What is the ph of a 0.20 m solution of nh4cl? [kb(nh3) = 1.8 x 10^-5]

"NH4+ <> NH3 + H+  

The constant of this equilibrium is: K = Kw / Kb = 1 x 10^-14 / 1.8 x 10^-5 =5.56 x 10^-10  

5.56 x 10^-10 = x^2 / 0.20-x  

x = [H+] =1.1 x 10^-5 M  

pH = 5.0"