Net Ionic Equation Calculator

Reduce a molecular equation to its net ionic form.

Handles aqueous, solid, liquid and gas states, plus bracketed groups.

Updated October 2, 2026

CreatorFrank Zhao

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What a net ionic equation tells you

Write a reaction as a molecular equation and almost every formula in it is clutter. Sodium chloride does not exist as molecules in water; it exists as sodium ions and chloride ions drifting around. Only some of those ions actually change during the reaction. The rest sit there unchanged before and after, watching it happen without taking part.

A net ionic equation strips that away. It shows only the species that genuinely change, and it conserves both mass and charge. That is why it is the form chemists care about: two reactions that look completely different can produce the same net ionic equation if the same chemistry is happening underneath.

This calculator takes the two sides of a reaction and returns four things: the balanced molecular equation, the complete ionic equation, the list of spectator ions, and the net ionic equation. You can start from an unbalanced equation with missing coefficients, because the first step balances it for you.

You want to find the spectators

The calculator names them explicitly, so you can check your own answer against a list rather than squinting at two long equations and spotting what matches.

Your equation is unbalanced

Start with whatever you were given. The balanced molecular equation is produced first, and everything downstream is built from that.

You are checking someone else's work

Watch the complete ionic equation. It shows exactly which formulas were split into ions and which were left whole, which is where most marking disputes start.

You need the coefficients afterwards

Once the equation is balanced you can go straight to the stoichiometric ratios with the molar ratio calculator.

How to use the calculator

1Type the reactant side into the left box and the product side into the right box, separating compounds with +. Spacing does not matter.
2Add a state symbol where you know it: (aq), (s), (l) or (g). If you leave it off, the compound is treated as dissolved in water.
3Do not bother balancing. Coefficients are optional and the calculator supplies them.
4Read the net ionic equation at the bottom, and the spectator list above it to see what was removed and why.

If you would rather start from something known, the five built-in example reactions fill both boxes for you. Picking one and then editing a character is the fastest way to see how the output reacts to a change.

A worked example

Silver nitrate meets sodium chloride. Both sides are already balanced here, which is not a requirement:

\mathrm{AgNO_3(aq)}

+

\mathrm{NaCl(aq)}

\longrightarrow

\mathrm{AgCl(s)}

+

\mathrm{NaNO_3(aq)}

Two of the four compounds dissolve completely, so each one is split into its ions. Silver chloride is a solid precipitate and stays whole:

\mathrm{Ag^{+}(aq)}

+

\mathrm{NO_3^{-}(aq)}

+

\mathrm{Na^{+}(aq)}

+

\mathrm{Cl^{-}(aq)}

\longrightarrow

\mathrm{AgCl(s)}

+

\mathrm{Na^{+}(aq)}

+

\mathrm{NO_3^{-}(aq)}

Nitrate and sodium appear on both sides in the same amount, so they are spectators and get cancelled. What is left is the chemical change itself:

\mathrm{Ag^{+}(aq)}

+

\mathrm{Cl^{-}(aq)}

\longrightarrow

\mathrm{AgCl(s)}

Check it two ways. Atoms: one silver and one chlorine on each side. Charge: +1 and -1 on the left, 0 inside a neutral solid on the right. Both balance, so this is a correct net ionic equation.

The rules behind every answer

Every result comes from the same three steps, applied in order. Once you can do them by hand, the calculator is just checking your work.

Balance it

Count every element on both sides and choose the smallest whole-number coefficients that make each count match. The calculator does this with exact rational arithmetic, so a coefficient is never rounded into the wrong ratio.

Split what dissolves

Every compound marked (aq) that is a strong electrolyte is broken into its ions. Everything else is left exactly as written.

Cancel the bystanders

Any ion with the same charge on both sides is removed in equal amounts. Those are the spectator ions.

Which compounds get split

This is the decision that matters, and it comes down to one question: is the substance actually dissolved and does it come apart completely? Only (aq) substances split. Solids, liquids and gases never do, because their particles are locked together and are not moving freely in solution.

Anion	Salt is soluble with	Insoluble exceptions
Nitrate, NO3-	Always soluble	none
Chloride, bromide, iodide	Soluble	Ag+, Pb2+, Hg2+
Sulfate, SO42-	Soluble	Ag+, Pb2+, Ca2+, Sr2+, Ba2+, Hg2+
Hydroxide, OH-	Group 1, NH4+, Ca2+, Sr2+, Ba2+	everything else
Carbonate, phosphate, chromate	Group 1, NH4+, Sr2+, Ba2+	everything else

Any salt built from a group 1 cation or ammonium dissolves completely, whatever the anion. Those are the standard solubility guidelines; the same table is set out in OpenStax Chemistry 2e, section 4.2, which also lists the strong acids and bases used below.

Strong acids and bases

Not everything that dissolves dissociates completely. Strong acids and strong bases do, which is why they appear as loose ions in the complete equation. Their weak counterparts stay as intact molecules.

Split into ions

Strong acids: HCl, HBr, HI, HNO3, HClO4, H2SO4
Strong bases: NaOH, KOH, Ca(OH)2, Sr(OH)2, Ba(OH)2
Every soluble salt, such as NaCl, KNO3, AgNO3

Left whole on purpose

Insoluble salts: AgCl, BaSO4, PbI2, CaCO3
Insoluble hydroxides: Mg(OH)2, Fe(OH)3, Cu(OH)2
Weak acids and bases: HNO2, H3PO4, HC2H3O2, NH3
Water, in any state

Ions you type yourself, such as $\mathrm{Cl^{-}}$ or $\mathrm{Fe^{3+}}$ , are already separated and are never split again.

Four reaction types, worked through

The same three rules produce very different-looking results depending on which kind of reaction you are dealing with. These four are the ones you will meet most often.

