Answer:
431.38 mg protein / mL
Explanation:
This is an example of the <em>Kjeldahl method</em>, for nitrogen determination. All nitrogen atoms in the protein were converted to NH₃ which then reacted with a <u>known excess of HCl</u>. This excess was later quantified via titration with NaOH.
First we calculate the <u>total amount of H⁺ moles from HCl</u>:
- 0.0388 M HCl * 10.00 mL = 0.388 mmol H⁺
Now we calculate the <u>excess moles of H⁺</u> (the moles that didn't react with NH₃ from the protein), from the <u>titration with NaOH</u>:
- 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 <u>moles of H⁺ that reacted with NH₃ from the protein</u>:
- 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 <u>calculate the mass of N</u> 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 <u>calculate the mass of protein</u>:
- 4.382 mg * 100/15.7 = 27.91 mg protein
Finally we <u>calculate the protein concentration in mg/m</u>L, <em>assuming your question is in 647 μL</em>, we first convert that value into mL:
- 647 μL *
0.647 mL
- 27.91 mg / 0.647 mL = 431.38 mg/mL