1) A sample of 0.5843 g of vegetable food was analyzed using the Kjeldahl method to find the nitrogen content. The liberated ammonia is collected in 50.00 mL of 0.1062M hydrochloric acid. The excess acid is titrated using 11.89 mL of 0.0925M NaOH. Calculate the results of this analysis in terms of N%, urea and protein?

Question

2) Titration of the dilute solution of an unknown organic acid with 0.1084 M NaOH consumes 28.62 mL from the base to the phenolphthalein end point. If the sodium salt of an acid is 0.2110 g, what is the molecular mass and formula of this acid?

Answer 1

NH3 + HCl ==> NH4Cl
millimoles HCl initially = mL x M = 50.00 x 0.1062 = 5.310
mmols NaOH to titrate xs = 11.89 x 0.0925 = 1.0998
Subtract 5.310-1.0998 = 4.210 mmols used to neutralize the NH3 liberated in the Kjeldahl process. grams NH3 = mols NH3 x molar mass NH3 0.004210 x 17 = ?
%N = (grams NH3/mas sample)*100 = ?
%protein = %N x factor = ? You need to look up the factor. I think it is 6.89 but you can find it on Google or in your text/notes. For urea the factor is (molar mass N2/molar mass urea)

Answer 2

I don't understand 2. If you titrated the acid where does the Na salt fit in?