AP Chem 4.2: Net Ionic Equations

One reaction, three equations

Lesson 2.7.4 covered the steps from a molecular equation to a net ionic equation. AP treats all three forms as equations for the same reaction and expects you to choose the one that fits the purpose:

The molecular equation化学方程式 writes every substance as a complete formula. It’s the one to use when deciding what to weigh out or make up as a solution.
The complete ionic equation完整的离子方程式 splits every substance that actually exists as ions in solution, so it is closest to what’s really in the beaker.
The net ionic equation净离子方程式 removes the spectator ions旁观离子, which appear unchanged on both sides, and keeps only the particles that actually change.

For the reaction of AgNO₃ solution with NaCl solution, the molecular and net ionic equations are:

\ce{AgNO3(aq) + NaCl(aq) -> AgCl(s) + NaNO3(aq)}

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

The net ionic equation shows that mixing any soluble solution containing Ag⁺ with any soluble solution containing Cl⁻ forms an AgCl precipitate, regardless of the Na⁺ and NO₃⁻.

What to split, what to keep

Split into ions (strong electrolytes in solution)	Keep the formula
Strong acids: HCl, HBr, HI, HNO₃, HClO₄, H₂SO₄	Weak acids: CH₃COOH, HF, H₂CO₃, and others
Strong bases: LiOH, NaOH, KOH, Ca(OH)₂, Sr(OH)₂, Ba(OH)₂	Weak bases: NH₃
Soluble salts	Solids (including precipitates), gases, liquid water
	Nonelectrolytes: glucose, ethanol, and others

AP solubility rule: the exam expects you to know just one rule: all salts containing Na⁺, K⁺, NH₄⁺, or NO₃⁻ are soluble. For other salts, the question tells you (or gives solubility information). The solubility table of lesson 2.7.3 helps with more cases. AP questions usually write acetic acid as HC₂H₃O₂; it’s the same substance as CH₃COOH.

Two conservation checks: in a net ionic equation, each kind of atom balances, and so does the total charge.

First, step through how a reaction is turned into its net ionic equation:

Try it: from molecular equation to net ionic equation

Strong versus weak: what’s really in the beaker

Strong acids and bases ionize completely in water and exist as ions, so they’re split; weak acids and bases stay almost entirely as molecules, with only a tiny fraction ionized, so they keep their formulas. Choose an acid and a base, and compare the particles before and after mixing with the three equations:

Try it: strong electrolytes split into ions; weak electrolytes keep their formulas

Acid:

Base:

HCl (strong acid)

NaOH (strong base)

After mixing

Molecular equation

\ce{HCl(aq)} + \ce{NaOH(aq)} \longrightarrow \ce{NaCl(aq)} + \ce{H2O(l)}

Complete ionic equation (spectator ions crossed out)

\ce{H+(aq)} + \cancel{\ce{Cl-(aq)}} + \cancel{\ce{Na+(aq)}} + \ce{OH-(aq)} \longrightarrow \cancel{\ce{Na+(aq)}} + \cancel{\ce{Cl-(aq)}} + \ce{H2O(l)}

Net ionic equation

\ce{H+(aq)} + \ce{OH-(aq)} \longrightarrow \ce{H2O(l)}

Spectator ions: Na⁺, Cl⁻.

HCl and NaOH are both strong electrolytes, completely ionized in water. The only species that actually react are H⁺ and OH⁻.

cationanionmolecule (not split)

Physical changes have ionic equations too

An ionic compound dissolving in water is a physical change, and it can be written as an ionic equation:

\ce{CaCl2(s) -> Ca^2+(aq) + 2Cl-(aq)}

Check both sides: the atoms balance (1 Ca, 2 Cl), and so does the charge (0 on the left; +2 + 2 × (−1) = 0 on the right).

Net ionic equations for other kinds of reactions

Forming a weak acid: mixing NaF solution with hydrochloric acid forms no precipitate and no gas, but it does form a weak electrolyte, HF. HF exists mostly as molecules, so it keeps its formula:

\ce{H+(aq) + F-(aq) -> HF(aq)}

An active metal and water: lithium in water forms LiOH (a strong base, so it’s split) and H₂:

\ce{2Li(s) + 2H2O(l) -> 2Li+(aq) + 2OH-(aq) + H2(g)}

A metal and the ions of another metal: a copper wire in AgNO₃ solution becomes coated with silver as the solution turns blue:

\ce{Cu(s) + 2Ag+(aq) -> Cu^2+(aq) + 2Ag(s)}

NO₃⁻ stays in solution throughout and is still in the filtrate after filtering, neither oxidized nor reduced: it’s a spectator ion. The last two are redox reactions (U4.9).

