Milk analysis? investigate the proteins

[CuriousSam]

For protein analysis, why is milk sample neutralised before the addition of formalin in Formol titration?

could we say that...

The proteins are too weak to be titrated directly with alkali,if formalin is added, it reacts with the -NH2 groups to form the methylene-amino(-N=CH2) group and the carboxyl group is then available for titration.

HOOC.CHR.NH2 + HCHO ----> HOOC.CHR.N=CH2 +H2O

HOOC.CHR.NH2(neutral) HCHO(formalin) HOOC.CHR.N=CH2(acidic)

Also , the proteins(which are acidic) in milk will react with NaOH will produce a higher amount
of volume used which would made higher protein content value thus giving an inaccurate
result.

Thank you very much and please help

 

[chemisttree]

I thought that a known amount of base was added to the neutralized protein and the excess base titrated after addition of formalin. Do I have the wrong method?

 

[CuriousSam]

The procedure is this
1) Add 0.5ml of 0.5% phenophthalein indictator and 0.4ml of neutral saturated potassium
oxalate.
2) Mix , allow to stand for a few minute and neutralise with 0.1M NaOH.
3) Add, 2ml formalin , allow to stand for a few minutes and titrate with the new acidity
produce with the 0.1N NaOH to the same pink colour(a ml) .Then carry out a blank titration
by replacing the milk sample with 10ml of water (b ml) .The protein content of the milk is
1.7 (a-b)%.


The question is , why must the milk sample is neutralised 1st?? is this base on the answers
i have derided in the 1st post or is it something else. pls enlighten me.

 

[chemisttree]

The procedure is this
1) Add 0.5ml of 0.5% phenophthalein indictator and 0.4ml of neutral saturated potassium
oxalate.
2) Mix , allow to stand for a few minute and neutralise with 0.1M NaOH.
3) Add, 2ml formalin , allow to stand for a few minutes and titrate with the new acidity
produce with the 0.1N NaOH to the same pink colour(a ml) .Then carry out a blank titration
by replacing the milk sample with 10ml of water (b ml) .The protein content of the milk is
1.7 (a-b)%.


The question is , why must the milk sample is neutralised 1st?? is this base on the answers
i have derided in the 1st post or is it something else. pls enlighten me.

What do you think you were neutralizing after you added the potassium oxalate? Would the amount of NaOH required be the same regardless of your sample? After you add the formalin, you have generated more acid. What does this new acidity correspond to? Would the acidity neutralized after you added the oxalate interfere with this measurement?