This is my acquired titration curve, using 0.4982 M NaOH as titrant and 0.803 g

Question

This is my acquired titration curve, using 0.4982 M NaOH as titrant and 0.803 g of the unknown amino acid. We are supposed to use this to calculate molecular weight, which in turn will be used to determine the identity of the amino acid . From looking at the graph, this appears to me to be a triprotic amino acid. I calculated the molecular weight of the amino acid to be 80.3 g/mol, which is closest to either alanine or glycine. However, neither of these are triprotic amino acids so I'm assuming my calculations are wrong. I could have gone wrong in calculating the volume of NaOH used (delta V). I got this value by subtracting Eq. pt 1 from Eq. pt 3, but I'm not sure if this is correct. For reference, here are the other points on the graph (which I calculated, so error could also exist here):

pKa1=2.35 pKa2=6.10 pKa3=9.25

Eq. pt1=5.35 mL Eq. pt2=15.50 mL Eq. pt3=26.25

So to obtain my delta V, I did (26.25-5.35) = 20.9 mL

Then to calculate moles of NaOH used, I did (0.4982 mol/L)(0.0209L) = 0.020 mol NaOH

If mol NaOH = mol amino acid, then the molecular weight of the amino acid should be:

(0.803g)/(0.010mol) = 80.3 g/mol

Explanation / Answer

pKa1=2.35 pKa2=6.10 pKa3=9.25

Eq. pt1=5.35 mL Eq. pt2=15.50 mL Eq. pt3=26.25

So to obtain my delta V, I did (26.25-5.35) = 20.9 mL

Then to calculate moles of NaOH used, I did (0.4982 mol/L)(0.0209L) = 0.010 mol NaOH (you did mistake here)

If mol NaOH = mol amino acid, then the molecular weight of the amino acid should be:

(0.803g)/(0.010mol) = 80.3 g/mol

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.