Write a balanced equilibrium equation and kb expression for the reaction of the hydrogen phosphate ion (HPO4^2-) with water.
Answer: HPO4^2-(aq) +H2O(l) <--> PO4^3-(aq) +H3O+ (aq)
kb= [PO4^3-][H3O+]/ [HPO4^2-]
is that right?
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)
Oh, you're really getting into some chemical clownery! I have to say, your equation and kb expression are almost spot on! But there's just one tiny thing missing to make it complete. Let me help you out:
The balanced equilibrium equation should be:
HPO4^2-(aq) + H2O(l) ⇌ PO4^3-(aq) + H3O+(aq)
And you're right, the kb expression is:
kb = ([PO4^3-][H3O+]) / [HPO4^2-]
It looks like you're on the right track! Keep up the good work, chemistry comedian!
Yes, you are correct. The balanced equilibrium equation for the reaction between the hydrogen phosphate ion (HPO4^2-) and water is:
HPO4^2-(aq) + H2O(l) <--> PO4^3-(aq) + H3O+(aq)
And the expression for the base dissociation constant (kb) is:
kb = [PO4^3-][H3O+]/[HPO4^2-]
Well done!