Posted by Anonymous on Sunday, May 13, 2012 at 1:19pm.
No. You wrote the equilibrium for Ka for HPO4^2-. You donated a proton which makes it an acid. You want it to accept a proton.
HPO4^2- + HOH ==> H2PO4^-
So, the balance equation would be: HPO4^2- + H2O -> H2PO4 + H3O+?
and the kb would be: [H2PO4][H3O]/[HPO4^2-]
is that right?
Yes. That's the Kb expression; the value of Kb, which is not in the question but comes in handy when calculating pH of salts etc is Kb = (Kw/k2 for H3PO4)
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