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diket
CH₃COOH = 100 mL , 0,15M => 15mmol
NaOH = 50 mL , 0,2 M => 10 mmol
Ka = 1,7 .10⁻⁵
ditanya : PH campuran
jawab:
CH₃COOH + NaOH -----> CH₃COONa + H₂O
15 mmol 10 mmol
10 mmol 10 mmol 10mmol 10mmol
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5 mmol - 10 mmol 10 mmol =>sisa asam lemah (Buffer asam)
PH = - log Ka [a]/[g]
PH = -log 1,7.10⁻⁵ (5/10)
PH = - log 0,85 . 10⁻⁵
PH = -log 85 .10⁻⁷
PH = 7 - log 85
PH = 7 - 1,93
PH = 5,07
mol NaOH = 50 x 0.2 = 10 mmol
[H+] = Ka x a/g
= 1.7 x 10^-5 x 15/10
= 1.7 x 1.5 x 10^-5
= 2.55 x 10^-5
pH = - log 2.55 x 10^-5
= 5 - log 2.55
= 5 - 0.4
= 4.6