1. Precipitation

Copper(II) sulfate meets sodium hydroxide. The solid precipitate is the product, and the ions it is made of are the whole story.

\mathrm{Cu^{2+}(aq)}

+

2\,\mathrm{OH^{-}(aq)}

\longrightarrow

\mathrm{Cu(OH)_2(s)}

Sulfate and sodium are spectators. Note that the copper(II) hydroxide is marked (s), which is exactly why it is never split into copper and hydroxide ions on the product side.

2. Strong acid neutralisation

Hydrochloric acid and sodium hydroxide. Both sides dissociate completely, so the result reduces to the simplest equation in chemistry.

\mathrm{H^{+}(aq)}

+

\mathrm{OH^{-}(aq)}

\longrightarrow

\mathrm{H_2O(l)}

Sodium and chloride are spectators. This single equation describes every strong acid plus strong base neutralisation, whatever salts are involved.

3. Weak acid neutralisation

Swap the hydrochloric acid for nitrous acid and the answer changes shape, because a weak acid does not dissociate.

\mathrm{HNO_2(aq)}

+

\mathrm{OH^{-}(aq)}

\longrightarrow

\mathrm{NO_2^{-}(aq)}

+

\mathrm{H_2O(l)}

Nitrous acid stays whole on the left but appears as nitrite on the right. That asymmetry is the signature of a weak electrolyte, and it is why the acid and its salt cannot be treated interchangeably.

4. Gas-forming reaction

Ammonium chloride plus sodium hydroxide releases ammonia gas. The escaping gas is what drives the reaction forward, and it stays intact.

\mathrm{NH_4^{+}(aq)}

+

\mathrm{OH^{-}(aq)}

\longrightarrow

\mathrm{NH_3(g)}

+

\mathrm{H_2O(l)}

Sodium and chloride are spectators. Ammonia here is a product that escapes as a gas, which is different from the ammonia you meet as a weak base in solution.

Mistakes that change the answer

These are the errors that turn a correct equation into a wrong net ionic one. Each one is about the input rather than the algebra.

Forgetting the state symbols

Anything you leave unmarked is treated as dissolved. The state symbol is the single most consequential thing you type, because it decides whether the compound splits at all.

\mathrm{AgCl(aq)}

\mathrm{AgCl(s)}

Splitting a precipitate

Silver chloride, barium sulfate and lead iodide all read like salts, but they are insoluble and must stay whole in the equation. Breaking them into ions is the classic error.

\mathrm{AgCl}\rightarrow\mathrm{Ag^{+}}+\mathrm{Cl^{-}}

\mathrm{AgCl(s)}

Splitting a weak acid

Nitrous acid and acetic acid are mostly undissociated in water. They are molecules in solution, not collections of ions, so they appear intact on the left and as an anion on the right.

\mathrm{H^{+}}+\mathrm{NO_2^{-}}

\mathrm{HNO_2(aq)}

Cancelling unequal amounts

Only ions appearing in equal amounts on both sides cancel. If one side holds twice as many, the excess is part of the answer and has to stay in the net ionic equation.

Lowercase element symbols

Element symbols are case-sensitive. A capital C with a lowercase L is chlorine; the same letters with a lowercase L in the wrong place is not a symbol at all, and the calculator will reject it.

\mathrm{h2o}

\mathrm{H_2O(l)}

A plus sign inside a charge

A plus sign both separates compounds and carries a cation charge. Write the charge tight against the formula with no space, and leave a clear plus between compounds so the two can never be confused.

\mathrm{Ag^{+}}

\mathrm{AgNO_3(aq)} + \mathrm{NaCl(aq)}

Frequently asked questions

Does the equation have to be balanced first?

No. The calculator balances it before anything else happens, so you can type an unbalanced equation with missing coefficients and it will supply them. If you do type coefficients, they are treated as part of the formula and the result is renormalised to the smallest ratio.

How do I check the answer is right?

Two checks, both quick. Count the atoms of each element on both sides and confirm they match. Then add up the charges on each side and confirm those totals match too. An equation can balance atoms and still be wrong if the ionic charges were assigned incorrectly, so do both. If you want the quantitative consequences next, the equilibrium constant calculator works from the balanced equation.

Why is the order of the ions different from mine?

It is not wrong. The ions come out in the order the compounds were written, with cations before anions inside each one. Order carries no chemical meaning in an ionic equation, so any consistent ordering that keeps each compound together is equally acceptable.

I only need the coefficients. Is there a simpler tool?

Yes. If cancellation and spectators are irrelevant to you, the chemical equation balancer does only that step.

Limitations

Hydrates are not supported. The dot in a hydrate such as $\mathrm{CuSO_4\cdot 5H_2O}$ is not accepted. Enter the anhydrous salt instead, and add the water yourself if you need it counted.
Coordination complexes are not supported. Square-bracket species such as $\mathrm{K_4[Fe(CN)_6]}$ are rejected because bracket notation is not part of the input format.
Solubility follows the classroom table, not full reference data. Uncommon ions such as silicate are not in the built-in table, so a salt built from one will stay whole when it should split. If a result looks as though a soluble salt refused to dissociate, this is usually why.
An unmarked state symbol means (aq). That is a deliberate shortcut, but it is wrong for a substance you meant as a solid or a gas, such as copper metal. Always mark it.
It balances by conservation, not by redox method. Oxidation and reduction are not identified separately, and no half-reactions are produced. The coefficients are still correct, but this is not a tool for working through electron bookkeeping.
It says nothing about how far a reaction goes. A net ionic equation describes what changes, not whether the change is complete or how much of it happens.

This is a study aid for building and checking net ionic equations. It does not model concentrations, temperature or equilibrium, so it should not be used to predict experimental outcomes.

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