Jawabanmu

2014-03-23T20:05:36+07:00
Mmol H3COOH = ml.M = 50 x 0,1 = 5
mmol NaOH = ml. M = 100 x 0,1 = 10

H3COOH + NaOH --> H3COONa + H2O
mmol sisa NaOH = 10-5 = 5

NaOH = mmol sisa/ml tot = 5/150 = 0,033 M

OH- = b. mb = 1. 0,033 = 0,033 (3,3.10^-2)
pOH = - log 3,3.10^-2 = 2- log 3,3

pH = 14- pOH
pH = 14-(2-log 3,3) = 12 + log 3,3

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2014-03-23T20:16:33+07:00
[H]=0,1.50.1            sebelum reaksi => [H]=0,1.1=10^-1  => pH=-log 10^-1=1
[H]=5 M mol
[OH]=0,1.100.1  sebelum reaksi => [OH]=0,1.1=10^-1  => pH=-log 10^-1=1=>pOH=13
[OH]=10 M mol
Sisa 10-5=5 M mol
[OH]=sisa/vol campuran
[OH]=(5)/(50+100)
[OH]= 5/150
[OH]=0,03333333333333333333333  => 0,03
setelah reaksi
pOH=-log10^-2 -log3
pOH= 2- log3      =====> pH=14-(2-log3) =12+log3

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