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A level Chemistry Titration

Please can someone explain how to do titration calculations with past paper examples?!!
Original post by amazedmatt
Please can someone explain how to do titration calculations with past paper examples?!!

Do you have an example of recent questions you struggled with? Could you post the question along with your attempts at working?

Titration questions vary and it's impossible to know what you're weak at without further information
Reply 2
This is from an exam paper. Normally I would do (19.95x0.08)/1000 and work from there, but the answer I got makes no sense.

A solution of nitric acid, HNO3, of concentration 100 g dm–3, can be used to artificially age wood.
A sample of nitric acid, thought to be suitable for this use, was diluted by pipetting 10.00 cm3 of this acid into a 250 cm3 volumetric flask, adding deionised water and making the solution up to the mark. The solution was thoroughly mixed.
A titration was carried out using this diluted solution of nitric acid. The burette was filled with 0.0800 mol dm–3 sodium hydroxide solution and 25.00 cm3 of the diluted nitric acid was pipetted into each of three conical flasks. The following results were obtained.

Titration 1
Titration 2
Titration 3

Final burette reading / cm3

Initial burette reading / cm3

Volume added / cm3
The equation for the reaction is
HNO3 + NaOH NaNO3 + H2O
(a) Select the appropriate titres and calculate the mean titre in cm3.
(b) Calculate the concentration of the undiluted nitric acid in g dm–3. Give your answer to one decimal place.
Deduce whether this nitric acid is suitable for use in artificially ageing wood.
A long winded question.

You are right, calculate moles of NaOH. This tell you the moles of HNO3 in 25cm3 does it not? Always check the chemical equation to see whether acid and base react 1:1.

There are two more stages. We know the moles of HNO3 in 25cm3. How many moles are there in 250cm3? Remember the volumetric flask bit.

What mass is that of acid is that? Then you can calculate the conc. of the original acid in g/dm3 and compare it to the ideal value.
Reply 4

your calculation is incorrect; (19.95* 10^-3) 0.08 = 1.596*10^-3; you have 1.596* 10^-2

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