# Neutralization reactions and salts

> Chemistry · IB
> Source: https://www.owlsprep.com/study/ib-chemistry-sl-u6-neutralization-reactions-and-salts/

We cover core neutralization definitions, balanced equation writing, salt prediction, solubility rules and standard exam calculation workflows for IB SL Chemistry.

**Prerequisites:** [Acid and base core properties](https://www.owlsprep.com/study/ib-chemistry-sl-u8-properties-of-acids-bases/); [Ionic compound formula writing](https://www.owlsprep.com/study/ib-chemistry-sl-u3-ionic-compounds/)

## Learning objectives

- Define neutralization in terms of H+ and OH- ion behaviour
- Write balanced full and net ionic equations for neutralization reactions
- Predict the identity and solubility of salts formed from common acid-base pairs
- Calculate molar enthalpy changes for standard neutralization reactions

## Core Definition of Neutralization

Neutralization is the defining reaction between aqueous Brønsted-Lowry acids and bases, and forms the basis of all acid-base titration work you will encounter later in the course. The reaction is exothermic, meaning it releases heat to the surrounding solution.

**Neutralization** — The reaction between aqueous hydrogen ions from the acid and aqueous hydroxide ions from the base to form neutral water, alongside a dissolved ionic salt.

*Notation:* $H^+(aq) + OH^-(aq) \rightarrow H_2O(l)$

> **info**
>
> All strong acid-strong base neutralizations have a standard molar enthalpy of reaction of approximately -57 kJ mol⁻¹, as the only net reaction is formation of water.

**Worked example:** Write the net ionic equation for the reaction between hydrochloric acid and sodium hydroxide, and state the expected enthalpy change per mole of water formed.

1. Write full balanced equation: $HCl(aq) + NaOH(aq) \rightarrow NaCl(aq) + H_2O(l)$
2. Split all fully dissociated strong electrolytes into individual ions: $H^+(aq) + Cl^-(aq) + Na^+(aq) + OH^-(aq) \rightarrow Na^+(aq) + Cl^-(aq) + H_2O(l)$
3. Cancel spectator ions that appear unchanged on both sides: Remove $Na^+$ and $Cl^-$
4. Final net ionic equation: $H^+(aq) + OH^-(aq) \rightarrow H_2O(l)$
5. Enthalpy change = -57 kJ mol⁻¹

> **Exam tip:** You will lose marks if you include spectator ions in your net ionic equation response for 'show that' neutralization questions.

## Naming and Predicting Salt Products

Salts are named using a standard two-part convention: the first part comes from the cation of the base (or metal/metal carbonate reactant), and the second part comes from the anion of the parent acid.

| Parent Acid | Salt Anion Name | Example Salt |
| --- | --- | --- |
| Hydrochloric acid (HCl) | Chloride | NaCl, CuCl₂ |
| Sulfuric acid (H₂SO₄) | Sulfate | K₂SO₄, (NH₄)₂SO₄ |
| Nitric acid (HNO₃) | Nitrate | Ca(NO₃)₂ |
| Ethanoic acid (CH₃COOH) | Ethanoate | CH₃COONa |

**Worked example:** Predict the name and formula of the salt formed when excess copper(II) oxide reacts with dilute sulfuric acid.

1. Identify the base cation: Copper(II) from CuO, charge +2
2. Identify the acid anion: Sulfate from H₂SO₄, charge -2
3. Combine ions to get neutral formula: $CuSO_4$
4. Full salt name: Copper(II) sulfate

**Check your understanding**

Test your salt naming knowledge:

1. What salt is formed from the reaction of potassium hydroxide and nitric acid?

   - Potassium nitrite
   - Potassium nitrate
   - Potassium nitride
   - Potassium nitric acid

   *Why:* Nitric acid produces nitrate salts, not nitrite or nitride.

## Practical Salt Preparation Methods

IB SL Chemistry exams frequently test the two main practical methods for producing pure, dry salt samples, depending on whether the target salt is soluble or insoluble in water.

**Comparing methods**

The two standard IB-required preparation methods are outlined below:

- **Titration (for soluble salts from alkali + acid)** — Use a pipette to measure a fixed volume of alkali into a conical flask, add indicator, titrate with acid to the end point, repeat without indicator, evaporate half the water, leave to crystallise, filter and dry crystals between filter paper.
  - Pros: Produces very pure, large uniform crystals
  - Cons: Only works for soluble salts with a soluble base (alkali)

- **Precipitation (for insoluble salts)** — Mix two separate aqueous solutions of soluble reactants that contain the two required ions for the insoluble salt, filter the precipitate, wash with distilled water, dry in a warm oven.
  - Pros: Fast, high yield for insoluble products
  - Cons: Does not work for soluble salt products

