CHEMISTRY · LESSON 08 OF 12

Reactions in solution

In water, many compounds exist as ions. Reactions in solution — precipitation, acid–base and redox — are written as net ionic equations, and their amounts are calculated through moles using balanced equations.

What this lesson explains

Most chemistry in industry, the environment and living things happens in water: water treatment removes ions by precipitation, titrations measure concentrations in quality control, neutralisation treats acidic waste, and redox reactions drive corrosion and batteries. Knowing which species actually react, and in what amounts, is the basis of process design and analysis.

Before you begin

The mole and molar mass

1 mol = 6.022 × 10²³ particles. Molar mass M (g/mol) is the sum of atomic masses in the formula: M(NaCl) = 22.99 + 35.45 = 58.44 g/mol. n = m/M.

Molarity

c = n / V in mol/L (M). Volumes in litres: 25.0 mL = 0.0250 L. See Properties of solutions.

Common ions

NO₃⁻ nitrate, SO₄²⁻ sulfate, CO₃²⁻ carbonate, OH⁻ hydroxide, NH₄⁺ ammonium, PO₄³⁻ phosphate.

The idea, made visible

A balanced equation conserves atoms and charge. Its coefficients give mole ratios, never mass ratios: in 2H₂ + O₂ → 2H₂O, 2 mol of H₂ react with 1 mol of O₂. To calculate amounts, always go through moles (see the figure).

When ionic compounds or strong acids dissolve, they separate into ions: they are strong electrolytes. NaCl(aq) is really Na⁺(aq) and Cl⁻(aq). Weak acids (such as acetic acid) and weak bases ionise only partly (weak electrolytes); sugar does not ionise at all (non-electrolyte).

Grams and litres cannot be compared directly across a reaction. Convert to moles, use the mole ratio from the balanced equation, then convert back.

Writing reactions as ions shows what really happens. Mixing AgNO₃(aq) with NaCl(aq) forms solid AgCl. In the complete ionic equation, Na⁺ and NO₃⁻ appear unchanged on both sides — they are spectator ions. Removing them gives the net ionic equation: Ag⁺(aq) + Cl⁻(aq) → AgCl(s). Solubility rules tell you which products are insoluble (precipitates).

The three main reaction types are: precipitation (an insoluble solid forms), acid–base neutralisation (H⁺ from an acid combines with OH⁻ from a base: H⁺ + OH⁻ → H₂O for strong acid–strong base), and oxidation–reduction (redox), in which electrons are transferred. Oxidation is loss of electrons (oxidation number increases); reduction is gain (oxidation number decreases). They always happen together.

When reactants are not in the exact mole ratio, the one that runs out first is the limiting reactant; it determines the maximum (theoretical) yield. The actual yield divided by the theoretical yield gives the percent yield. For solutions, moles = molarity × volume, which makes titration possible: measure the volume of a solution of known concentration needed to react exactly with an unknown.

Assigning oxidation numbers
species	oxidation numbers	reason
O₂, Fe, Cl₂	0	uncombined elements
Fe³⁺	+3	monatomic ion: its charge
H₂O	H +1, O −2	standard values; sum 0
SO₄²⁻	S +6, O −2	S + 4(−2) = −2
MnO₄⁻	Mn +7, O −2	Mn + 4(−2) = −1
H₂O₂	H +1, O −1	peroxide: exception for O

Key terms

Electrolyte: A substance that produces ions in water, so that the solution conducts electricity. Strong: fully ionised; weak: partly ionised.
Precipitate: An insoluble solid formed when two solutions are mixed.
Spectator ion: An ion present on both sides of the ionic equation, unchanged.
Net ionic equation: The equation showing only the species that actually change.
Acid / base (Arrhenius and Brønsted–Lowry): An acid produces H⁺ in water (proton donor); a base produces OH⁻ (proton acceptor).
Oxidation number: A bookkeeping charge: elements 0; monatomic ions their charge; O usually −2; H usually +1 (−1 in metal hydrides); the sum equals the overall charge.
Limiting reactant: The reactant that is completely consumed first; it fixes the maximum amount of product.
Equivalence point: The point in a titration at which the reactants have been mixed in exactly the stoichiometric ratio.

The formulas and what they mean

Symbols, meanings and units
Symbol	Meaning	Unit
n	amount of substance	mol
m	mass	g
M	molar mass	g/mol
c (or [X])	molar concentration	mol/L (M)
V	volume of solution	L
(s), (l), (g), (aq)	solid, liquid, gas, dissolved in water	—

Moles

n = m / M; n = cV

Conditions and limits: m in grams, M in g/mol; c in mol/L, V in litres.

