Titration Calculations (Cambridge (CIE) IGCSE Chemistry)
Revision Note
Written by: Caroline Carroll
Reviewed by: Stewart Hird
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Titration calculations
Extended tier only
Titrations are a method of analysing the concentration of solutions
Acid-base titrations are one of the most important kinds of titrations
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
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 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/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
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 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 = 0.040 dm3 x 1000 = 40 cm3
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