Write a balanced equilibrium equation and kb expression for the reaction of the hydrogen phosphate ion (HPO4^2-) with water.

Question

Answer: HPO4^2-(aq) +H2O(l) <--> PO4^3-(aq) +H3O+ (aq)

kb= [PO4^3-][H3O+]/ [HPO4^2-]

is that right?

Answer 1

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^-

Answer 2

So, the balance equation would be: HPO4^2- + H2O -> H2PO4 + H3O+?

and the kb would be: [H2PO4][H3O]/[HPO4^2-]

is that right?

Answer 3

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)