BaSO4(s)}
\]

The second white solid has a different chemical composition from the first.

The first filtration removes silver chloride particles. It doesn’t remove the dissolved barium ions, because those ions aren’t part of an insoluble compound at that stage.

The tempting prediction is that adding a precipitating reagent removes all metal ions from a solution.

But chloride is selective here: it precipitates silver while leaving barium dissolved. Sulfate then provides a different insoluble combination.

The sequence shows why a clear filtrate can contain ions capable of forming a completely new precipitate.

Question 5

Four unlabelled containers, A, B, C, and D, each hold one of the following clear aqueous solutions:

• Magnesium nitrate
• Sodium carbonate
• Sodium sulfate
• Barium chloride

Each solution is used once. All six possible pairs are mixed separately using clean equipment.

The observations are:

The student then filters and thoroughly washes the precipitate from A + B.

When dilute nitric acid is added to this isolated solid, gas bubbles form and the solid disappears.

Assume the usual solubility behaviour of these simple salts at room temperature, with concentrations sufficient to produce the expected precipitates. The gas formation is due to the reaction of an insoluble carbonate with acid.

Determine the identity of A, B, C, and D.

Write balanced molecular equations with state symbols for the three precipitation reactions and the reaction between the isolated solid and nitric acid.

Finally, explain why the original six mixing observations, without the acid test, do not uniquely identify the four solutions.

Solution 5

The solutions are A = magnesium nitrate, B = sodium carbonate, C = barium chloride, and D = sodium sulfate.

The three precipitates are magnesium carbonate, barium carbonate, and barium sulfate, respectively. The acid test is essential because the six pairwise mixing observations alone allow two different assignments.

Step 1: Identify which pairs of the named salts can precipitate.

From the usual solubility rules:

• Magnesium carbonate, \(\ce{MgCO3}\), is poorly soluble.
• Barium carbonate, \(\ce{BaCO3}\), is poorly soluble.
• Barium sulfate, \(\ce{BaSO4}\), is poorly soluble.

The other pairings of these four solutions produce no insoluble product under the stated conditions.

Therefore, there are exactly three precipitating pairs.

An important pattern follows.

Magnesium nitrate can precipitate with sodium carbonate, but not with sodium sulfate or barium chloride.

Sodium sulfate can precipitate with barium chloride, but not with magnesium nitrate or sodium carbonate.

Sodium carbonate can precipitate with both magnesium nitrate and barium chloride.

Barium chloride can precipitate with both sodium carbonate and sodium sulfate.

Step 2: Use the pattern of observations.

The observed precipitations form a chain:

\[
\mathrm{A-B-C-D}
\]

Each connection represents a pair that gives a precipitate.

A and D each precipitate with only one other solution, so they must be magnesium nitrate and sodium sulfate, in some order.

B and C each precipitate with two other solutions, so they must be sodium carbonate and barium chloride, in some order.

At this stage, two assignments are possible.

Both assignments reproduce all six mixing observations.

This is the central trap: even a complete table of precipitation results doesn’t necessarily identify every unknown.

The pattern is symmetric. Reversing the chain gives the same three positive tests.

Step 3: Use the acid test to distinguish the assignments.

Under Assignment 1, A + B forms magnesium carbonate.

Under Assignment 2, A + B forms barium sulfate.

These solids can both appear white, but their reactions with dilute nitric acid are different.

Magnesium carbonate reacts with acid, releasing carbon dioxide:

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

The solid disappears as soluble magnesium nitrate forms, and carbon dioxide escapes as bubbles.

Barium sulfate doesn’t produce carbon dioxide when treated with dilute nitric acid.

Therefore, the bubbling solid from A + B is magnesium carbonate, not barium sulfate.

Assignment 1 is correct.

Washing the isolated solid was useful because it removed soluble material carried over from the original mixture. Without washing, bubbles might have come from a dissolved carbonate contaminating a different solid.

Step 4: Write the three complete precipitation equations.

For A + B, magnesium nitrate reacts with sodium carbonate:

\[
\begin{aligned}
&\ce{Mg(NO3)2(aq) + Na2CO3(aq)}\\
&\quad\ce{-> MgCO3(s) + 2NaNO3(aq)}
\end{aligned}
\]

For B + C, sodium carbonate reacts with barium chloride:

\[
\begin{aligned}
&\ce{Na2CO3(aq) + BaCl2(aq)}\\
&\quad\ce{-> BaCO3(s) + 2NaCl(aq)}
\end{aligned}
\]

For C + D, barium chloride reacts with sodium sulfate:

\[
\begin{aligned}
&\ce{BaCl2(aq) + Na2SO4(aq)}\\
&\quad\ce{-> BaSO4(s) + 2NaCl(aq)}
\end{aligned}
\]

Every state symbol has a chemical reason.

The three carbonate or sulfate products identified as poorly soluble are solids. The sodium salts remain aqueous because they are soluble.

The nitric acid reaction also includes \(\ce{CO2(g)}\), because a gas is released, and \(\ce{H2O(l)}\), because water is produced.

Step 5: Explain why the acid test was necessary.

The six original observations established which pairs could form solids, but not which of two chemically consistent arrangements was correct.

Both arrangements predict the same pattern:

• Three white precipitates.
• Three mixtures that remain clear.
• Two solutions that each precipitate with one partner.
• Two solutions that each precipitate with two partners.

The observation of a white solid from A + B could represent either a carbonate or a sulfate under those two assignments.

Only the additional reaction with acid distinguishes the two possibilities.

A strong answer must recognise that the original observations were insufficient before using the additional evidence.

Simply choosing an assignment that fits the precipitation table isn’t enough. The other assignment fits just as well.

This is the same principle used in experimental chemistry: a convincing explanation must account for the observations and rule out plausible alternatives.

07What to do when the solubility rules aren’t enough

Most introductory precipitation equations can be handled by writing the formulas correctly and using the usual solubility rules.

But a compound described as insoluble is not literally incapable of dissolving. At very low concentrations, a mixture may contain ions that could form an insoluble salt without producing any precipitate.

That means the full decision can depend on how much of each ion is present, not just the names of the compounds.

The next useful concept is the solubility product, \(K_{sp}\), which helps you decide whether a solid is expected to form at specified concentrations. See Predicting Precipitation with Ksp for HSC Chemistry for that quantitative method.

For state-symbol questions, keep the basic distinction clear: \((aq)\) means dissolved in water, \((s)\) means a solid phase, and the equation must describe the substances at the stage of the process you are actually analysing.