Titration Calculations (Oxford AQA IGCSE Chemistry)

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