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15.0,1=20 . M2
M2=0,075
2. 50.0,1/50.0,1 x 1,8.10^-5=1,8.10^-5
pH = 5 -log 1,8
NaOH: b=1 Vb=15ml Mb=0,1M
dit: Ma?
jwb: axVaxMa = bxVbxMb
1x25xMa = 1x15x0,1
25xMa=1,5
Ma=1,5/25
Ma=0,06M
2.dik: CH3COOH: V=50ml=0,05L M=0,1M
CH3COONa: V=50ml=0,05L M=0,1M
Ka=1,8x10^-5
dit: pH?
jwb: nCH3COOH=MxV=0,1x0,05=5x10^-3mol
nCH3COONa=MxV=0,1x0,05=5x10^-3mol
[H+]=ka x mol asam/mol garam
=1,8x10^-5 x 5x10^-3 / 5x10^-3
=1,8x10^-5 x 1 = 1,8x10^-5
pH=-log[H+]
=-log 1,8x10^-5
=5 - log 1,8