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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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[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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