Titration Calculations (OCR GCSE Chemistry A (Gateway))

Example 1:

A solution of 25.0 cm³ of hydrochloric acid was titrated against a solution of 0.100 mol / dm³ NaOH and 12.1 cm³ were required for 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 ÷ 1000) x concentration

Moles of NaOH = 0.0121 dm³ x 0.100 mol / dm³ = 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

Step 4: Find the concentration of the acid

Concentration = moles ÷ volume

Concentration of HCl =1.21 x 10^-3 mol  ÷ 0.025 dm³ = 0.0484 mol / dm³

Example 2:

25.0 cm³ of a solution of 0.05 mol / dm³  sodium carbonate was completely neutralised by 20.00 cm³ of dilute hydrochloric acid.

Calculate the concentration of the hydrochloric acid in mol / dm³.

Answer:

Make sure that you have a balanced equation and that your volumes have been converted to dm³ before you start the concentration calculations.

You are not given the concentration-moles formula triangle in exams so you have to learn it. It is a good idea to write it down before you start a problem, so you get all the parts in the correct place.

Finally, check that your answer makes sense! The concentrations will usually be similar orders of magnitude, so double check if the values seem very different for the two solutions.