A 0.1103 M KOH solution was used to titrate a 0.552 g sample of ascorbic acid (V
ID: 899574 • Letter: A
Question
A 0.1103 M KOH solution was used to titrate a 0.552 g sample of ascorbic acid (Vitamin C) that was dissolved in 20 mL of water. The equivalence point occurred after 28.42 mL of the base was added. When only 10.0 mL of the base was added the pH was 3.72. Determine the molar mass and Ka for vitamin C.
Using the information from the problem above sketch a titration curve by calculating the pH: a. at the beginning of the titration b. at one-half of the equivalence point c. at the equivalence point d. at 5.0 mL beyond the equivalence point e. Pick a suitable indicator for this titration
Explanation / Answer
Moles NaOH = 0.1103 x 0.02842 L = 0.003135
= moles ascorbic acid
Molar mass = 0.552 g / 0.003135 mol =176.1 g/mol
Moles ascorbic acid = 0.552 / 176.1 =0.00313
moles NaOH = 0.0100 L x 0.1103 =0.001103
C6H8O6 + OH- >> C6H7O6- + H2O
Moles ascorbic acid in excess = 0.00313 - 0.001103 = 0.002027
Moles C6H7O6- = 0.001103
total volume = 20 + 10 = 30 mL = 0.030 L
concentration acorbic acid = 0.002027 / 0.030 =0.0676 M
concentration C6H7O6- = 0.001103 / 0.030 =0.0368 M
pH = pKa + log [C6H7O6-] / [C6H8O6]
3.72 = pKa + log 0.0368 / 0.0676
3.72 = pKa - 0.264
pKa =3.984
Ka = 10^-3.984 =1.03 x 10^-4
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