Percentage deprotonation

Erin Jannusch
Posts: 14
Joined: Fri Sep 29, 2017 7:07 am

Percentage deprotonation

Why is it that when calculating percentage deprotonation we can substitute [H30+] with [A-]?

Andrea Grigsby 1I
Posts: 60
Joined: Fri Sep 29, 2017 7:03 am

Re: Percentage deprotonation

because a representative equation for acid reactions is
HA --> H3O+ + A-
where Kc=[H3O+][A-]/[HA]
if you were to use an ice table to calculate molarities, Kc would become: Kc= x^2 / [HA]-x as both have the same amount of moles
and because the concentration of [H3O+] and [A-] are the same, it can be substituted for each other

nathansalce 3e
Posts: 52
Joined: Thu Jul 27, 2017 3:01 am

Re: Percentage deprotonation

Yes i agree with above, both [H3O+] and [A-] start out with having a zero concentration. But after dissociation of the reactant in water, both [H3O+] and [A-] increase by x amount, as seen in a drawn ice table. If they both increase by x, you can equate them since their values are equal, allowing you to substitute one for the other.

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