Question

A protein was previously determined to contain 15.7 wt% nitrogen. A 647 m 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 0.0388 M HCI. 3.83 mL of 0.0196 M NaOH was required to react with the excess, unreacted HCl. Calculate the protein concentration of the solution in units of milligrams per milliliter. Number 0 mg protein mL solution

Answer 1

Answer:

431.38 mg protein / mL

Explanation:

This is an example of the Kjeldahl method, for nitrogen determination. All nitrogen atoms in the protein were converted to NH₃ which then reacted with a known excess of HCl. This excess was later quantified via titration with NaOH.

First we calculate the total amount of H⁺ moles from HCl:

• 0.0388 M HCl * 10.00 mL = 0.388 mmol H⁺

Now we calculate the excess moles of H⁺ (the moles that didn't react with NH₃ from the protein), from the titration with NaOH:

• HCl + NaOH → H₂O + Na⁺ + Cl⁻

• 0.0196 M * 3.83 mL = 0.075068 mmol OH⁻ = 0.0751 mmol H⁺

Now we substract the moles of H⁺ that reacted with NaOH, from the total number of moles, and the result is the moles of H⁺ that reacted with NH₃ from the protein:

• HCl + NH₃ → NH₄⁺ + Cl⁻

• 0.388 mmol H⁺ - 0.0751 mmol H⁺ = 0.313 mmol H⁺ = 0.313 mmol NH₃

With the moles of NH₃ we know the moles of N, then we can calculate the mass of N present in the aliquot:

• 0.313 mmol NH₃ = 0.313 mmol N

• 0.313 mmol N * 14 mg/mmol = 4.382 mg N

From the exercise we're given the concentration of N in the protein, so now we calculate the mass of protein:

• 4.382 mg * 100/15.7 = 27.91 mg protein

Finally we calculate the protein concentration in mg/mL, assuming your question is in 647 μL, we first convert that value into mL:

• 647 μL * $\frac{1L}{10^{6}uL} *\frac{1000mL}{1L} =$ 0.647 mL

• 27.91 mg / 0.647 mL = 431.38 mg/mL