Titration Calculations (Oxford AQA IGCSE Chemistry)
Revision Note
Written by: Philippa Platt
Reviewed by: Stewart Hird
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
Worked Example
A solution of 25.0 cm3 of hydrochloric acid was titrated against a solution of 0.100 mol/dm3 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 = () x concentration
Moles of NaOH = 0.012 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 (25.0 cm3 = 0.025 dm3)
Step 4: Find the concentration of the acid
Concentration =
Concentration of HCl = = 0.0484 mol/dm3
Worked Example
Calculating concentration
25.00 cm3 of 0.15 mol/dm3 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)
Answer:
Step 1: Calculate the number of moles of barium hydroxide
Moles of barium hydroxide = concentration x volume (dm3) = 0.15 x 0.025 = 3.75 x 10–3 mols
Step 2: Using the equation, calculate the number of moles of nitric acid
Moles of nitric acid = 3.75 x 10–3 x 2 = 7.5 x 10-3
The number of moles must be multiplied by 2 due to the 1:2 ratio
Step 3: Calculate the concentration of nitric acid
Concentration of nitric acid = = 0.59 mol/dm3 to 2 dp
Remember to convert cm3 to dm3 by dividing by 1000
Worked Example
Calculating the volume
Calculate the volume of 0.50 mol/dm3 nitric acid, HNO3, required to neutralise 25.00 cm3 of 0.80 mol/dm3 potassium hydroxide, KOH. Give your answer in cm3.
KOH (aq) + HNO3 (aq) → KNO3 (aq) + H2O (l)
Answer:
Step 1: Calculate the number of moles of potassium hydroxide
Moles of potassium hydroxide = concentration x volume (dm3) = 0.80 x 0.025 = 0.02 mols
Step 2: Using the equation, calculate the number of moles of nitric acid
Moles of nitric acid = 0.02 mols
The ratio is 1:1 so the number of moles of nitric acid is the same
Step 3: Calculate the volume of nitric acid in cm3
Volume of nitric acid == = 0.040 dm3
Volume of nitric acid = 40 cm3
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