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*n CH3COOH = 0,5 mol
*n NaOH = 0,1 mol
*Ka = 10^-5
*pH ?
CH3COOH + NaOH --> CH3COONa + H2O
m 0,5 0,1
b -0,1 -0,1 +0,1
s 0,4 - 0,1
Bersisa asam lemah dan garamnya, sehingga:
[H+] = Ka x a/g
= 10^-5 x 0,4/0,1
= 4 x 10^-5
pH = 5 - log 4 (C)