A protein was previously determined to contain 16.2 wt% nitrogen. A 691 μL aliquot of a solution containing the protein was digested in boiling sulfuric acid. The solution was made basic and the liberated NH3 was collected in 10.00 mL of 0.0390 M HCl . A volume of 4.16 mL of 0.0155 M NaOH was required to react with the excess, unreacted HCl . Calculate the protein concentration of the solution in units of milligrams per milliliter.

6.60mg N / mL

Explanation:

All nitrogen is converted in NH₃ that react with the HCl, thus:

NH₃ + HCl → NH₄⁺ + Cl⁻

In the problem, the excess of HCl reacts with NaOH, thus:

HCl + NaOH → NaCl + H₂O

The moles of NaOH = Moles of HCl in excess is:

4.16mL = 4.16x10⁻³L * (0.0155mol / L) = 6.448x10⁻⁵mol HCl in excess

Initial moles of HCl are:

10x10⁻³L * (0.0390mol / L) = 3.9x10⁻⁴moles HCl

That means the moles of HCl that reacts = Moles of NH3 = Moles of N are:

3.9x10⁻⁴ moles - 6.448x10⁻⁵moles = 3.2552x10⁻⁴ moles N.

To convert these moles to grams we need to use molar mass of N = 14.01g/mol:

3.2552x10⁻⁴ moles N * (14.01g/mol) = 4.56x10⁻³g * (1000mg / g) =

4.56mg of N

And volume in mL is:

691 μL * (1mL /  1000μL) = 0.691mL

Concentration in milligrams per mililiter is:

4.56mg N / 0.691mL =

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