Explain the estimation of nitrogen of an organic compound by
b) Kjeldahl's method

Explanation

Eslimation of `N_(2)` by Kjeldahl's method : In this method the organic compound is heated with conc, `H_(2)SO_(4)` in presence of a small amount of `CuSO_(4)`. Then `N_(2)` is quantitively converted into ammonium sulphate. The contents of the flask are transferred into another flask and heated with excess of `NaOH` solution to liberated is passed and absorbed in a knwon vol. of known conc. `H_(2)SO_(4)` (excess). Now the excess of acid remained after the neutrallsation by `NH_(3)` is lirated against a slandard solution of akali. From this, the amount of `H_(2)SO_(4)` used to neutralise `NH_(3)` is calculated. From this the mass of ammonia formed is calculated and from the `%` of `N_(2)` gas is calculated.

Organic compound `+H_(2)SO_(4)rarr(CH_(4))_(2)SO_(4)`
`(NH_(4))_(2)SO_(4)+2NaOH rarr Na_(2)SO_(4)+2H_(2)O+2NH_(3)`
`2NH_(3)+_(2)SO_(4)rarr (NH_(4))_(2)SO_(4)`.
Calculations :
Mass of organic compound = a g
Vol. of `H_(2)SO_(4)` Initially taken = V ml.
Molarity of `H_(2)SO_(4)=M`
Vol. of `NaOH` consumed after complete neutrallsation `=V_(1)ml`.
Molarity of `NaOH =M`
`2NaOH +H_(2)SO_(4) rarr Na_(2)SO_(4)+2H_(2)O`
Then from the formula,
`(NaOH)=(MV_(1))/(n_(1))=(MV_(2))/(n_(2))(H_(2)SO_(4))`
We get, `(MV_(1))/(2)=(MV_(2)/(1) or V_(2)=(V_(1))/(2)" ml. "`
`therefore" Vol. of "H_(2)SO_(4)" neutralilsed by "NH_(3)`
`=(V-(V_(1))/(2))ml`
`=(2(V-(V_(1))/(2))ml` of M molar `NH_(3)` solution.
`"1000 ml. of 1 M"NH_(3)` solution.
1000 ml. of `1 M NH_(3)` solution contains 17 g of `NH_(3)` of 14 g of `N_(2)`.
`therefore 2(V-(V_(1))/(2))` ml of M molar `NH_(3)` solution .............?
`=(14xxM xx2(V-(V_(1))/(2)))/(1000)g" of "N_(2)`
`therefore % " of Nitrogen"`
`=(14xxMxx2(V-(V_(1))/(2)))/(1000)xx(100)/(a)`