Asked by Hannah

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?
Answered by DrBob222
No. You go through the hydrolysis equation.
.........NH4^+ + H2O ==> H3O^+ + NH3
initial..0.144............0.......0
change....-x...............x......x
equil...0.144-x............x......x

Ka for NH4^+ = (Kw/Kb for NH3) = (NH3)(H3O^+)/(NH4^+)
Substitute for NH3 and H3O^+ in the above. Substitute 0.144-x for NH4 and solve for x = (H3O^+), then convert to pH.
Answered by Hannah
so the values for both NH3 and H3O are 0?
Answered by DrBob222
No. They are x. And you're solving for x. And you convert x to pH.
Answered by Hannah
For the Ka of NH4^+^ I did 1.0e-14/1.83-5 = 5.55e-10. What do I do with this value?

So for (NH3)(H3O) / NH4 it would be

x^2/0.144-x
Answered by DrBob222
Kb = 5.55E-10 so
5.55E-10 = x^2/0.144-x
Answered by Hannah
oh ok I get it, thank you!!
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