Study Vault
All PostsFlashcardsResourcesAI Chat
  1. Home
  2. /↳All Posts
  3. /↳Chemistry
  4. /↳Salts and Titration
Study VaultStudy Vault

Free, comprehensive study notes for CSEC students.

matthewlloydw@gmail.com

Navigate

  • Home
  • All Posts
  • Flashcards
  • Resources
  • AI Chat

Sciences

  • Biology
  • Chemistry
  • Physics
  • Additional Mathematics
  • Mathematics
  • Information Technology

Humanities

  • English Language
  • English Literature
  • Spanish
  • Economics
  • Principles of Business
  • Principles of Accounting

Community

  • Contributors
  • Changelog
  • Suggest a Feature
  • My Suggestions
  • Bookmarks

© 2026 Matthew Williams. Made with other contributors for all.

Chemistry

Salts and Titration

PDF
Matthew Williams
|May 15, 2026|6 min read
CSEC ChemistryPaper 01Paper 02Salt PreparationSaltsSBASection ASolubilityTitrationVolumetric Analysis

Normal and acid salts, basicity of acids, solubility rules, methods of preparing soluble and insoluble salts, uses and dangers of salts, thermometric titration, indicator choice, and volumetric analysis calculations.

A salt is the ionic product of an acid-base reaction. Knowing which salts are soluble, which preparation method to apply, and how to perform a volumetric calculation are among the most tested skills in CSEC Chemistry.

Salts: Types and Basicity

A salt is the ionic compound formed when the replaceable hydrogen of an acid is fully or partially replaced by a metal ion or ammonium ion.

A normal salt is formed when all replaceable hydrogen atoms are replaced:

H2SO4+2NaOH→Na2SO4+2H2O\text{H}_2\text{SO}_4 + 2\text{NaOH} \rightarrow \text{Na}_2\text{SO}_4 + 2\text{H}_2\text{O}H2​SO4​+2NaOH→Na2​SO4​+2H2​O

An acid salt is formed when only some of the replaceable hydrogen is replaced — the salt still contains ionisable hydrogen:

H2SO4+NaOH→NaHSO4+H2O\text{H}_2\text{SO}_4 + \text{NaOH} \rightarrow \text{NaHSO}_4 + \text{H}_2\text{O}H2​SO4​+NaOH→NaHSO4​+H2​O

The basicity of an acid is the number of replaceable hydrogen atoms per molecule:

  • HCl — monobasic (one replaceable H)
  • H₂SO₄ — dibasic (two replaceable H)
  • H₃PO₄ — tribasic (three replaceable H)

Solubility Rules

The preparation method depends on whether the target salt is soluble or not, so check solubility first.

Knowing which salts dissolve in water determines which preparation method to use:

Salt typeSolubilityExceptions
NitratesAll solubleNone
ChloridesMost solubleAgCl (insoluble), PbCl₂ (insoluble)
SulfatesMost solubleBaSO₄, PbSO₄ (insoluble); CaSO₄ (slightly soluble)
CarbonatesMost insolubleNa₂CO₃, K₂CO₃, (NH₄)₂CO₃ (soluble)
HydroxidesMost insolubleNaOH, KOH (soluble); Ca(OH)₂ (slightly soluble)
Sodium, potassium, ammonium saltsAll solubleNone

Methods of Salt Preparation

There are three preparation methods. Which one applies depends on whether the salt is soluble or insoluble, and whether the base used to make it is soluble or not.

Soluble Salts — Using Excess Insoluble Base

Used when the base (metal oxide, hydroxide, or carbonate) is insoluble.

  1. Add excess base to the acid until no more dissolves and the solid remains.
  2. Filter off the excess base.
  3. Evaporate the filtrate to concentrate it.
  4. Allow to crystallise; filter and dry the crystals.

Excess base ensures all the acid reacts. It can be filtered off because it is insoluble and the salt is in solution.

