Percentage Deprotonated

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Rachel Lipman
Posts: 45
Joined: Fri Sep 25, 2015 3:00 am

Percentage Deprotonated

Postby Rachel Lipman » Sun Nov 29, 2015 3:16 pm

When calculating the percentage, it is the concentration of A- divided by the initial concentration of HA times 100%. In the textbook it says it is also H3O+ divided by initial concentration. I am confused because when making the ice block chart, H3O+ is different than A-. Clarification please...

Anne Cam 3A
Posts: 71
Joined: Fri Sep 25, 2015 3:00 am

Re: Percentage Deprotonated

Postby Anne Cam 3A » Sun Nov 29, 2015 5:44 pm

The textbook does this assuming that [H3O+]=[A-], because according to the standard weak acid dissociation equation, HA(aq) + H2O(l) <--> H3O+(aq) + A-(aq), the mole ratio of H3O+ and A- is 1:1. Thus H3O+ and A- would be interchangeable in the deprotonation equation.

Rachel Lipman
Posts: 45
Joined: Fri Sep 25, 2015 3:00 am

Re: Percentage Deprotonated

Postby Rachel Lipman » Mon Nov 30, 2015 7:57 pm

Great thank you. But in other cases the ratio is not always one-to-one? And sometimes A- is not equivalent to H3O+?

Anne Cam 3A
Posts: 71
Joined: Fri Sep 25, 2015 3:00 am

Re: Percentage Deprotonated

Postby Anne Cam 3A » Wed Dec 02, 2015 6:14 pm

You would need to use [A-] for the deprotonation equation in that case, because the definition of deprotonation is the conversion of the acid HA to its conjugate base A-.


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