The pH of 0.01 M NaOH(aq) solution will be [2019]
(3)
NaOH0.01 M→Na++OH-0.01 M
∴ [OH-]=0.01 M
∴ pOH=-log[OH-]=-log(0.01)=2
∴ pH=14-pOH=14-2=12