kishka77 - 9-7-2011 at 11:07
Hi!
I'm having trouble solving this problem, even though I know it shouldn't be too hard:
If 10.7mL of 0.105 M NaOH is required to reach the first equivalence point of a solution of citric acid (H3C6H5O7), how many mL of NaOH are required
to completely neutralize this solution?
What I have so far:
10.7mL of 0.105 M NaOH works out to be 0.00112 mol NaOH, which I think is equal to mol citric acid at the equivalence point. But I have no clue what
to do next.
Help??
Cheers,
K
barley81 - 9-7-2011 at 11:23
Citric acid has three carboxy groups, so three equivalents of sodium hydroxide are needed to completely neutralize the solution (even though pH 7 will
be reached before all the NaOH is added). Since one equivalent of sodium hydroxide is in 10.7mL of the solution, you need 21.4 mL more of the
solution.
Hope this helps.
[Edited on 9-7-2011 by barley81]
kishka77 - 9-7-2011 at 11:39
Thanks!! I don't understand the connection between having 3 carboxy groups and needing 3 NaOH equivalents for neutralization though... Could you
explain why that is the case?
barley81 - 9-7-2011 at 11:49
The three
O
||
-C-OH
groups of citric acid dissociate in water (incompletely though) to form
O
||
-C-O (-)
and H<sub>3</sub>O<sup>+</sup> (hydronium ions).
Each hydronium ion reacts with one hydroxide ion (from the sodium hydroxide) to form two molecules of water.
kishka77 - 9-7-2011 at 11:51
Thank you so much! I understand it now.