Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: remeday86 on February 07, 2009, 05:11:51 AM
-
Ascorbic acid (vitamin C) is a diprotic acid having the formula H2C6H6O6. A sample of a vitamin supplement was analyzed by titrating a 0.5054 g sample dissolved in water with 0.0713 M NaOH. A volume of 10.80 mL of the base was required to completely neutralize the ascorbic acid. What was the percentage by mass of ascorbic acid in the sample? ____% asorbic acid
--My work
0.0713 M NaOH = mole NaOH/ 0.0108 L = 7.7004 E -4 mol NaOH
7.7004 E -4 mol NaOH x ( 1 mol asorbic acid / 1 mol NaOH ) = 7.7004 E -4 mol ascorbic acid
7.7004 E -4 mol ascorbic acid x 176.1248 g/mol ascorbic acid = 0.1356 g ascorbic acid
(0.1356 g ascorbic acid / 0.5054 g sample) x 100% = 26.8 % ascorbic acid
but...its wrong. I really tried... :-\ Where did I go wrong? Please explain. Thanks in advance!
-
Ascorbic acid (vitamin C) is a diprotic acid having the formula H2C6H6O6.
7.7004 E -4 mol NaOH x ( 1 mol asorbic acid / 1 mol NaOH ) = 7.7004 E -4 mol ascorbic acid
-
This is only a suggestion, instead of working with moles and molarilty, it would be very helpful if you understood the concept of equivalents and normality.
-
Ascorbic acid (vitamin C) is a diprotic acid having the formula H2C6H6O6.
7.7004 E -4 mol NaOH x ( 1 mol asorbic acid / 1 mol NaOH ) = 7.7004 E -4 mol ascorbic acid
Thanks!
7.7004 E -4 mol NaOH x ( 2 mol asorbic acid / 1 mol NaOH ) = 1.5401 E -3 mol ascorbic acid
1.5401 E -3 mol ascorbic acid x 176.1248 g/mol ascorbic acid = 0.2712 g ascorbic acid
(0.2712 g ascorbic acid / 0.5054 g sample) x 100% = 53.7% ascorbic acid
....but i still got it wrong. :-[
-
2 mol asorbic acid / 1 mol NaOH
Think again, every 1 mole of acid has 2 moles of H+. So if you use 2 moles of acid, you have 4 moles of H+ but you use only 1 mole of base which has only 1 mole of OH-. It can't neutralise.