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

Concentration, moles, volume formula triangle

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

Worked example

Calculating concentration

A solution of 25.0 cm³ of hydrochloric acid was titrated against a solution of 0.100 mol/dm³ NaOH.

12.1 cm³of 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 = ( $\frac{volume}{1000}$ ) 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 = $\frac{moles}{volume ({dm}^{3})}$
- Concentration of HCl = $\frac{1.21 \times 10^{- 3} mol}{0.025 {dm}^{3}}$ = 0.0484 mol/dm³

Worked example

Calculating concentration

25.00 cm³of 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^–3mols
Step 2: Using the equation, calculate the number of moles of nitric acid
- Moles of nitric acid = 3.75 x 10^–3x 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 = $\frac{7.5 \times 10^{- 3}}{0.0128}$ = 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 cm³of 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 = 0.02mols
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 = $\frac{moles}{concentration}$ = $\frac{0.02}{0.50}$ = 0.040 dm³
- Volume of nitric acid = 0.040 dm³ x 1000 = 40 cm³

Titration Calculations (CIE IGCSE Chemistry)

Revision Note