December 02, 2021, 03:08:59 AM
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### Topic: Diprotic Buffer pKa2 when capacity is exceeded assistance  (Read 303 times)

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#### Calebowns

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##### Diprotic Buffer pKa2 when capacity is exceeded assistance
« on: October 10, 2021, 12:11:33 PM »
Hi all!
So today I ran into an issue with an acid/base buffer problem I was given. The data points are as follows:

Compound X has a pKa of 3.1, and a second pKa between 3-5.
152 mL of 0.14M KOH was added to 29 mL of 0.22M of compound X. This resulted in a pH increase to 4.2 (final pH). What is the pKa of the second group (second pKa). The initial pH is 3.1.

I "Think" the answer is the second pKa being 3.75, but I don't know how to get there because of the excess strong base. Any help would be GREATLY appreciated.

(My work shows the confusion, I run out of HA- when I do my math, so how can i use HH to find pKa2?)
« Last Edit: October 10, 2021, 12:35:56 PM by Calebowns »

#### Borek

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##### Re: Diprotic Buffer pKa2 when capacity is exceeded assistance
« Reply #1 on: October 10, 2021, 12:44:24 PM »
Something is wrong with the question. Excess of the base (152 mL × 0.14 M - 2 × 29 mL × 0.22M = 8.52×10-3 moles) is so large acid is completely neutralized and its Ka values are irrelevant. Final pH should be around 12.5.
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#### Calebowns

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##### Re: Diprotic Buffer pKa2 when capacity is exceeded assistance
« Reply #2 on: October 10, 2021, 12:49:55 PM »
Something is wrong with the question. Excess of the base (152 mL × 0.14 M - 2 × 29 mL × 0.22M = 8.52×10-3 moles) is so large acid is completely neutralized and its Ka values are irrelevant. Final pH should be around 12.5.

Thank you so much for the reply, Borek. That's what I was trying to tell my professor. The prof was adamant that the students should be able to get a second pKa of 3.75. However, I did the calculations a few times, calculated the pH to be what you got as well. Again much appreciated.
« Last Edit: October 10, 2021, 04:15:33 PM by Calebowns »