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pH=-log[H+]=-log4.10^-3=3-log4
b.[H+]=M.val=0,01.1=0,01
pH=-log[H+]=-log10^-2=2
c.[OH-]=M.val=0,03.2=0,06
pOH=-log[OH-]=-log6.10^-2=2-log6
pH=pkw-pOH=14-(2-log6)=12+log6