What is the Molecular Weight and Ionization Constant of Ascorbic Acid?

AI Thread Summary
The discussion focuses on determining the molecular weight and ionization constant of ascorbic acid through titration with NaOH. The student successfully calculated the molecular weight using the relationship between moles of acid and base at the equivalence point. However, challenges arose in calculating the acid ionization constant (K_a) and the equilibrium constant due to the need for the concentrations of the acid and its conjugate base at a specific pH. The use of an ICE table and the Henderson-Hasselbalch equation was suggested to find the necessary concentrations for these calculations. The thread also includes a light-hearted commentary on the delayed responses and the concept of "necroposting."
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1) In an experiment to determine the molecular weight and the ionization constant for ascorbic acid (vitamin C), a student dissolved 1.3717 grams of the acid in water to make 50.00 milliliters of solution. The entire solution was titrated with a 0.2211-molar NaOH solution. The pH was monitored throughout the titration. The equivalence point was reached when 35.23 milliliters of the base had been added. Under the conditions of this experiment, ascorbic acid acts as a monoprotic acid that can be represented as HA.

(a) From the information above, calculate the molecular weight of ascorbic acid.

(b) When 20.00 milliliters of NaOH had been added during the titration, the pH of the solution was 4.23. Calculate the acid ionization constant for ascorbic acid.

(c) Calculate the equilibrium constant for the reaction of the ascorbate ion, A¯, with water.

(d) Calculate the pH of the solution at the equivalence point of the titration.

This was an AP Chemistry question in 1989.

I had no problems doing the first part. At the equivalence point, the number of moles of acid is equal the number of moles of base. I used this relationship to find out the number of moles, and grams / moles = molecular weight.

I have problems with part b, which consequently affects the correct answer to part c and d.

Because the pH is 4.23, I know the hydrogen ion concentration must be 10^{-4.23}. I know that the equilibrium expression is K_a = \frac{(A^{-})(H^{+})}{HA}. So, I know the numerator is (10^{-4.23})^2. But, what is the denominator?
 
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you should be able to use the henderson-hasselbach equation (assuming this is a buffer region), the sole information that you need now is the ratio of the acid and its conjugate base. You know how much base was added, and thus you should easily be able to calculate how much of the acid had reacted as well as how much of its conjugate base was produced.
 
this is easy.
use your ICE table to find the equilibrium concentration. then just minus off the concentration of H+ ions that reacted. you should get your concentration of ascorbic acid at equilibrium.
 
ZELEZNY said:
this is easy.
use your ICE table to find the equilibrium concentration. then just minus off the concentration of H+ ions that reacted. you should get your concentration of ascorbic acid at equilibrium.

I appreciate your solution, but I'm surprised that you were able to find this thread, since I created it last year when I was in AP Chemistry.

Speaking of which, damn - that was a long time ago :bugeye:
 
This has to be the first time I've seen someone return after several months to respond to a necropost on a really old thread.

Damn!
 
Gokul43201 said:
This has to be the first time I've seen someone return after several months to respond to a necropost on a really old thread.

Damn!

Had I not checked "e-mail notification," I would never have known.
 
Gokul43201 said:
This has to be the first time I've seen someone return after several months to respond to a necropost on a really old thread.

Damn!

'Necropost'... I love it!

(it is typically done by first time posters)

Welcome to the Forum, ZELEZNY!
 
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