GCSEChemistryWJECRevision Notes2. Chemical Bonding, Application Of Chemical Reactions & Organic Chemistry2.2 Acids, Bases & SaltsNeutralisation Calculations

Neutralisation Calculations (WJEC GCSE Chemistry)

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Neutralisation Calculations

Higher Tier

  • Titrations involve a neutralisation reaction and is a method of analysing the concentration of solutions
  • Acid-base titrations are one of the most important kinds of titrations
  • They 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
  • Once a titration is completed and the average titre has been calculated, you can now proceed to calculate the unknown variable using the formula triangle as shown below

Concentration moles formula triangle, downloadable IB Chemistry revision notes

Formula triangle showing the relationship between concentration, number of moles and volume of liquid

Worked example

A solution of 25.0 cm3 of hydrochloric acid was titrated against a solution of 0.100 mol dm–3 NaOH. 

12.1 cm3 of NaOH was required for a complete reaction.

Determine the concentration of the acid.

Answer

Step 1: Write the equation for the reaction:

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

Step 2: Calculate the number of moles of the NaOH

  • Moles = (volume over 1000) x concentration
  • Moles of NaOH = 0.012 dm3 x 0.100 mol dm–3 = 1.21 x 10–3 mol

Step 3: Deduce the number of moles of the acid

  • Since the acid reacts in a 1:1 ratio with the alkali,  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 dm3)

Step 4: Find the concentration of the acid

  • Concentration = fraction numerator moles over denominator volume space open parentheses dm cubed close parentheses end fraction 
  • Concentration of HCl = fraction numerator 1.21 space cross times space 10 to the power of negative 3 end exponent space mol over denominator 0.025 space dm cubed end fraction= 0.0484 mol dm–3

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    Philippa

    Author: Philippa

    Expertise: Chemistry

    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.