a level chemistry

nnn121337

i don't understand back titration question what the general way to do it.

Reply 1

Tulipbloom

Original post by nnn121337

i don't understand back titration question what the general way to do it.

Hey can you give me a question to solve and I will try my best to explain it to you

Reply 2

Pigster

Original post by nnn121337

i don't understand back titration question what the general way to do it.

Imagine trying to find out how much CaCO3 there is in a bit of limestone.

You might think you could add just the right amount of acid to neutralise the CaCO3 present. The problem is, the CaCO3 is solid, so the reaction will be slow. You might add some acid and the colour changes only for the 'excess' acid to be neutralised by the solid and the colour changes back again.

So... what you do it add a known, excess, amount of acid. That way, all the CaCO3 reacts and leaves behind some unreacted acid. You can do a titration to find out how many mol of acid there are in excess and then do some sums.

e.g. you have 1 g of limestone (if it were pure CaCO3 it would have 1/100.1 mol which I'll just round to 0.01)

CaCO3 + 2HCl -> CaCl2 + H2O + CO2

So, lets add 0.03 mol of HCl (we need a max of 0.02 mol, so this is an excess)

Some of the HCl will react (but since we don't know how many actual mol of CaCO3 there are, we don't know how many actually do)

So, we do a titration of the partially neutralised acid against 1 mol dm-3 NaOH and find that 12.55 cm3 of NaOH is required.

n(NaOH) = c x V = 1 x 12.55/1000 = 0.01255 mol = n(HCl) in excess

Since we started with 0.03 mol of HCl, then 0.01745 mol of HCl must have reacted with CaCO3

Since that is a 2:1 ratio, then 0.008725 mol of CaCO3 must have reacted with the acid. Which is 0.873373 g of CaCO3, i.e. 87.3% purity.

Reply 3

nnn121337

Original post by Tulipbloom

Hey can you give me a question to solve and I will try my best to explain it to you

student added 627 mg of hydrated sodium carbonate (Na2CO3.xH2O) to 200
cm3 of 0.250 mol dm–3 hydrochloric acid in a beaker and stirred the mixture.
After the reaction was complete, the resulting solution was transferred to a
volumetric flask, made up to 250 cm3 with deionised water and mixed thoroughly.
Several 25.0 cm3 portions of the resulting solution were titrated with 0.150 mol
dm–3 aqueous sodium hydroxide. The mean titre was 26.60 cm3 of aqueous
sodium hydroxide.
Calculate the value of x in Na2CO3.xH2O
Show your working.
Give your answer as an integer.
i get the question i done it but i would only get 5/7 marks. the only bit i not to sure about is dividing by 2 is it beacuse there two sodium compared to hcl even though normally sodium and hcl would react in one to one ration. the final bit i don't understand -H2O = 0.627/5.05 × 10–3
-106.0 = 18 (.16) x=1 is it using molecular formular or not.