Question

The Kb of ammonia is 1.8 x 10–5. What is the percent ionization of a solution of ammonia, 0.50 mol L–1 NH3, that contains 0.020 mol L–1 of ammonium nitrate, NH4NO3?

when i use the ICE table i don't end up with an unknown so im not sure what to do. Am i to try and make the equation with NH3 + NH4NO3 on the same side?

Answer 1

----NH3 + HOH => NH4OH(fm NH4NO3)
-----NH4OH <=> NH4^+ + OH^-
I:-->0.50M --------0.20M-----0
C:---(-x)---------- (+x)----(+x)
E:--(0.5-x)--------(0.2+x)---(x)
---->(~0.5)-------->(~0.2)---(x)

Kb=[NH4^+][OH^-]/[NH4OH]
1.8X10^-5= (0.2)(X)/(0.5)
Solve for 'x'
=>x=(0.50)(1.8x10^-5)/(0.2)
=>x=4.5x10^-5M (This is the fraction of (OH^-)ion delivered into solution on ionization of NH4OH)
%Ionization=[(4.5x10^-5)/(0.50)]x100%
= 0.009% Ionized