Nitrogen content in fertiliser using back titration?

[Malinn]

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 which means the excess OH- ions were titrated against 0.1 M HCl. I think?

NaOH- +HCl ---> NaCl +H2O

Is this equation correct? If so are the following calculations correct too?

n (HCl)= 0.1 x 0.0039

n(HCl)=0.0039

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

Edited March 20, 2012 by Malinn

[Bioc]

Hi, in case you still need this: you added an ammount of OH^- ions that can be summarized like this:

 

Total OH^- ions = (OH^- ions that reacted with NH₄⁺ to yield water) + (OH^- ions in excess that did not react)

 

So, when you add HCl, you can determine the excess of OH^- ions. The total OH^- ions is the amount of NaOH that you used to titrate the fertilizer.

 

You now have two of three variables

 

PD: Remember that the ratio of the reactions is 1:1, for example 1 mole of HCl reacts exactly with 1 mole of NaOH, and 1 mole of NaOH reacts exactly with 1 mole of NH₄⁺.