4.2 - Net Ionic Equations

Overview

A chemical equation is chemistry's shorthand. It tells you which substances react, what they make, and in what amounts. This topic covers how to balance equations, how to write them for physical changes too, and the three ways to write a reaction that happens in water: molecular, complete ionic, and net ionic. Writing a correct net ionic equation is one of the most common tasks on the AP Chemistry free-response section.

Reading a chemical equation

\mathrm{2H_2(g) + O_2(g) \rightarrow 2H_2O(l)}

Reactants are on the left and products are on the right. The arrow means "react to form."

Coefficients, the numbers in front, count particles: 2 molecules of $\mathrm{H_2}$ react with 1 molecule of $\mathrm{O_2}$ . They count moles too (Topic 4.5). A coefficient of 1 isn't written.

Subscripts are part of each formula and tell you what the substance is. $\mathrm{2H_2O}$ means two water molecules, but $\mathrm{H_2O_2}$ is hydrogen peroxide, a different substance. Never change a subscript to balance an equation.

Phase labels: (s) solid, (l) liquid, (g) gas, and (aq) aqueous, meaning dissolved in water.

Every change can be written as a balanced equation

All physical and chemical processes can be represented by balanced equations. That covers three situations:

Physical changes. The formula stays the same and only the phase changes, as in $\mathrm{H_2O(l) \rightarrow H_2O(g)}$ . Dissolving counts too: $\mathrm{C_6H_{12}O_6(s) \rightarrow C_6H_{12}O_6(aq)}$ for glucose, and $\mathrm{NaCl(s) \rightarrow Na^{+}(aq) + Cl^{-}(aq)}$ for an ionic solid, which breaks up into its ions.

Reactions described in words. "Solid sodium reacts with chlorine gas to form solid sodium chloride" becomes $\mathrm{2Na(s) + Cl_2(g) \rightarrow 2NaCl(s)}$ . Write correct formulas first, and remember the seven diatomic elements: $\mathrm{H_2}$ , $\mathrm{N_2}$ , $\mathrm{O_2}$ , $\mathrm{F_2}$ , $\mathrm{Cl_2}$ , $\mathrm{Br_2}$ , and $\mathrm{I_2}$ .

Ions in solution. Reactions between dissolved ions are written as ionic and net ionic equations, covered below.

Balanced means atoms and charge are conserved

A chemical change is a rearrangement of atoms into new combinations. No atom is created or destroyed, so a balanced equation has the same number of atoms of each element on both sides. Atoms keep their mass, so this is also why mass is conserved: the total mass of the reactants equals the total mass of the products.

Charge is conserved too. Add up the charges of all the ions on each side (coefficient times charge). The two totals must match. In a molecular equation every species is neutral, so this check is automatic, but in ionic equations it catches a lot of mistakes.

How to balance by inspection

Write the correct formula for every reactant and product. From here on, only coefficients change.

Start with an element that appears in just one reactant and one product.

Treat a polyatomic ion that stays intact on both sides, like $\mathrm{SO_4^{2-}}$ or $\mathrm{NO_3^{-}}$ , as a single unit.

Save elements that appear by themselves, like $\mathrm{O_2}$ or $\mathrm{Fe}$ , for last. Changing their coefficient doesn't disturb anything else.

Reduce to the smallest whole-number coefficients, then check every element and the total charge.

Example 1: Balancing a combustion reaction

Balance the combustion of propane:

\mathrm{C_3H_8(g) + O_2(g) \rightarrow CO_2(g) + H_2O(g)}

Carbon: there are 3 C on the left, so put a 3 in front of

\mathrm{CO_2}

Hydrogen: there are 8 H on the left, so put a 4 in front of

\mathrm{H_2O}

Oxygen: the right side now has

3(2) + 4(1) = 10

O atoms, so put a 5 in front of

\mathrm{O_2}

\mathrm{C_3H_8(g) + 5O_2(g) \rightarrow 3CO_2(g) + 4H_2O(g)}

Element	Left side	Right side
C	3	3
H	8	$4 \times 2 = 8$
O	$5 \times 2 = 10$	$3(2) + 4(1) = 10$

Example 2: Clearing a fraction

Balance the combustion of ethane,

\mathrm{C_2H_6}

Balancing carbon and hydrogen gives

\mathrm{C_2H_6 + O_2 \rightarrow 2CO_2 + 3H_2O}

. The right side has

2(2) + 3 = 7

O atoms, which would take

\frac{7}{2}

\mathrm{O_2}

. Fractions aren't allowed in the final answer, so multiply every coefficient by 2:

\mathrm{2C_2H_6(g) + 7O_2(g) \rightarrow 4CO_2(g) + 6H_2O(g)}

Check: 4 C, 12 H, and 14 O on each side.

