National 5ChemistrySQARevision NotesChemical Changes & StructureAcids & BasesAcid-base Titrations

Acid-base Titrations (SQA National 5 Chemistry): Revision Note

Exam code: X813 75

Philippa Platt

Written by: Philippa Platt

Reviewed by: Richard Boole

Updated on

Acid-base titration

  • Titrations are a analytical method of analysing the concentration of solutions

  • Titrations can determine exactly how much alkali is needed to neutralise a quantity of acid – and vice versa

  • You may be asked to calculate:

    • The moles present in a given amount

    • The concentration or volume required to neutralise an acid or a base

  • A few drops of a suitable indicator are added to show the end-point:

    • The end-point in a titration is when the indicator changes colour showing that the acid and base have completely reacted with each other

  • Specific apparatus must be used both when preparing the standard solution and when completing the titration, to ensure that volumes are measured precisely

Required apparatus

Five labelled laboratory items: 1. Beaker with liquid, 2. Burette, 3. Pipette, 4. Conical flask with liquid, 5. Volumetric flask with liquid.
Some key pieces of apparatus used to prepare a volumetric solution and perform a simple titration

  1. Beaker

  2. Burette

  3. Volumetric pipette

  4. Conical flask

  5. Volumetric flask

The apparatus set up

Diagram of a titration setup with a burette on a stand above a conical flask, showing labelled components and process description for mixing solutions.
A burette is used to add an accurate, variable volume of one solution to a precise, fixed volume of another solution (measured with a pipette) in a conical flask.

Acid-base titration calculations

  • Once a titration is completed and the average titre has been calculated, you can calculate the unknown variable using the formula triangle as shown below:

Formula triangle for acid-base titration calculations

Triangle chart illustrating the relationship of moles, concentration (mol l⁻¹), and volume (l) in equations.
A chemistry triangle diagram showing the relationship between moles, concentration, and volume in litres, with equations for calculations.

  • Once a titration is completed and the average titre is known, you can calculate the unknown concentration or volume

  • There are two ways to do this:

    1. The "step-by-step" method

    2. The "SQA formula" method

The step-by-step method

  • It always follows three steps:

  1. Calculate the moles of the 'known' solution:

    • Use the solution for which you know both the concentration and the volume

    • Use the formula n = c × v

  2. Deduce the moles of the 'unknown' solution, using the mole ratio:

    • Look at the balancing numbers in the balanced chemical equation

    • Use the ratio to work out the number of moles of the 'unknown' solution that must have reacted

  3. Calculate the unknown concentration or volume:

    • Now that you know the moles of the 'unknown' solution, you can use a rearranged formula to find what you're looking for:

    • c = n / v (to find concentration)

    • v = n / c (to find volume)

Worked Example

A solution of 25.0 cm3 of hydrochloric acid was titrated against a solution of 0.100 mol l-1 NaOH. 

12.1 cm3 of NaOH was required for a complete reaction.

Determine the concentration of the acid.

[3]

Answer:

  • The balanced chemical equation is:

HCl (aq) + NaOH (aq) → NaCl (aq) + H2O (l)

  1. Calculate the number of moles of the 'known' solution, NaOH:

moles = open parentheses volume over 1000 close parentheses x concentration

moles of NaOH = 0.012 l x 0.100 mol l-1 

moles of NaOH = 1.21 x 10–3 mol [1 mark]

  1. Deduce the number of moles of the 'unknown' solution, HCl:

    • The acid reacts in a 1:1 ratio with the alkali

    • Therefore, the number of moles of HCl is also 1.21 x 10–3 mol

This is present in 25.0 cm3 of the solution (25.0 cm3 = 0.025 l) [1 mark]

  1. Calculate the unknown concentration:

concentration = moles over volume 

concentration of HCl = fraction numerator 1.21 cross times 10 to the power of negative 3 end exponent space mol over denominator 0.025 space straight l end fraction

concentration of HCl = 0.0484 mol l-1 [1 mark]

The SQA formula method

  • The SQA Data Booklet provides a single formula that combines all three of the steps above into one equation

    • This is often a faster method to use in an exam

fraction numerator c subscript 1 v subscript 1 over denominator n subscript 1 end fraction = fraction numerator c subscript 2 v subscript 2 over denominator n subscript 2 end fraction

  • Values in the equation:

    • c = concentration (in mol l-1)

    • v = volume (in litres, l)

    • n = number of moles from the balanced equation (the balancing number)

    • '1' and '2' just stand for the two different solutions (e.g., acid and base)

How to use the SQA formula method

  • Write the balanced chemical equation for the reaction

    • This step may not be necessary if you are given the chemical equation

  • List all six variables (c1, v1, n1, c2, v2, n2), identifying the one you need to find

    • Convert all volumes to litres (l) by dividing by 1000.

