Calculate the amount of N added (mg) in the urea addition.

Divide the mg N added by 700 and multiply by 1000 = mg N/kg soil

Paul Milham sir, the formula u explained will tell me the N mg/kg but I wanted to calculate ammonia emission. My first day reading was 3.26 ml of sulphuric acid used for back titration of boric acid in which ammonia was trapped

Dear Sir,

1 ml of a 0.01 N H2SO4 releases 0.01 mmol H+ which neutralize the same amount of NH3/NH4 trapped in the boric acid. Thus, the amount of titrated H2SO4 in ml multiplied by 0.01 gives you the amount of NH3 trapped in the boric acid in mmol. If you only have titrated an aliquote of the boric acid you have to take this in consideration. Finally, you multiply your result with the molar weight of N to get the amount of emmitted Nitrogen in mg.

Kind regards.

Dear Dieter Lohr

Thanks so much for your answer, you help will resolve my issue

Muhammad Rashid Sir can you provide me the calculation of ammonia emission using urea.

For the correct final calculation you have to consider the "Equivalent concentration" . See in the attached pdf file under 'Examples' : There is shown that with sulfuric acid H2SO4 only 0.5 mol of H2SO4 are needed to neutralize 1 mol OH-.

The equivalent concentration is alreay considered in the calculation above because a 0.01 N H2SO4 and not a 0.01 M is used. However, if concentration of the acid is given in mol/l you have to consider their valency.

Although this is correct it of course easily can be mixed up.

Please help me also in the calculation of Ammonia Emission. I have used 2% Boric Acid to trap ammonia and titrated against 0.005M H2SO4 (instead of 0.01N).. I have applied N @ 250 mg kg-1 in glass jar containing 100 g soil.

Kind Regard@ Dieter Lohr

As mentioned by Dr. Zerche sulfuric acid is bivalent, thus a 0.005 M sulfuric acid is the same as 0.01 N one. So you just can use the formula given above. For the calculation you only need to know, how many mol H+ you have in your titer solution. This amount is equivalent to the amount of ammonia trapped in the boric acid. The concentration of the boric acid itself is meaningless.

Kind regards

Dieter Lohr

thank you so much dear sir,

the answer is very comprehensive now. and one more thing I want to ask is that I have to find out % N losses as NH3 Emission out of total applied N in the soil. So need your guidance about the formula that how can I convert calculated answer (from the above formula) to NH3 emission/100 g soil.

Best Regard@Dieter Lohr

according to your first post, you added 250 mg N per kg soil and put 100 g of this soil to your glas jar. Thus, the total amount of nitrogen in your jar (= 100 %) is 25 mg N. For example, if you need 10 ml of your 0.005 M sulfuric acid for titration, this equivalent to 10 ml x 0.01 mmol H+/ml => 0.1 mmol NH3 => 0.01 mmol N x 14 mg/mmol => 1.4 mg N/jar. Thus, for this example N losses are = 1.4/25 * 100 = 5.6 % of applied N.

Kind regards

Dieter Lohr