Three ways to write a reaction in water

When ionic compounds react in solution, the same reaction can be written three ways. Take mixing solutions of lead(II) nitrate and potassium iodide, which makes a yellow precipitate of lead(II) iodide.

Molecular equation: every compound is written as a complete formula, as if nothing had broken apart.

\mathrm{Pb(NO_3)_2(aq) + 2KI(aq) \rightarrow PbI_2(s) + 2KNO_3(aq)}

Complete ionic equation: every strong electrolyte that's dissolved (aq) is split into the ions it actually exists as in water. Solids, liquids, and gases stay as written.

\mathrm{Pb^{2+}(aq) + 2NO_3^{-}(aq) + 2K^{+}(aq) + 2I^{-}(aq) \rightarrow PbI_2(s) + 2K^{+}(aq) + 2NO_3^{-}(aq)}

Net ionic equation: cross out the spectator ions, the ions that appear unchanged on both sides. What's left is the chemistry that actually happens.

\mathrm{Pb^{2+}(aq) + 2I^{-}(aq) \rightarrow PbI_2(s)}

Here

\mathrm{K^{+}}

and

\mathrm{NO_3^{-}}

are spectators. They start dissolved and end dissolved, untouched.

\mathrm{Pb^{2+}}

and

\mathrm{I^{-}}

are not spectators, because they go from dissolved ions to part of a solid.

In the particle view, the gray spectator ions sit in the same places before and after. Only the

\mathrm{Pb^{2+}}

and

\mathrm{I^{-}}

ions change, joining into solid

\mathrm{PbI_2}

: 2 lead ions and 4 iodide ions, the 1 to 2 ratio in the formula.

Which form to use

Form	Best for
Molecular	Stoichiometry with the compounds you actually measure out, like grams of $\mathrm{KI}$ (Topic 4.5)
Complete ionic	Showing every species in the solution, for example to draw a particle diagram or to find which ions are left after mixing
Net ionic	Showing the reaction itself. Any soluble lead(II) compound mixed with any soluble iodide gives this same net ionic equation.

When an AP question asks for a net ionic equation, spectator ions in your answer cost you the point.

What breaks into ions and what stays together

The whole skill comes down to knowing which species to write as separate ions.

Write as separate ions (when aqueous)	Keep together as written
Soluble ionic compounds, like $\mathrm{NaCl(aq)}$ or $\mathrm{Pb(NO_3)_2(aq)}$	Solids (s), including precipitates
Strong acids: $\mathrm{HCl}$ , $\mathrm{HBr}$ , $\mathrm{HI}$ , $\mathrm{HNO_3}$ , $\mathrm{HClO_4}$ , $\mathrm{H_2SO_4}$	Liquids (l), such as $\mathrm{H_2O}$
Strong bases: Group 1 hydroxides (like $\mathrm{NaOH}$ and $\mathrm{KOH}$ ), plus $\mathrm{Ca(OH)_2}$ , $\mathrm{Sr(OH)_2}$ , and $\mathrm{Ba(OH)_2}$	Gases (g), such as $\mathrm{CO_2}$ and $\mathrm{H_2}$
	Weak acids and weak bases, such as acetic acid $\mathrm{CH_3COOH}$ , $\mathrm{HF}$ , and $\mathrm{NH_3}$
	Molecular compounds that dissolve without forming ions, such as sugar

Polyatomic ions stay in one piece. $\mathrm{Pb(NO_3)_2(aq)}$ becomes $\mathrm{Pb^{2+}(aq)}$ and $\mathrm{2NO_3^{-}(aq)}$ , never separate N and O atoms.

Hydrogen ion. A strong acid's hydrogen ion can be written as $\mathrm{H^{+}(aq)}$ or $\mathrm{H_3O^{+}(aq)}$ . Both are accepted. Topic 4.8 and Unit 8 cover acids and bases in depth.

Solubility. The only solubility rule the AP exam expects you to know is that all sodium, potassium, ammonium, and nitrate salts are soluble in water (Topic 4.7). Any other solubility information will be given to you, often by telling you that a precipitate forms.

Example 3: A precipitation reaction

Aqueous calcium chloride is mixed with aqueous sodium phosphate, and a white precipitate of calcium phosphate forms. Write the molecular, complete ionic, and net ionic equations.