Example 1: Forming a precipitate

Solutions of $\ce{Na2CO3}$ and $\ce{CaCl2}$ are mixed, and a white precipitate forms. $\ce{CaCO3}$ is insoluble in water. Write the molecular, complete ionic, and net ionic equations.

Step-by-step solution

Write the balanced molecular equation. The two salts swap partners, forming insoluble $\ce{CaCO3}$ :
$\begin{aligned} &\ce{Na2CO3(aq) + CaCl2(aq)} \\ &\ce{-> CaCO3(s) + 2NaCl(aq)} \end{aligned}$
Decide what to split. Only soluble strong electrolytes are written as ions:
- $\ce{Na2CO3}$ , NaCl: sodium salts are soluble, split
- $\ce{CaCl2}$ : soluble, split
- $\ce{CaCO3}$ : precipitate, keep the formula
Complete ionic equation.
$\begin{aligned} &\ce{2Na+(aq) + CO3^2-(aq)} \\ &\ce{+ Ca^2+(aq) + 2Cl-(aq)} \\ &\ce{-> CaCO3(s)} \\ &\ce{+ 2Na+(aq) + 2Cl-(aq)} \end{aligned}$
Remove the spectator ions. $\ce{Na+}$ and $\ce{Cl-}$ appear unchanged on both sides, so they drop out:
$\boxed{ \begin{aligned} &\ce{Ca^2+(aq) + CO3^2-(aq)} \\ &\ce{-> CaCO3(s)} \end{aligned} }$
Check. Atoms: 1 Ca, 1 C, and 3 O on each side. Charge: (+2) + (−2) = 0 on the left, 0 on the right.

Example 2: A weak acid and a strong base

Hydrofluoric acid (HF, a weak acid) reacts with KOH solution. Write the net ionic equation, and explain why it isn’t $\ce{H+(aq) + OH-(aq) -> H2O(l)}$ .

Step-by-step solution

Write the balanced molecular equation.
$\begin{aligned} &\ce{HF(aq) + KOH(aq)} \\ &\ce{-> KF(aq) + H2O(l)} \end{aligned}$
Decide what to split.
- HF: a weak acid, only slightly ionized in water, keep the formula
- KOH: a strong base, split
- KF: a soluble potassium salt, split
- $\ce{H2O}$ : keep the formula
Complete ionic equation.
$\begin{aligned} &\ce{HF(aq) + K+(aq) + OH-(aq)} \\ &\ce{-> K+(aq) + F-(aq) + H2O(l)} \end{aligned}$
Remove the spectator ion, $\ce{K+}$ .
$\boxed{ \begin{aligned} &\ce{HF(aq) + OH-(aq)} \\ &\ce{-> F-(aq) + H2O(l)} \end{aligned} }$
Check. Atoms: 2 H, 1 F, and 1 O on each side. Charge: −1 on both sides.
Why not $\ce{H+(aq) + OH-(aq) -> H2O(l)}$ ? HF is a weak acid, so the solution is almost all HF molecules with very little free $\ce{H+}$ . What actually reacts with $\ce{OH-}$ is HF molecules, so HF stays in the net ionic equation.

Practice: split into ions or keep the formula in an ionic equation?

NaNO₃(aq)
AgCl(s)
HCl(aq)
CH₃COOH(aq)
H₂O(l)
NH₃(aq)
KOH(aq)
CO₂(g)
HF(aq)
(NH₄)₂SO₄(aq)
C₆H₁₂O₆(aq)
Ba(OH)₂(aq)
SrSO₄(s)
H₂SO₃(aq)
LiOH(aq)

Common mistakes

Mistake 1: “Every acid is written as H⁺.” Only strong acids ionize completely. Weak acids such as acetic acid and hydrofluoric acid are mostly molecules in solution and are written as CH₃COOH and HF.

Mistake 2: “A precipitate is ionic, so it should be split too.” What you split depends on the form a substance actually takes in the beaker. A precipitate is a solid with its ions locked in a crystal, so it keeps its formula.

Mistake 3: “A net ionic equation only has to balance atoms.” Charge must balance too. $\ce{Fe^3+(aq) + Cu(s) -> Fe^2+(aq) + Cu^2+(aq)}$ balances the atoms, but the charge is +3 on the left and +4 on the right; the correct equation is $\ce{2Fe^3+(aq) + Cu(s) -> 2Fe^2+(aq) + Cu^2+(aq)}$ .

