Titration calculations

I'm so terrible at these- I never manage all the steps necessary. It'd be great if you can help me out!

A fertiliser contains ammonium sulphate, (NH4)2SO4. A sample of the fertiliser weighing 3.8g was dissolved in water and volume made to 250cm3. To 25cm3 portions of this solution about 5cm3 (an excess) of aqueous methanal was added:
4NH+4 + 6HCHO ------> C6H12N4 + 4H+ + 6H2O
The liberated acid was titrated directly with 0.1moldm-3 aqueous sodium hydroxide. 28cm3 was required. Calculate % of ammonium sulphate in fertiliser.

Thanks so much!

Reply 1

Hawk

Moles of acid produced
Moles = Vol x Conc
= 0.028 x 0.1
= 0.0028

Moles of Ammonium ions in 25cm3 sample
Ratio of H+ : NH4+ = 1:1
Therefore moles of NH4+ = 0.0028

Moles of Ammonium ions in 250cm3
x 10 dilution factor = 0.0028 x 10 = 0.028

Moles of Ammonium Sulphate

.......................((aq)
(NH4)2SO4 ------------> 2(NH4)+ + (SO4)2-

Ratio of NH4+ to (NH4)2SO4 is 2:1
therefore moles of ammonium sulpahte = 0.028 / 2
= 0.014

Mass of Ammonium Sulphate

RMM = 132

Mass = Moles x RMM
= 0.014 x 132
= 1.848g

Percentage Purity

Total mass = 3.8g
Mass of Ammonium Sulphate = 1.848g

% Purtiy = 1.848/3.8 x100
= 48.6%

Hopefully someone can check this, it was a bit rushed.

Looks right.

Thanks a lot, Hawk, I really appreciate your help (as you have done many a times!)

I have a another question in continuation- any help would be great. (boy, do I hate these questions)

in a second determination of the ammonium ion content the same mass fertiliser (i.e. 3.8g) was treated with excess HaOH and heated. The ammonia liberated was passed into a known excess of HCl. The unreacted HCl was titrated with aqeous NaOH.
Calculation of % composition of fertiliser gave a value that was 5% lower than the value obtained by the method in previous question. Suggest reasons for this error other than those from measurements.

thanks again!

Reply 4

Mimo

Anyone? No?... :frown:

Reply 5

supreme

Hawk

I get the same.

Reply 6

supreme

Mimo

•

loss of solutions through washings

Reply 7

Hawk

Mimo

in a second determination of the ammonium ion content the same mass fertiliser (i.e. 3.8g) was treated with excess HaOH and heated. The ammonia liberated was passed into a known excess of HCl. The unreacted HCl was titrated with aqeous NaOH.
Calculation of % composition of fertiliser gave a value that was 5% lower than the value obtained by the method in previous question. Suggest reasons for this error other than those from measurements.
!

The Ammonia produced will not be passing through the delivery tube under any great pressure and so as the reaction begins to complete, the final few cm3 of ammonia will remain in the airspaces, as opposed to reacting with the HCl.

Reply 8

Mimo

Hawk

But it is not stated that a delivery tube is used, does it?

Reply 9

Hawk

Mimo

But it is not stated that a delivery tube is used, does it?

Mimo

The ammonia liberated was passed into a known excess of HCl

Well it says it is passed through the HCl, so even if they don't use a delivery tube there will still be some airspace that will be occupied by Ammonia.

Reply 10

Mimo

Hawk

Well it says it is passed through the HCl, so even if they don't use a delivery tube there will still be some airspace that will be occupied by Ammonia.

So it does. Right, thanks again! :smile:

Reply 11

arif

Hawk

hoe the hell n the world do u get a grade '' A " in bio give me some tips.
the teacher in our class cannoy teach her previous students got only c's and d's