A typical protein contains 16.2wt%nitrogen. A 0.500 ml aliquot of protein solution was digested, and the liberated NH3 was distilled into 10.00 ml of 0.0214 M HCI. The unreacted HCI required 3.26 ml of 0.0198 M NAOH for complete titration. Find the concentration of protein (mg protein /ml) in the original sample.

The 4 similar figures carried out is as follows

Explanation:

[HCL] initial = (10.00ml) (0.02140M) = 0.2140mmol

NaOH = (3.26ml) (0.0198M) = 0.0645mmol

NH3 produced in HCl will the difference = 0.2140mmol minus 0.0645mmol = 0.1495mmol

Since 1mol N in protein produces 1 mol of NH3 there's 0.1495mmol of N

MW N = 14.0067mg by mol

(0.1495mmol) (14.0067mg by mmol) = 2.094mg N

Since the protein contains 16.2% N = 2.094mg N by 0.162 mg N = 12.93mg protein

12.95mg protein by .500l = 25.85mg protein by lit

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