What is the equilibrium constant for the reaction of ammonium ion with formate ion?
I know that the ionization constant for ammonium ion is Ka=5.6e-10 and for formate Kb=5.6e-11.
I am not sure what to do next. Should I divide ka by kb?
I know that the ionization constant for ammonium ion is Ka=5.6e-10 and for formate Kb=5.6e-11.
I am not sure what to do next. Should I divide ka by kb?