Percent Ionization

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Rachel Yu 1G
Posts: 113
Joined: Fri Aug 09, 2019 12:15 am

Percent Ionization

Postby Rachel Yu 1G » Mon Jan 20, 2020 10:19 pm

HA + H2O <--> A- + H3O+

Is the equation for % ionization %=[A-]/[HA] x 100% or % = [H3O+]/[HA]? Most of the time I found that [A-] and [H3O+] tend to have the same concentration because they have the same stoichiometric coefficients. However, if there is a reaction where this isn't the case, which concentration would be used as the numerator and why?

Ryan Lee 1E
Posts: 50
Joined: Sat Aug 17, 2019 12:16 am

Re: Percent Ionization

Postby Ryan Lee 1E » Mon Jan 20, 2020 10:37 pm

I think it would be the H30+ since that measures the actual amount of H+ that comes away from the acid.

Renee Grange 1I
Posts: 56
Joined: Fri Aug 30, 2019 12:16 am
Been upvoted: 1 time

Re: Percent Ionization

Postby Renee Grange 1I » Wed Jan 22, 2020 8:29 pm

From what I have in my notes I think its %=[A-]/[HA] x 100% , but I'm also confused about this.

Ashley Tran 2I
Posts: 108
Joined: Thu Jul 11, 2019 12:17 am

Re: Percent Ionization

Postby Ashley Tran 2I » Wed Jan 22, 2020 8:35 pm

It's [H3O+]/[HA] x 100 because the percent ionization is measuring how much the acid [HA] gave up protons [H3O+] the [A-] is not involved.

205405339
Posts: 77
Joined: Thu Jul 11, 2019 12:16 am

Re: Percent Ionization

Postby 205405339 » Wed Jan 22, 2020 8:52 pm

its [A-]/[HA] but if [H30+]=[A-] then [H30+]/[HA] will yield the same % value


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