Quick check

Pb(NO₃)₂ solution and KI solution are mixed, forming a yellow PbI₂ precipitate. Which is the correct net ionic equation?

Acetic acid reacts with NaOH solution. Which net ionic equation is correct, and why?

What are the spectator ions in the reaction of $\ce{BaCl2(aq)}$ with $\ce{Na2SO4(aq)}$ ?

Which net ionic equation does not conserve charge?

Which substance should you know to be soluble in water without being told, as the AP exam expects?

Hydrochloric acid is dripped onto solid calcium carbonate, and bubbles form. Which net ionic equation is correct?

Sulfurous acid (H₂SO₃, a weak acid) reacts with Ba(OH)₂ solution to form a BaSO₃ precipitate and water. Which is the correct net ionic equation?

NaF solution is mixed with hydrochloric acid. Which statement is correct?

A copper wire is placed in AgNO₃ solution; silver forms on the wire, and the solution turns blue. After filtering, what has happened to the NO₃⁻ ions in the filtrate?

A small piece of lithium is dropped into water, forming H₂ and LiOH solution. What is the net ionic equation?

AP-style free response

A student mixes 50.0 mL of 0.10 M $\ce{CuSO4}$ with 50.0 mL of 0.20 M NaOH, and a blue precipitate of $\ce{Cu(OH)2}$ forms.

(a) Write the molecular equation for the reaction.

(b) Write the net ionic equation, and identify the spectator ions.

(c) After the reaction is complete, which ions remain in solution? Draw (or describe) a particle diagram of the mixture in the beaker (leave out the water molecules).

(d) The student repeats the experiment with aqueous ammonia instead of NaOH, and Cu(OH)₂ precipitates again. A classmate writes the net ionic equation as $\ce{Cu^2+(aq) + 2OH-(aq) -> Cu(OH)2(s)}$ , claiming it’s identical to (b). Write the correct net ionic equation, with ammonia written as NH₃.

Model answer and scoring

(a) $\ce{CuSO4(aq) + 2NaOH(aq) -> Cu(OH)2(s) + Na2SO4(aq)}$

(b) $\ce{Cu^2+(aq) + 2OH-(aq) -> Cu(OH)2(s)}$ ; the spectator ions are Na⁺ and SO₄²⁻.

(c) The Cu²⁺ and OH⁻ react exactly (0.0050 mol Cu²⁺ with 0.010 mol OH⁻), leaving Na⁺ and SO₄²⁻ in solution in a 2 : 1 ratio. Diagram: solid Cu(OH)₂ at the bottom of the beaker and, in solution, Na⁺ and SO₄²⁻ ions (for example, 4 Na⁺ and 2 SO₄²⁻), with no Cu²⁺ or OH⁻.

(d) NH₃ is a weak base, present in solution mostly as molecules, so it keeps its formula: $\ce{Cu^2+(aq) + 2NH3(aq) + 2H2O(l) -> Cu(OH)2(s) + 2NH4+(aq)}$ .

Scoring: (a) balanced molecular equation; (b) correct net ionic equation and both spectator ions; (c) only Na⁺ and SO₄²⁻ remain, in a 2 : 1 ratio, with the precipitate shown; (d) NH₃ kept as a formula, with atoms and charge balanced.

Summary

Molecular, complete ionic, and net ionic equations describe the same reaction; choose by purpose. The net ionic equation keeps only the particles that change.
Split: strong acids, strong bases, soluble salts. Keep: solids (precipitates), gases, liquid water, weak acids, weak bases, nonelectrolytes.
AP expects you to know only that salts of Na⁺, K⁺, NH₄⁺, and NO₃⁻ are always soluble.
Both atoms and charge must balance in a net ionic equation; physical changes such as dissolving can be written as ionic equations too.

AP tip: Before writing a net ionic equation, ask what form each substance takes in the beaker. The most common lost points come from splitting a weak acid (HF, CH₃COOH) or a weak base (NH₃), and from splitting a precipitate into ions. Always check the charge at the end.

Net Ionic Equations净离子方程式

Learning goals

Before you start

One reaction, three equations

What to split, what to keep

Strong versus weak: what’s really in the beaker

Physical changes have ionic equations too

Net ionic equations for other kinds of reactions

Example 1: Forming a precipitate

Example 2: A weak acid and a strong base

Common mistakes

Quick check

AP-style free response

Summary

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