Mole ratio

n_B = n_A × coefficient of B / coefficient of A

Conditions and limits: Only from a balanced equation. To predict how much product forms, start from the limiting reactant.

Percent yield

% yield = actual yield / theoretical yield × 100%

Conditions and limits: Theoretical yield computed from the limiting reactant.

Titration at equivalence

c_AV_A × b / a = c_BV_B for aA + bB → products

Conditions and limits: At the equivalence point. For a 1:1 reaction (HCl + NaOH), c_AV_A = c_BV_B.

Solubility guidelines (summary)

soluble: Na⁺, K⁺, NH₄⁺, NO₃⁻ compounds (all); most Cl⁻, Br⁻, I⁻ (except Ag⁺, Pb²⁺, Hg₂²⁺); most SO₄²⁻ (except Ba²⁺, Pb²⁺, Ca²⁺ slightly). Insoluble: most CO₃²⁻, PO₄³⁻, OH⁻, S²⁻ (except with Group 1 and NH₄⁺)

Conditions and limits: Rules of thumb at room temperature; “insoluble” means very slightly soluble.

Why it works: From a molecular equation to a net ionic equation

Silver nitrate solution is mixed with sodium chloride solution.

Molecular: AgNO₃(aq) + NaCl(aq) → AgCl(s) + NaNO₃(aq).
Balanced; AgCl is insoluble (solubility rules).
Complete ionic: Ag⁺(aq) + NO₃⁻(aq) + Na⁺(aq) + Cl⁻(aq) → AgCl(s) + Na⁺(aq) + NO₃⁻(aq).
Split strong electrolytes into ions; keep the solid together.
Cancel the spectators Na⁺ and NO₃⁻.
They appear unchanged on both sides.
Net ionic: Ag⁺(aq) + Cl⁻(aq) → AgCl(s).
Check: atoms balanced; charge 0 = 0.

Any soluble silver salt and any soluble chloride give the same net reaction.

A first worked example

Solving a reaction-in-solution problem

Write and balance the molecular equation; identify the reaction type.
If asked, write the complete ionic and net ionic equations (split only soluble strong electrolytes).
Convert every given quantity to moles (n = m/M or n = cV).
Find the limiting reactant: divide each reactant’s moles by its coefficient; the smallest result is limiting.
Use the mole ratio from the limiting reactant to find moles of product, then convert to the requested unit.
Check charge and atom balance, units and significant figures.

Balancing an equation

Problem. Balance the combustion of propane: C₃H₈ + O₂ → CO₂ + H₂O.

Carbon: 3 on the left, so write 3 CO₂.
Balance elements that appear in only one substance on each side first.
Hydrogen: 8 on the left, so write 4 H₂O.
Each water molecule has 2 H.
Oxygen on the right: 3 × 2 + 4 × 1 = 10 atoms, so write 5 O₂ on the left.
Leave the free element (O₂) until last; it can be adjusted without upsetting the others.
Check: C 3 = 3; H 8 = 8; O 10 = 10.
Every element must balance.

Result: C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O.

What it means: Only coefficients were changed. Changing a subscript (writing H₂O₂) would describe a different substance.

A different case

Moles from a balanced equation

Problem. For 2H₂ + O₂ → 2H₂O, how much water forms from 3 mol H₂ with excess O₂? How much O₂ is used?

n(H₂O) = 3 × 2 / 2 = 3 mol.
Ratio H₂O : H₂ = 2 : 2.
n(O₂) = 3 × 1 / 2 = 1.5 mol.
Ratio O₂ : H₂ = 1 : 2.

Result: 3 mol H₂O; 1.5 mol O₂ consumed.

What it means: In grams: 3 mol × 18.02 g/mol = 54.1 g of water.

More worked cases

Each case below uses a different skill. Every step and result is shown.

Limiting reactant in a precipitation

Problem. 0.20 mol Ag⁺ is mixed with 0.12 mol Cl⁻ (Ag⁺ + Cl⁻ → AgCl). Find the maximum amount of AgCl and what is left over.

The ratio is 1 : 1, and there is less Cl⁻.
Divide each by its coefficient (1): 0.20 vs 0.12.
Cl⁻ is limiting: n(AgCl) = 0.12 mol.
The limiting reactant fixes the product.
Ag⁺ left: 0.20 − 0.12 = 0.08 mol.
Excess reactant remains in solution.

Result: 0.12 mol AgCl (≈ 17.2 g, M = 143.32 g/mol); 0.08 mol Ag⁺ remains.

What it means: The reactant present in the larger amount is not automatically in excess; compare moles divided by coefficients.

