0.100 molar NaUH. We find that at the equivalence point, when we have added 25.0 mL of 0.100 molar NaOH, the pH is 9.5. We want to find the pka. The first step is to fill in this ICE chart. What do we put in for x in this chart?

Chemistry
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Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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Chapter15: Acid-base Equilibria
Section: Chapter Questions
Problem 19Q: Consider the titration of 100.0 mL of 0.10 M H3AsO4 by 0.10 M NaOH. What are the major species...
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We are titrating 25.0 mL 0.100 molar weak acid HA with
0.100 molar NaOH.
We find that at the equivalence point, when we have added
25.0 mL of 0.100 molar NaOH,
the pH is 9.5.
We want to find the pka. The first step is to fill in this ICE
chart.
What do we put in for x in this chart?
Hint, find pOH (14- pH) Then, [OH-] is 10^-pOH
[OH-] is x
(You do NOT have to do it, but would you know how to get pka
now?? Hint, you would need to find pKb)
C
E
A- (+ H2O)
0.050 M
© 2.5x106
O 1.5 x 10-3
○ 2.9 x 10-4
3.1 x 10-5
0.012
0.050 M-X
<=>
0
X
HA
OH-
0
X
Transcribed Image Text:We are titrating 25.0 mL 0.100 molar weak acid HA with 0.100 molar NaOH. We find that at the equivalence point, when we have added 25.0 mL of 0.100 molar NaOH, the pH is 9.5. We want to find the pka. The first step is to fill in this ICE chart. What do we put in for x in this chart? Hint, find pOH (14- pH) Then, [OH-] is 10^-pOH [OH-] is x (You do NOT have to do it, but would you know how to get pka now?? Hint, you would need to find pKb) C E A- (+ H2O) 0.050 M © 2.5x106 O 1.5 x 10-3 ○ 2.9 x 10-4 3.1 x 10-5 0.012 0.050 M-X <=> 0 X HA OH- 0 X
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