Nitrogen content in fertiliser via titration

[iwfc87]

Hey all, I think I need a little help cause I'm a little confused.

 

Well, heres the case: I'm trying to find the % content of Nitrogen in fertiliser by mixing an ammonium sulphate(NH4)2 SO2 fertiliser solution with a known volume of NaOH and then titrating it with a known concentration of HCl.

 

From the equation, if I'm not mistaken, the number of moles of NH4 OH produced would be the same as the NaOH (calculated from a known concentration)

 

2NaOH + (NH4)2SO2 => Na2SO2 + 2NH4 OH

 

This solution was heated and cooled, and then titrated with HCl and the volume required to change the colour of the indicator was recorded. I'm assuming the important part of the equation above is 2NH4 OH. Hence the equation during the titration would be:

 

HCl + NH4 OH => NH4Cl + H20

 

As the number of moles of HCl required to neutralise the solution is less then the number of moles of NH4 OH, I assume that there is an excess of NH4 OH. So I then used this smaller number of moles in my calculations. To me..I think this is the stuff up, as if there were an excess, the solution would still generally be basic.

 

Would this be right or have I made a stuff up? Please do guide me..I think I'm really lost..

[budullewraagh]

ammonium sulfate isn't [math](NH_4)_2SO_2[/math] it's [math](NH_4)_2SO_4[/math]

the reason why you dont need equal amounts of ammonium hydroxide and hydrogen chloride is that hydrogen chloride dissociates much easier than ammonium hydroxide does. as a result, it takes more moles ammonium hydroxide to bring the pH of a solution up than, say potassium hydroxide.

[iwfc87]

Whoops..my error..but yeah. So how am I going to solve the problem of calculating the % content of nitrogen in the fertiliser if I have so many different volumes and numebr of moles and all??