[chaneliman]

Lawn fertiliser contains nitrogen. A 1.3g sample of lawn fertiliser was weighed and carefully transferred to a 250ml volumetric flask. The weighing bottle was washed with a little de-ionised water and added to the content, until it reached the calibration line.  A 20ml aliquot of this solution was added to a flask containing 20ml of .1M sodium hydroxide solution. 50ml of de-ionised water was than added. The flask wash heated until the reaction NH4+ (aq) +OH-(aq) = NH3 (aq) +H20) (l) was complete.  The burette was than titrated with .1M hydrochloric acid, using methyl red as indicator. The end point was reached when 44.3ml had been added.

Calculated the amount of NH4+ ions in the 1.3g fertiliser sample and the percentage by mass of nitrogen in the fertiliser, assuming all the nitrogen is present as ammonium ions.

(Can u plz show working?) thanks

[Arkcon]

OK, lets try to start this one.  Forget, for a little while, that it's a back titration to determine ammonia.  Let's find out first how much NaOH was there to be titrated by the HCl

[chaneliman]

i think its .002mol  to your question

[Borek]

Honestly - I don't understand the procedure. It doesn't make sense to me. Are you sure you have copied the procedure correctly?

Could be I have not woke up yet, but if the ammonia is left in the solution, you should use 20 mL of 0.1M acid for titration - you will be titrating both NaOH left and ammonia. 44.3 mL means there were 24.3 meq of other base present in the sample. It doesn't tell anything about the ammonia content.

.002 mol is OK.

[Arkcon]

Good, Borek has checked your math.  That saves me some time and trouble, I really don't care for titration calculations that I don't have to do, i.e. I'm not getting paid for.  Heh. B)

Now, Borek is also worried that the procedure isn't correct -- that NH3 remains in solution, and will reform NH4OH.  I just figure, if the procedure is done correctly, an excess of NaOH, and thorough heating to drive off all NH3 vapor, the only thing left is excess NaOH.  If you, chaneliman, think this might not be the case, then we can just stop.  But if this is a textbook problem, or a procedure you've performed we can run with it.  Although I think Borek and I would prefer to see the reaction: NH₄⁺ + OH^- --> NH₃(g)↑ + H₂O

OK, now, how much NaOH was added, in total to the flask, before the HCl titration, that tells us how much ammonia was consumed.

[Borek]

if the procedure is done correctly, an excess of NaOH, and thorough heating to drive off all NH3 vapor, the only thing left is excess NaOH

Excess... that's where the problem lies. 2 mmol of NaOH added, 4.43 mmol of HCl used in titration.

[Arkcon]

I had half a mind to sit down, and work the whole thing out by hand, so I wouldn't be talking out of my arse for this entire thread.

I'd also time myself, and write up a bill for chaneliman:, at $26.50 an hour.

Or at least bill him a percentage of the cost of a copy of ChemBuddy.