Question

A 7.279 gram sample of meat was analyzed for its nitrogen content using kjeldahl method. Upon digestion, the ammonia liberated was collected in 250 ml of 0.855 M H3BO3. The resulting solution was titrated with 37.25 ml of 0.3122 M HCl using mixed indicator. Determine the % protein in the sample using 6.25 as a factor for meat product ? Answer: 13.98%

Answer 1

You only need to use the following equation:

%N = MHCl * (Vs/mg sample) * AM * 100

Now, why are we using this equation?. Mainly because the titration with HCl, gives us the number of moles of ammonia and then, the nitrogen contain. Because of this reactions:

After digestion of sample, the ammonia reacts with H3BO3:

NH₃ + H₃BO₃ (boric acid) -----> NH₄⁺ + H₂BO₃^-

Then, reacts with HCl:

H₂BO₃^- + H⁺ -----> H₃BO₃

All of thiese reactions have 1:1 relation, so, when you calculate the moles of HCl, you are actually calculating the number of moles of H3BO3 and then, the moles of ammonia, and when you use the atomic weight of N gives you the quantity of N in the ammonia. With this, and the data provided, let's calculate the %N and then, the %protein:

%N = 0.3122 mol/L * 37.25 mL * 14 g/mol * (1. L /1000 mL) / 7.279 g

%N = 2.2367%

%Protein = F * %N

%Protein = 6.25 * 2.2367 = 13.9794% or 13.98%

Hope this helps