Soluble Sodium, Potassium, or Ammonium Salts — By Titration

Used when the base is NaOH, KOH, or NH₃(aq) — all soluble, so they cannot be filtered off.

  1. Use an indicator to find the exact volume of acid that neutralises a known volume of alkali.
  2. Repeat the titration using the same volumes but without the indicator (the indicator would contaminate the product).
  3. Evaporate and crystallise the pure salt.

Insoluble Salts — By Precipitation

  1. Mix solutions of two soluble salts, each containing one of the required ions.
  2. The insoluble salt precipitates immediately.
  3. Filter, wash with distilled water, and dry.

Pb(NO3)2(aq)+Na2SO4(aq)→PbSO4(s)↓+2NaNO3(aq)\text{Pb(NO}_3)_2\text{(aq)} + \text{Na}_2\text{SO}_4\text{(aq)} \rightarrow \text{PbSO}_4\text{(s)}\downarrow + 2\text{NaNO}_3\text{(aq)}Pb(NO3​)2​(aq)+Na2​SO4​(aq)→PbSO4​(s)↓+2NaNO3​(aq)

Uses and Dangers of Salts

Salts are not only laboratory products. Many are everyday substances in food, medicine, and construction, though some carry health risks at high doses.

SaltUse
NaCl (sodium chloride)Food seasoning, preservation, de-icing roads
NaHCO₃ (baking powder)Leavening agent in baking, antacid
CaCO₃ (calcium carbonate)Cement and concrete manufacture
CaSO₄·½H₂O (plaster of Paris)Casts, moulds, wall plaster
MgSO₄·7H₂O (Epsom salts)Laxative, bath soaks
NaNO₃ (sodium nitrate)Fertiliser, food preservative
NaNO₂ (sodium nitrite)Cured meat preservation

Dangers: sodium nitrate and nitrite, when consumed in excess, are potentially carcinogenic and have been implicated in brain damage in infants. Their use in food preservation is regulated.

Neutralisation and Titration

Making a salt by titration requires knowing exactly when neutralisation is complete, and calculating concentrations from the volumes used. That is what volumetric analysis covers.

Neutralisation is always exothermic — the formation of water from H⁺ and OH⁻ releases heat. This can be used in a thermometric titration: the temperature peaks at the equivalence point. Plotting temperature against volume gives two straight lines whose intersection marks the end point.

In standard titrations, an indicator signals the end point by colour change:

  • Phenolphthalein: colourless (acid) to pink (alkali)
  • Methyl orange: red (acid) to yellow (alkali)

Volumetric Analysis Calculations

n=c×V(V in dm3,  1 dm3=1000 cm3)n = c \times V \quad (V \text{ in dm}^3, \; 1 \text{ dm}^3 = 1000 \text{ cm}^3)n=c×V(V in dm3,1 dm3=1000 cm3)

Example

25.0 cm³ of 0.10 mol dm⁻³ NaOH was titrated with HCl. 20.0 cm³ of acid was required. Find the concentration of the HCl.

Equation: HCl + NaOH → NaCl + H₂O (1 : 1 ratio)

Moles NaOH = 0.10 × 0.025 = 0.0025 mol
Moles HCl = 0.0025 mol (1 : 1 ratio)
Concentration HCl = 0.0025 / 0.020 = 0.125 mol dm⁻³

Example

20.0 cm³ of H₂SO₄ was neutralised by 30.0 cm³ of 0.20 mol dm⁻³ NaOH. Find the concentration of the H₂SO₄.

Equation: H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O (1 : 2 ratio)

Moles NaOH = 0.20 × 0.030 = 0.006 mol
Moles H₂SO₄ = 0.006 / 2 = 0.003 mol (1 : 2 ratio)
Concentration H₂SO₄ = 0.003 / 0.020 = 0.15 mol dm⁻³

Previous in syllabus order
Acids and Bases
Next in syllabus order
Oxidation and Reduction