12.81

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Jose_Arambulo_2I
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Joined: Wed Sep 21, 2016 2:59 pm

12.81

Postby Jose_Arambulo_2I » Wed Nov 30, 2016 1:41 am

Hello! So in this question we're asked to find the pH of a diprotic acid, like H2CO3. I understand that two steps take place in which a proton is donated in each step, but I do not understand why answer book says that "the second ionization can be ignored" since the the second ka is smaller. I thought that Ka2 is by nature leas that ka1 because it's harder to lose that photon, but does that mean that every second ionization in a diprotic acid can be ignored? Or is there some specific indicator to figure out when to ommit the second ionization?

Maggie Bui 1H
Posts: 35
Joined: Fri Jul 22, 2016 3:00 am
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Re: 12.81

Postby Maggie Bui 1H » Wed Nov 30, 2016 9:41 am

In this specific problem, the answer book probably says to ignore the second ionization because it is too small take into account. I don't think there's a specific cutoff of when to omit the second ionization, but I think if it doesn't contribute to changing the significant figures of the answer, it's okay to ignore the second ionization.


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