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Chemistry-acid base equilibria

by chromeX
Tags: base, chemistryacid, equilibria
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chromeX
#1
Nov15-06, 10:12 PM
P: 4
p-Hydroxybenzoic acid, HOC6H4COOH, is a weak diprotic acid. A 25.00 mL aliquot of a dilute solution of HOC6H4COOH is titrated with a 0.0200 M NaOH solution. The first equivalence point was reached after 16.24 mL of the NaOH solution was added.
a. If the values of the pH after 8.12 and 16.24 mL of base added were 4.57 and 7.02, respectively determine Ka1 and Ka2 for p-hydroxybenzoic acid.
b. Calculate the pH of the solution at the beginning of the titration.
c. Calculate the pH of the solution at the second equivalent point of the titration.
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physics girl phd
#2
Nov15-06, 11:18 PM
physics girl phd's Avatar
P: 936
What things have you tried so far?
chromeX
#3
Nov16-06, 09:38 AM
P: 4
Here's what i've tried- the moles of H2A at the beginning is .01624L times 0.0200M. You also need 32.48mL of base to reach 2nd equiv point. the part that trips me up is that HA-is an amphoteric species.
I used ICE with HA- + H20 -> A2- + H3O+, finding the conc. of [H3O+] as 10^-7.02 with .012992M as conc. of HA-. I ended up with 7.02 * 10 ^-13 as my answer for Ka1, but I don't think that's right. I also don't know what to do for Ka2.

chromeX
#4
Nov16-06, 02:22 PM
P: 4
Chemistry-acid base equilibria

i figured out how to get ka1, but i'm still having trouble with ka2.

how i got ka1-
use ICE with H2A + H2O -> HA- + H3O+, using the values for halfway to the first equivalence point.
Ka1=([HA-][H3O+]/[H2A])
[H2A]=[HA-]=.0049. (cancels out), leaving Ka1 = [H3O+] = 10^-4.57=2.69*10^-5.

Any tips for solving Ka2?
chromeX
#5
Nov16-06, 02:32 PM
P: 4
found ka2 to be 3.39*10^-10.


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