Tentukan pH dari larutan 500 mL amonia 0,1 M (Kb = 4 x 10^-5)?

[OH-] = √ Kb x M basa = √ 4×10^-5 x 0,1 = √ 4 x 10^-6 
pOH = – log [OH-] = – log √ 4 x 10^-6  

OH=(kb.M)^1/2
OH=(4.10^-5.0,1)^1/2
OH=2.10^-3
pOH=3-log2
pH=14-pOH
pH=14-(3-log2)
pH=11+log2