Mass of precipitate from solution volumes

Problem. 50.0 mL of 0.100 M BaCl₂ is mixed with excess Na₂SO₄. What mass of BaSO₄ precipitates? (M(BaSO₄) = 233.4 g/mol)

Net ionic: Ba²⁺(aq) + SO₄²⁻(aq) → BaSO₄(s).
BaSO₄ is insoluble; Na⁺ and Cl⁻ are spectators.
n(Ba²⁺) = 0.100 × 0.0500 = 5.00 × 10⁻³ mol.
n = cV with V in litres.
n(BaSO₄) = 5.00 × 10⁻³ mol (1 : 1); m = 5.00 × 10⁻³ × 233.4 = 1.17 g.
Mole ratio, then mass.

Result: 1.17 g BaSO₄.

What it means: This gravimetric method is used to determine sulfate or barium content.

Acid–base titration

Problem. 25.0 mL of HCl is neutralised by 18.6 mL of 0.150 M NaOH. Find the HCl concentration.

HCl + NaOH → NaCl + H₂O (1 : 1). Net ionic: H⁺ + OH⁻ → H₂O.
Strong acid, strong base.
n(NaOH) = 0.150 × 0.0186 = 2.79 × 10⁻³ mol = n(HCl).
At equivalence the moles match.
c(HCl) = 2.79 × 10⁻³ / 0.0250 = 0.112 M.
c = n/V.

Result: [HCl] = 0.112 M.

What it means: For H₂SO₄, which gives two H⁺ per formula, the NaOH needed would double for the same concentration.

Identify oxidation and reduction

Problem. In Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s), which species is oxidised, which reduced, and what are the oxidising and reducing agents?

Zn: 0 → +2 (loses 2 e⁻): oxidised.
Oxidation number increases.
Cu: +2 → 0 (gains 2 e⁻): reduced.
Oxidation number decreases.
Oxidising agent: Cu²⁺ (it takes electrons). Reducing agent: Zn (it gives electrons).
Each agent is the reactant that causes the other change.

Result: Zn is oxidised (reducing agent); Cu²⁺ is reduced (oxidising agent).

What it means: Half-reactions: Zn → Zn²⁺ + 2e⁻ and Cu²⁺ + 2e⁻ → Cu. This reaction powers the Daniell cell (Electrochemistry).

Percent yield

Problem. A reaction should give 12.0 g of product (theoretical yield) but 10.2 g is collected. Find the percent yield.

10.2 / 12.0 × 100% = 85.0%.
Actual over theoretical.

Result: 85.0%.

What it means: A percent yield above 100% signals an error, such as a wet (not fully dried) product.

Common misunderstandings

Misunderstanding: Using mass ratios from the coefficients.
Correct idea: Coefficients are mole ratios. Convert grams to moles first.
Misunderstanding: Changing subscripts to balance an equation.
Correct idea: Only coefficients may change; subscripts define the substance.
Misunderstanding: Splitting solids, weak acids or water into ions.
Correct idea: Only soluble strong electrolytes are written as separate ions.
Misunderstanding: Using mL in n = cV.
Correct idea: Convert to litres.
Misunderstanding: Assuming the reactant with the smaller mass is limiting.
Correct idea: Compare moles divided by coefficients.
Misunderstanding: Oxidation = gaining oxygen only.
Correct idea: Oxidation is loss of electrons (increase in oxidation number), whether or not oxygen is involved.

Keep in mind

n = m / M; n = cVMoles
n_B = n_A × coefficient of B / coefficient of AMole ratio
% yield = actual yield / theoretical yield × 100%Percent yield
c_AV_A × b / a = c_BV_B for aA + bB → productsTitration at equivalence
soluble: Na⁺, K⁺, NH₄⁺, NO₃⁻ compounds (all); most Cl⁻, Br⁻, I⁻ (except Ag⁺, Pb²⁺, Hg₂²⁺); most SO₄²⁻ (except Ba²⁺, Pb²⁺, Ca²⁺ slightly). Insoluble: most CO₃²⁻, PO₄³⁻, OH⁻, S²⁻ (except with Group 1 and NH₄⁺)Solubility guidelines (summary)

Scope of this lesson

Balancing redox equations by the half-reaction method is shown in Electrochemistry (acidic solution).
Acid–base equilibria (pH of weak acids, buffers) are not covered in this lesson; check your outline.

Next: Properties of solutions. A solution’s composition is described by concentration units (molarity, molality, mass percent, mole fraction). Dilution conserves solute. Dissolved particles lower vapour pressure and freezing point and raise boiling point and osmotic pressure.

Original study text. Sources and credits.

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