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HW 12.3 c

Posted: Mon Dec 04, 2017 11:41 pm
by Lindsay H 2B
ex 12.3 c is to write the proton transfer equilibria for H2PO4^- and identify the conjugate acid-base pairs. in the case of H2PO4^- reacting with H2O, it seems like this ion could behave either as a base (by accepting a H+ from H2O and resulting in H3PO4), or as an acid (by donating a H+ and resulting in HPO4^2-). The solutions manual says that the equilibrium reaction produces HPO4^2- + H3O+ and therefore H2PO4^- is acting as an acid in this situation, and I'm not quite sure why it's this reaction and not the other one. Is there a simple way to tell whether a molecule like H2PO4^- would be more likely to donate a proton or accept one?

Re: HW 12.3 c

Posted: Tue Dec 05, 2017 6:43 am
by Mitch Mologne 1A
I believe that is because H2PO4^2- is a more stable ion than H3PO4 is. Thus, the atom would be more likely to donate, so it would gain more stability.

Re: HW 12.3 c

Posted: Tue Dec 05, 2017 10:11 am
by Michelle Dong 1F
In the question, it says to write the equilibria for the following acids, suggesting that H2PO4^- is acting as an acid when you write it in this case.