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*n C6H5COOH = 100 mL x 0,2 M = 20 mmol
*n NaOH = 100 mL x 0,1 M = 10 mmol
*Ka = 6 x 10^-5
C6H5COOH + NaOH --> C6H5COONa + H2O
m 20 10
b -10 -10 +10
s 10 - 10
[H+] = Ka x a/g
= 6 x 10^-5 x 10/10
= 6 x 10^-5
pH = 5 - log 6
mula2 20 mol 10 mol - -
terurai 10 mol 10 mol 10 mol 10 mol
______________________________-_______________________+
sisa 10 mol - 10 mol 10 mol
= 6 ×
= 6 ×
pH = - log 6 ×
= 5 - log 6