Determine nitrogen content in lawn fertiliser using volumetric analysis

[Malinn]

Homework Statement

I undertook a practical experiment where a back titration was used to determine the amount of nitrogen in fertiliser.My titres are 3.9, 3.7 and 4.2 So I know this reaction occurred:
NH4+ + OH- ---> NH3 +H2O
I then boiled the solution to expel the Ammonia and the excess OH- ions were titrated against 0.1 M HCl.

OH- +HCl ---> Cl- +H2O
Is this equation correct? If so are the following calculations correct too?
n (HCl)= 0.1 x 0.0039
n(HVl)=0.0039


If this calculation is correct, what do I have to do next?

 

[Borek]

OH- +HCl ---> Cl- +H2O
Is this equation correct?

Yes and no. Yes, it will give the correct result, no there is no HCl in solution. It is completely dissociated to H⁺ and Cl^-.

If so are the following calculations correct too?
n (HCl)= 0.1 x 0.0039
n(HVl)=0.0039

Sorry, but neither makes sense to me. I can only guess in the first you are trying to calculate number of moles of HCl, I can only guess 0.1 is molarity of the acid used, but I don't understand what is 0.0039. You should use average volume, but 0.0039 is not the average of volumes you listed. And I can't even guess what you mean by n(HVl)=0.0039, unless you made a typo and math error in one line.

If this calculation is correct, what do I have to do next?

http://www.titrations.info/back-titration

 

[Malinn]

I was attempting to calculate the amount of mol of HCl was used and then using the mol ratio from the equation, calculate how many mole of nitrogen was present. The molarity of the HCl used was 0.1M and I was supposed to multiply it by 0.003933 (as this is the average of the titres) to find the mol (as the formula states n= concentration x volume (in litres)).

By doing this I hoped to find the mol of HCl used which is what was meant by n(HVl)=0.0039
the (HVl) was a typo for (HCl)

So what is the proper equation? Would this work in its place?
NaOH +HCl ---> NaCl +H2O

Sorry about the mistakes.