Answer to Question #151770 in Physical Chemistry for olalekan Segun

Question

1cm^3 of 0.1M of NaOH=0.011g of casein. calculate the g/100cm^3 of the casein in milk.
values got from my results are below;
mass of the milk= 5.04g
mass of the casein precipitate=7.89g
volume(cm^3) of the acid used=43.50 of Hcl
Molar conc of Hcl = 0.1
Molar conc of NaOH=0.1

2) Also calculate the amount of casein in mg/L

Answer 1

Method is based on the precipitation of casein at the isoelectric point when titrated with acid.

V(NaOH, theor for solubilization) = V(HCl for precipitation) = 43.5 ml (because molar conc of HCl=molar conc of NaOH)

Proportion:

1 ml - 0.011 g

43.5 - x

x = 0.4785 g

d(milk) = 1 g/ml

Proportion:

0.4785 g of casein - 5.04 ml milk

x - 100

x = 9.49405 g/(100 ml) = 94941 mg/L

Answer:

9.49405 g/(100 ml)

94941 mg/L