  • Substitute the known values into the formula

  • Rearrange the formula for the unknown value

  • Solve the formula for the unknown value

Examiner Tips and Tricks

A common mistake is forgetting to convert volumes

  • Whichever method you use, the volume (v) must be in litres (l) for the calculation

  • You must divide any volumes given in cm3 by 1000

Worked Example

Calculating concentration

25.00 cm3 of 0.15 mol l-1 barium hydroxide, Ba(OH)2, was required to neutralise 12.80 cm3 of nitric acid, HNO3 , during a titration.

Calculate the concentration of HNO3 that was used. Give your answer to 2 decimal places.

Ba(OH)2 (aq) + 2HNO3 (aq) → Ba(NO3)2 (aq) + 2H2O (l)

[3]

Answer:

  1. List all variables:

    • c1 (HNO3) = ?

    • v1 (HNO3) = 12.80 cm3 = 0.0128 l

    • n1 (HNO3) = 2 (from 2HNO3)

    • c2 (Ba(OH)2) = 0.15 mol l-1

    • v2 (Ba(OH)2) = 25.00 cm3 = 0.025 l

    • n2 (Ba(OH)2) = 1 (from 1Ba(OH)2)

[1 mark]

  1. Substitute the values into the equation:

fraction numerator c subscript 1 v subscript 1 over denominator n subscript 1 end fraction = fraction numerator c subscript 2 v subscript 2 over denominator n subscript 2 end fraction

fraction numerator c subscript 1 cross times 0.0128 over denominator 2 end fraction = fraction numerator 0.15 cross times 0.025 over denominator 1 end fraction

  • This simplifies to:

fraction numerator c subscript 1 cross times 0.0128 over denominator 2 end fraction = 0.15 x 0.025

  1. Rearrange the equation:

c1 (HNO3) = fraction numerator 0.15 cross times 0.025 cross times 2 over denominator 0.0128 end fraction

[1 mark]

  1. Solve the equation:

c1 (HNO3) = 0.59 mol l-1 to 2 dp

[1 mark]

Worked Example

Calculating volume

Calculate the volume of 0.50 mol l-1 nitric acid, HNO3, required to neutralise 25.00 cm3 of 0.80 mol l-1 potassium hydroxide, KOH. Give your answer in cm3.

KOH (aq) + HNO3 (aq) → KNO3 (aq) + H2O (l)

[3]

Answer:

  1. List all variables:

    • c1 (HNO3) = 0.50 mol l-1

    • v1 (HNO3) = ?

    • n1 (HNO3) = 1 (from HNO3)

    • c2 (KOH) = 0.80 mol l-1

    • v2 (KOH) = 25.00 cm3 = 0.025 l

    • n2 (KOH) = 1 (from KOH)

[1 mark]

  1. Substitute the values into the equation:

fraction numerator c subscript 1 v subscript 1 over denominator n subscript 1 end fraction = fraction numerator c subscript 2 v subscript 2 over denominator n subscript 2 end fraction

fraction numerator 0.50 cross times v subscript 1 over denominator 1 end fraction = fraction numerator 0.80 cross times 0.025 over denominator 1 end fraction

  • This simplifies to:

0.50 x v1 = 0.80 x 0.025

  1. Rearrange the equation:

v1 (HNO3) = fraction numerator 0.80 cross times 0.025 over denominator 0.50 end fraction

[1 mark]

  1. Solve the equation:

v1 (HNO3) = 0.040 l

v1 (HNO3) in cm3 = 0.040 x 1000 = 40 cm3

[1 mark]

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    Author
    Reviewer
    Philippa Platt

    Author: Philippa Platt

    Expertise: Chemistry Content Creator

    Philippa has worked as a GCSE and A level chemistry teacher and tutor for over thirteen years. She studied chemistry and sport science at Loughborough University graduating in 2007 having also completed her PGCE in science. Throughout her time as a teacher she was incharge of a boarding house for five years and coached many teams in a variety of sports. When not producing resources with the chemistry team, Philippa enjoys being active outside with her young family and is a very keen gardener

    Richard Boole

    Reviewer: Richard Boole

    Expertise: Chemistry Content Creator

    Richard has taught Chemistry for over 15 years as well as working as a science tutor, examiner, content creator and author. He wasn’t the greatest at exams and only discovered how to revise in his final year at university. That knowledge made him want to help students learn how to revise, challenge them to think about what they actually know and hopefully succeed; so here he is, happily, at SME.