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40+1=500M
41/500=M
M=0,082=8,2.10^-2
pH= 2 - log 8,2
ph=2=molarnya 0,01
masukin rumus molar campuran.
Mcampur=M1.v1+M2.v2/volume total
n=M.v
Mcampuran = (0,1.400+0,01.100)/500
=(40+1)/500
=41/500
=0,082
=0,08 kurang lbh
ph= -log[M]
ph campuran = 2-log8