Molecular. Calcium phosphate,

\mathrm{Ca_3(PO_4)_2}

, is the solid (the question tells you). Sodium chloride stays dissolved because it's a sodium salt. Balance the ions first: 3

\mathrm{Ca^{2+}}

for every 2

\mathrm{PO_4^{3-}}

\mathrm{3CaCl_2(aq) + 2Na_3PO_4(aq) \rightarrow Ca_3(PO_4)_2(s) + 6NaCl(aq)}

Complete ionic. Split every (aq) compound into its ions. Keep the coefficients: 3

\mathrm{CaCl_2}

gives 3

\mathrm{Ca^{2+}}

and 6

\mathrm{Cl^{-}}

\mathrm{3Ca^{2+}(aq) + 6Cl^{-}(aq) + 6Na^{+}(aq) + 2PO_4^{3-}(aq) \rightarrow Ca_3(PO_4)_2(s) + 6Na^{+}(aq) + 6Cl^{-}(aq)}

Net ionic.

\mathrm{Na^{+}}

and

\mathrm{Cl^{-}}

appear unchanged on both sides, so cancel them.

\mathrm{3Ca^{2+}(aq) + 2PO_4^{3-}(aq) \rightarrow Ca_3(PO_4)_2(s)}

Check: 3 Ca, 2 P, and 8 O on each side. The charge is

3(+2) + 2(-3) = 0

on the left and 0 on the right.

Example 4: Strong acid versus weak acid

Write the net ionic equation for (a) nitric acid with potassium hydroxide and (b) acetic acid with sodium hydroxide.

(a) Nitric acid is a strong acid and

\mathrm{KOH}

is a strong base, so both split into ions.

\mathrm{K^{+}}

and

\mathrm{NO_3^{-}}

are spectators, and water is a liquid:

\mathrm{H^{+}(aq) + OH^{-}(aq) \rightarrow H_2O(l)}

Written with hydronium, the same equation is

\mathrm{H_3O^{+}(aq) + OH^{-}(aq) \rightarrow 2H_2O(l)}

(b) Acetic acid is a weak acid. It stays mostly as whole molecules in water, so it's written as

\mathrm{CH_3COOH(aq)}

. The only spectator is

\mathrm{Na^{+}}

\mathrm{CH_3COOH(aq) + OH^{-}(aq) \rightarrow CH_3COO^{-}(aq) + H_2O(l)}

Check: 2 C, 5 H, and 3 O on each side, and a charge of

-1

on each side. Writing

\mathrm{H^{+}}

plus

\mathrm{OH^{-}}

for a weak acid is a classic mistake.

Example 5: Atoms balance but charge doesn't (AP-style)

A copper wire is placed in silver nitrate solution. Silver metal forms on the wire, and the solution turns blue from

\mathrm{Cu^{2+}}

ions. A student writes this net ionic equation:

\mathrm{Cu(s) + Ag^{+}(aq) \rightarrow Cu^{2+}(aq) + Ag(s)}

Is it balanced? If not, fix it.

Atoms: 1 Cu and 1 Ag on each side, so the atoms are fine.

Charge:

+1

on the left but

+2

on the right, so the equation is not balanced. Put a 2 in front of

\mathrm{Ag^{+}}

and

\mathrm{Ag}

\mathrm{Cu(s) + 2Ag^{+}(aq) \rightarrow Cu^{2+}(aq) + 2Ag(s)}

Now there are 1 Cu and 2 Ag on each side, and the charge is

+2

on both sides. Topic 4.9 shows how to build equations like this from half-reactions.

Example 6: Checking a lab equation

The AP lab that finds the percent of hydrogen peroxide in a drugstore bottle can do it by reacting the peroxide with permanganate ion in acid. Verify that this net ionic equation is balanced:

\mathrm{2MnO_4^{-}(aq) + 5H_2O_2(aq) + 6H^{+}(aq) \rightarrow 2Mn^{2+}(aq) + 5O_2(g) + 8H_2O(l)}

Check	Left side	Right side
Mn	2	2
O	$2(4) + 5(2) = 18$	$5(2) + 8(1) = 18$
H	$5(2) + 6 = 16$	$8(2) = 16$
Charge	$2(-1) + 6(+1) = +4$	$2(+2) = +4$

Every element and the total charge match, so the equation is balanced.

Common mistakes

Changing subscripts to balance an equation. That changes the substance.

Splitting solids, liquids, gases, or weak acids into ions. Only dissolved strong electrolytes split.

Breaking up polyatomic ions, or leaving off their charges.

Checking atoms but not charge.

Leaving spectator ions in a net ionic equation, or canceling an ion that changes phase, like $\mathrm{Pb^{2+}(aq)}$ , which ends up in $\mathrm{PbI_2(s)}$ .

Writing oxygen or chlorine as single atoms instead of $\mathrm{O_2}$ and $\mathrm{Cl_2}$ .

Leaving coefficients that can be reduced, like $\mathrm{2Ba^{2+} + 2SO_4^{2-} \rightarrow 2BaSO_4}$ . Divide by 2.

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