In a experiment to determine the molecular weight and the Ka for ascorbic acid (vit. c.) a student dissolved 1.3713g of the monoprotic acid in water to make 50 mL of solution. The pH was monitored throughout the titration. The equivalence point was reached when 35.23 mL of the base was added.
1) Calculate the molecular mass of ascorbic acid.
2) When 20mL of NaOH was added during the titration the pH was 4.23. Calculate the Ka for ascorbic acid.
3) What was the pH at the equivalence point?
To find molecular mass of an acid in a titration experiment, one must use this formula:
Ca*Va=Cb*Vb
therefore
Ca=(Cb*Vb)/Va
C=concentration (molarity)
V=volume
a=acid
b=base
Thanks for the answer, but I haven't yet seen that formula yet, can you tell me the name for the formula?
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