**Worked example:** State the correct method to prepare a pure dry sample of insoluble barium sulfate.

1. Select two soluble starting reactants: Barium nitrate (soluble) and sodium sulfate (soluble)
2. Mix the two aqueous solutions in a beaker to form white barium sulfate precipitate
3. Filter the mixture to collect the solid precipitate
4. Wash the solid on the filter paper with small volumes of distilled water to remove soluble spectator ion impurities
5. Leave the washed precipitate in a low temperature oven to dry completely

## Enthalpy of Neutralization Calculations

Simple enthalpy of neutralization calculations are a very common 3-4 mark question on IB SL Paper 2, using standard calorimetry data. The core assumption is that the dilute aqueous solution has the same specific heat capacity as pure water, 4.18 J g⁻¹ °C⁻¹.

**Derivation:** Calculate molar enthalpy of neutralization

*Starting from:* $q = mc\Delta T$

1. Calculate total mass of combined acid and base solution, assume 1 cm³ = 1 g
2. Use $q = mc\Delta T$ to find total heat energy released in joules
3. Convert q to kJ, divide by the number of moles of water formed in the reaction
4. Add a negative sign to indicate the reaction is exothermic

*Conclusion:* Final value is $\Delta H_{neut}$ in kJ mol⁻¹

**Worked example:** 50 cm³ of 1.0 mol dm⁻³ HCl is mixed with 50 cm³ of 1.0 mol dm⁻³ NaOH, temperature rises from 22.0 °C to 28.5 °C. Calculate the molar enthalpy of neutralization.

1. Total mass of solution = 50 + 50 = 100 g
2. Temperature change $\Delta T$ = 28.5 - 22.0 = 6.5 °C
3. $q = 100 * 4.18 * 6.5 = 2717 J = 2.717 kJ$
4. Moles of water formed = moles of limiting reactant = 0.05 * 1.0 = 0.05 mol
5. $\Delta H_{neut} = -2.717 / 0.05 = -54 kJ mol^{-1}$ (2 s.f.)

## Common pitfalls

- **Wrong:** Forgetting to balance ionic charges when writing salt formulas, e.g. writing NaSO₄ instead of Na₂SO₄
  - Why it fails: Sulfate has a -2 charge, sodium only has +1, so two sodium ions are required for a neutral compound
  - Correct: Always cross the ion charges to find the correct subscript for each ion in the salt
- **Wrong:** Omitting the negative sign from enthalpy of neutralization final answers
  - Why it fails: Neutralization is always exothermic, so a positive value will lose you a mark automatically
  - Correct: Add a negative sign to your final ΔH value immediately after calculating the magnitude of heat released
- **Wrong:** Trying to use titration to prepare an insoluble salt
  - Why it fails: Precipitation is the only valid method for insoluble salts, as titration will leave the dissolved salt in solution
  - Correct: Check the IB solubility rules first to select the correct preparation method for your target salt
- **Wrong:** Including spectator ions in net ionic equations for neutralization
  - Why it fails: The IB mark scheme explicitly awards marks for removing unchanged spectator species
  - Correct: Write out all full dissociated ions first, then cross out any ions that appear on both sides of the reaction arrow
- **Wrong:** Using the volume of only one reactant instead of total combined volume for q=mcΔT calculations
  - Why it fails: Both the acid and base contribute to the total mass of solution that is being heated
  - Correct: Add the volumes of both reactants together to get the total mass of the solution (1 cm³ = 1 g)

## Cheatsheet

| Reaction Type | General Equation | ΔH_neut Approx |
| --- | --- | --- |
| Strong acid + strong alkali | HA + MOH → MA + H₂O | -57 kJ mol⁻¹ |
| Strong acid + weak base | HA + BOH → BA + H₂O | -50 to -55 kJ mol⁻¹ |
| Weak acid + strong alkali | HA + MOH → MA + H₂O | -50 to -55 kJ mol⁻¹ |
| Weak acid + weak base | HA + BOH → BA + H₂O | -45 to -50 kJ mol⁻¹ |

## What's next

Mastering neutralization reactions and salts is the critical foundation for your upcoming study of acid-base titrations, buffer systems and pH calculations, which make up 15-20% of the weight of IB Chemistry SL Paper 2. You will apply the net ionic equation and enthalpy calculation skills you learned here directly to longer, multi-mark exam questions on volumetric analysis, and to explain the behaviour of buffer solutions in industrial and biological contexts. Make sure you practice writing 3-4 different neutralization equations from memory before moving on.

---

From [OwlsPrep](https://www.owlsprep.com) — free study guides for A-Level, IB, AP and IGCSE, written against the official syllabus. Canonical page: https://www.owlsprep.com/study/ib-chemistry-sl-u6-neutralization-reactions-and-salts/