J.R. S. answered • 04/07/22
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H3PO4(aq) + H2O(l) ==> H3O+(aq) + H2PO4-(aq) ... Ka1
H2PO4-(aq) + H2O(l) ==> H3O+(aq) + HPO42-(aq) ... Ka2
HPO42-(aq) + H2O(l) ==> H3O+(aq) + PO43-(aq) ... Ka3
H3PO4(aq) ==> H+(aq) + H2PO4-(aq) ... Ka1
H2PO4-(aq) ==> H+(aq) + HPO42-(aq) ... Ka2
HPO42-(aq) ==> H+(aq) + PO43-(aq) ... Ka3
J.R. S.
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I did show each successive ionization losing an H+, but again, since this is a weak acid, they are probably looking for the equations without the H2O. I can go back and change it and see if that works.
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04/07/22
Christine T.
Hello again, the homework is saying that is wrong and this is the feedback. Each successive ionization reaction removes an H+ ion, so the final reaction should produce the unprotonated PO3−4 ion. Pay close attention to the charge on each species. For example, the diprotic acid H2CO3 ionizes to form H+ and HCO−3. Then, HCO−3 ionizes to form H+ and CO2−3.04/07/22