Chemistry(Please help)
posted by Hannah on .
What is the pH of 0.144M aqueous solution of ammonium nitrate?
I am not sure what to do. Would I do 10^0.144^ = 0.71?

No. You go through the hydrolysis equation.
.........NH4^+ + H2O ==> H3O^+ + NH3
initial..0.144............0.......0
change....x...............x......x
equil...0.144x............x......x
Ka for NH4^+ = (Kw/Kb for NH3) = (NH3)(H3O^+)/(NH4^+)
Substitute for NH3 and H3O^+ in the above. Substitute 0.144x for NH4 and solve for x = (H3O^+), then convert to pH. 
so the values for both NH3 and H3O are 0?

No. They are x. And you're solving for x. And you convert x to pH.

For the Ka of NH4^+^ I did 1.0e14/1.835 = 5.55e10. What do I do with this value?
So for (NH3)(H3O) / NH4 it would be
x^2/0.144x 
Kb = 5.55E10 so
5.55E10 = x^2/0.144x 
oh ok I get it, thank you!!