The pH of a 0.02M NH4Cl solution will be [given Kb(NH4OH)=10^−5 and log 2=0.301] The pH of a 0.02M NH4Cl solution will be [given Kb(NH4OH)=10^−5 and log 2=0.301] November 17, 2020 Category: Chapter 7 - Equilibrium , Chemistry , JEE Adv and Mains Chemistry 41 Years Solved Video Solutions 1979 - 2019 , Facebook Messenger WhatsApp Share this:TwitterFacebook Related