Mix an acid with a base and you get a salt and water. That much you have known for years. What is worth knowing now is that underneath all the different-looking equations, one single reaction is doing the work every time.
📚 What you need to know
acid + base → salt + water is the general pattern.
The reaction that actually happens is H+(aq) + OH–(aq) → H2O(l).
The ions that take no part are spectator ions, and they make up the salt.
The acid decides the salt: hydrochloric gives chlorides, nitric gives nitrates, sulfuric gives sulfates, ethanoic gives ethanoates.
Acids react with metal oxides, metal hydroxides, carbonates, hydrogencarbonates and ammonia.
Carbonates and hydrogencarbonates also give carbon dioxide.
The enthalpy of neutralisation for a strong acid and strong base is always about –57 kJ mol–1.
The reaction underneath
Take hydrochloric acid and sodium hydroxide. Both are fully ionised, so the flask really contains four separate ions in solution. Write them all out and something obvious jumps out.
Evaporate the water afterwards and the sodium and chloride ions are still there, now packed into a lattice. They never reacted — they just came along.
Spectator is a well-chosen word. Those ions were in solution before the reaction and are in solution after it, entirely unchanged. Calling the product “a salt” is really just naming whichever pair of spectators you happen to be left with.
Naming the salt
You can predict the salt without thinking hard: the metal comes from the base and the rest comes from the acid.
Acid
Ion it provides
Salt produced
Example
hydrochloric, HCl
Cl–
chloride
sodium chloride, NaCl
nitric, HNO3
NO3–
nitrate
copper(II) nitrate, Cu(NO3)2
sulfuric, H2SO4
SO42–
sulfate
magnesium sulfate, MgSO4
ethanoic, CH3COOH
CH3COO–
ethanoate
sodium ethanoate, CH3COONa
any acid + ammonia
NH4+ from the base
ammonium salt
ammonium chloride, NH4Cl
The reactions an acid will do
Four of these five produce water, so four of them are neutralisations. The reaction with a metal produces hydrogen instead, which is a redox reaction rather than a proton transfer to a base.
Metal oxide
2HCl(aq) + CaO(s) → CaCl2(aq) + H2O(l)
Metal hydroxide
H2SO4(aq) + Mg(OH)2(s) → MgSO4(aq) + 2H2O(l)
Ammonia is the odd one out and worth remembering separately. It has no hydroxide to give, so no water is produced — the proton simply transfers to the nitrogen lone pair: HCl(aq) + NH3(aq) → NH4Cl(aq). The salt is all you get.
Why the enthalpy change is always the same
Measure the heat released when any strong acid neutralises any strong alkali and you get roughly the same figure per mole of water formed: about –57 kJ mol–1. That looks like a coincidence until you remember the ionic equation. The spectators do nothing, so every one of these reactions is H+ + OH– → H2O. Same reaction, same energy.
With a weak acid the value comes out slightly less exothermic. Most of the acid is still undissociated at the start, and breaking those molecules apart to release H+ costs energy, which eats into the heat given out.
WORKED EXAMPLE
Write balanced equations for the following, and name the salt formed. (a) sulfuric acid + potassium hydroxide (b) nitric acid + calcium carbonate (c) hydrochloric acid + zinc oxide
(a) an acid and a hydroxideH₂SO₄ + 2KOH → K₂SO₄ + 2H₂Opotassium sulfateSulfuric acid supplies two protons, so it needs two hydroxides.(b) an acid and a carbonate2HNO₃ + CaCO₃ → Ca(NO₃)₂ + H₂O + CO₂calcium nitrateCarbonates always give a gas as well — do not leave the CO₂ out.(c) an acid and a metal oxide2HCl + ZnO → ZnCl₂ + H₂Ozinc chlorideA metal oxide gives a salt and water only. There is no hydrogen gas here, however much it looks like a metal.
WORKED EXAMPLE
You need to prepare magnesium nitrate. State an acid and two different bases that would work, and write one balanced equation.
Step 1 — split the salt name“Nitrate” comes from the acid, so use nitric acid. “Magnesium” comes from the base.acid: HNO₃Step 2 — pick the basesMgO, Mg(OH)₂ or MgCO₃Step 3 — one equation2HNO₃ + MgO → Mg(NO₃)₂ + H₂OMagnesium metal would also give the salt, but with hydrogen rather than water, so it is not a neutralisation.
WORKED EXAMPLE
Calculate the volume needed to exactly neutralise 25.0 cm3 of 0.100 mol dm–3 sodium hydroxide, using (a) 0.200 mol dm–3 hydrochloric acid (b) 0.200 mol dm–3 sulfuric acid
Step 1 — moles of basen(NaOH) = 0.0250 × 0.100 = 2.50 × 10⁻³ mol(a) HCl reacts 1 : 1n(HCl) = 2.50 × 10⁻³ molV = 2.50 × 10⁻³ ÷ 0.200 = 0.0125 dm³12.5 cm³(b) H₂SO₄ supplies two protonsn(H₂SO₄) = 2.50 × 10⁻³ ÷ 2 = 1.25 × 10⁻³ molV = 1.25 × 10⁻³ ÷ 0.200 = 0.00625 dm³6.25 cm³Exactly half, as you would expect. Always check the ratio in the balanced equation before dividing.
💡 Exam tip
Name the salt by splitting it: metal from the base, everything else from the acid.
Do not forget the CO2 when a carbonate or hydrogencarbonate is involved.
Balance carefully with diprotic acids and with hydroxides such as Ca(OH)2.
If asked for the ionic equation, cancel the spectators and write H+ + OH– → H2O.
Quote the enthalpy of neutralisation as per mole of water formed, not per mole of acid.
Include state symbols when the question asks for a full equation.
⚠️ Common mix-up
Calling acid + metal a neutralisation. No water is formed, so it is not.
Leaving out the carbon dioxide in a carbonate reaction, and then wondering why the equation will not balance.
Forgetting that H2SO4 needs two hydroxides, halving every subsequent volume.
Expecting water from ammonia. It has no OH– to give, so only the salt forms.
Cancelling the wrong ions in an ionic equation — only ions unchanged on both sides are spectators.
Assuming the enthalpy of neutralisation is identical for weak acids. It is slightly less exothermic.
Up next: pH Titration Curves — you have just calculated the exact volume needed to neutralise a solution. Now watch what the pH does on the way there, because it does not do what most students expect.
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