Titration to calculate the concentration of acids and alkalis

Category Interesting QuestionsPosted on:

You will meet a very wide variety of titrations on the A Level Chemistry course. Titration calculations can sometimes be quite complex (and worth 6, 7 or 8 marks!) but the basic procedure will always be the same:

  • First, use the data in the question to find the moles of one substance.
  • Then, use the balanced symbol equation, and information on sampling and dilution, to find the moles of another substance.
  • Finally, do something with your answer!

Before looking at more complex situations, let’s look at a much more straightforward example: using a titration to determine the concentration of an acid or a base.

(These types of titration questions sometimes appear as 6 mark questions on the GCSE Chemistry exams. At A Level, you’ll need to work a lot harder for 6 marks!)

Here is an example calculation to determine the concentration of an acid:

Q1: 25.00 cm3 of a H2SO4 solution is titrated with 40.70 cm3 of 0.1 mol dm-3 NaOH. Calculate the concentration of the H2SO4 solution.

2 NaOH + H2SO4 –> Na2SO4 + 2 H2O

Step 1

We know that moles = concentration x volume:

n NaOH = 0.1 x (40.70 / 1000) = 4.07 x 10-3 mol.

Step 2

NaOH and H2SO4 react in a 2:1 ratio so we can calculate the moles of H2SO4 reacted.

n H2SO4 = 0.5 x 4.07 x 10-3 = 2.04 x 10-3 mol

Step 3

We need to re-arrange the n = cv to make c the subject and also convert the 25 cm3 into 0.025 dm3 by dividing by 1000.

c H2SO4 = n / v = 2.04 x 10-3 / 0.025 = 0.081 mol dm-3

Here is another calculation to determine the concentration of a base. You will see that the structure of the calculation is identical:

Q2: 25.00 cm3 of a Ba(OH)2 solution is titrated with 20.50 cm3 of 0.01 mol dm-3 HCl. Calculate the concentration of the Ba(OH)2 solution

Ba(OH)2 + 2 HCl –> BaCl2 + 2 H2O

Step 1

We know that moles = concentration x volume:

n HCl = 0.01 x (20.50 / 1000) = 2.05 x 10-4 mol.

Step 2

Ba(OH)2 and HCl react in a 1:2 ratio, so we can calculate the moles of Ba(OH)2 reacted.

n Ba(OH)2 = 0.5 x 2.05 x 10-4 = 1.03 x 10-4 mol

Step 3

We need to re-arrange the n = cv to make c the subject and also convert the 25 cm3 into 0.025 dm3 by dividing by 1000.

c Ba(OH)2 = n / v = 1.03 x 10-4 / 0.025 = 0.0041 mol dm-3