Hitunglah pH dari larutan 100ml NH4OH 0,1M (kb=1,8 x 10^-5

100 mL NH4OH 0,1 M
Kb = 1,8 × 10^-5

pH ... ?

[OH^-] = √( Kb × Mb )
[OH^-] = √( 1,8 × 10^-5 × 0,1 )
[OH^-] = √( 1,8 × 10^-6 )
[OH^-] = 10^-3 × √1,8

pOH = - log [OH^-]
pOH = - log 10^-3 × √1,8
pOH = 3 - log √1,8

pH = 14 - pOH
pH = 14 - (3 - log √1,8)
pH = 14 - 3 + log √1,8
pH = 11 + log √1,8
pH = 11 + log 1,34