Double-Replacement and Neutralization Reactions

In a double-replacement (double-displacement) reaction, the positive and negative ions of two compounds exchange partners, AB + CD → AD + CB, and the reaction proceeds in solution when it forms a precipitate, a gas or a weak electrolyte such as water; a neutralization reaction is a double replacement between an acid and a base that gives a salt and water. In brief: HCl + NaOH → NaCl + H₂O is a neutralization, and its net ionic equation is H⁺ + OH⁻ → H₂O.

These reactions are the main ones in aqueous chemistry. They are used in titrations, water treatment, antacids and in the preparation of salts. This article explains the pattern and the three driving forces, shows how to write molecular, complete ionic and net ionic equations, covers neutralization reactions and their energy, works through calculations with volumes and concentrations, and gives some history.

What is the pattern?

Two ionic compounds in solution swap their cations (or anions). The cation of the first compound combines with the anion of the second, and vice versa. If the original compounds are AB and CD, the new compounds are AD and CB. Charges are used to write the correct formulas: the formulas of the products must be balanced by charge, for instance when calcium chloride reacts with sodium carbonate, the products are calcium carbonate (CaCO₃) and sodium chloride (NaCl). When the ions only mix and nothing happens, the result is no reaction, since all the ions remain in solution.

What drives a double replacement?

• Formation of a precipitate: an insoluble solid forms, removing ions from solution, as in AgNO₃ + NaCl → AgCl↓ + NaNO₃.
• Formation of a gas: a product decomposes to a gas that escapes, as in Na₂CO₃ + 2HCl → 2NaCl + H₂O + CO₂↑.
• Formation of a weak electrolyte or molecular compound: most often water, as in neutralization, but also weak acids such as acetic acid.

If none of these occurs, all ions remain free in solution, and there is no net reaction.

How do you write ionic equations?

1. Molecular equation: write and balance the complete formulas, with state symbols. Example: Pb(NO₃)₂(aq) + 2KI(aq) → PbI₂(s) + 2KNO₃(aq).
2. Complete ionic equation: split all strong electrolytes (soluble ionic compounds, strong acids and strong bases) into ions. Leave solids, gases, water and weak electrolytes as molecules. Pb²⁺(aq) + 2NO₃⁻(aq) + 2K⁺(aq) + 2I⁻(aq) → PbI₂(s) + 2K⁺(aq) + 2NO₃⁻(aq).
3. Net ionic equation: cancel the spectator ions that appear unchanged on both sides (K⁺ and NO₃⁻). Pb²⁺(aq) + 2I⁻(aq) → PbI₂(s).

What is a neutralization reaction?

An acid reacts with a base to produce a salt and water: HX + MOH → MX + H₂O. For a strong acid and a strong base, the net ionic equation is always the same: H⁺(aq) + OH⁻(aq) → H₂O(l). The spectator ions form the salt, which remains in solution unless it is insoluble or the water is evaporated. If the acid is weak, it is written as a molecule in the ionic equation: CH₃COOH(aq) + OH⁻(aq) → CH₃COO⁻(aq) + H₂O(l).

The enthalpy change of neutralization for a strong acid with a strong base is about −57 kJ per mole of water formed. The solution is neutral (pH 7 at 25 °C) at the equivalence point only for a strong acid with a strong base. Neutralization of a weak acid with a strong base gives a basic solution, since the anion of the weak acid reacts with water.

Examples of neutralization reactions

Worked examples

1. Volume of base needed. What volume of 0.200 mol/L NaOH neutralizes 25.00 mL of 0.100 mol/L HCl? n(HCl) = 0.02500 × 0.100 = 2.50 × 10⁻³ mol. The ratio is 1 : 1, so n(NaOH) = 2.50 × 10⁻³ mol. V = 2.50 × 10⁻³ ÷ 0.200 = 0.01250 L = 12.50 mL.
2. Diprotic acid. What volume of 0.200 mol/L NaOH neutralizes 25.00 mL of 0.100 mol/L H₂SO₄? n(H₂SO₄) = 2.50 × 10⁻³ mol. The ratio is 1 : 2, so n(NaOH) = 5.00 × 10⁻³ mol and V = 5.00 × 10⁻³ ÷ 0.200 = 25.0 mL.
3. Mass of salt. In example 1, the mass of NaCl formed is 2.50 × 10⁻³ × 58.44 = 0.146 g.
4. Temperature rise. The heat released in example 1 is 2.50 × 10⁻³ mol × 57.1 kJ/mol = 0.143 kJ = 143 J. The total volume is 37.50 mL (about 37.5 g). ΔT = 143 ÷ (37.5 × 4.18) = 0.91 °C.
5. Net ionic equation for a precipitate. Write the net ionic equation for the reaction of barium chloride with sodium sulfate: BaCl₂(aq) + Na₂SO₄(aq) → BaSO₄(s) + 2NaCl(aq). Net: Ba²⁺(aq) + SO₄²⁻(aq) → BaSO₄(s). Spectators: Na⁺ and Cl⁻.
6. Antacid calculation. A tablet contains 500 mg of Mg(OH)₂ (58.32 g/mol). It neutralizes acid: Mg(OH)₂ + 2HCl → MgCl₂ + 2H₂O. n = 0.500 ÷ 58.32 = 8.57 × 10⁻³ mol, which neutralizes 2 × 8.57 × 10⁻³ = 0.0171 mol of HCl.
7. No reaction. NaCl(aq) + KNO₃(aq): the possible products are NaNO₃ and KCl, which are both soluble, so all ions remain in solution and there is no reaction.

Where are neutralization reactions used?

• Antacids: magnesium hydroxide, aluminum hydroxide and calcium carbonate neutralize excess stomach acid.
• Titration: the volume of a standard base needed to neutralize an acid measures its concentration.
• Agriculture: lime (calcium oxide or hydroxide) is spread on acidic soil.
• Industry: acidic waste water is neutralized before release, and flue gases are treated with lime to remove sulfur dioxide.
• Everyday life: toothpaste (a mild base) neutralizes acids in the mouth, and vinegar neutralizes baking soda.
• Safety: small acid spills in the laboratory are neutralized with sodium bicarbonate.

How were these reactions studied?

Early chemists such as Robert Boyle in the 17th century described the ability of acids and bases to neutralize one another and the change of color of plant extracts. Jeremias Benjamin Richter showed in the 1790s that fixed masses of acids and bases neutralize each other, and in doing so laid the foundation of stoichiometry. Svante Arrhenius explained neutralization in 1884 as the combination of H⁺ and OH⁻ to form water.

Which mistakes are common?

Mistake 1: Splitting insoluble compounds or weak electrolytes into ions

Write solids, gases, water and weak acids as molecules in ionic equations.

Mistake 2: Writing the wrong formula for the products

Use the charges of the ions. For aluminum sulfate the formula is Al₂(SO₄)₃.

Mistake 3: Forgetting that a double replacement needs a driving force

If all of the products are soluble and no gas or water forms, there is no reaction.

Mistake 4: Assuming that every neutralization gives pH 7

The pH at the equivalence point depends on the strengths of the acid and base.