Titration Calculations (AQA GCSE Chemistry): Revision Note

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 = $\frac{7.5\times {10}^{-3}}{0.0128}$ = 0.59 mol/dm³ to 2 dp

• Remember to convert cm³ to dm³ by dividing by 1000