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m= 0,2v (0,1x4) - -
r = 0,2v 0,2v 0,2v
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s= - (4 - 0,2v) 0,2v
pH = 5, [H+] = 10^-5
[H+] = ka . n (asam) / n(garam)
10^-5 = 10^-5 . (4 -0,2v) / (0,2v)
0,2v = 4 - 0,2v
0,4v = 4
v = 10ml
NaOH + CH3COOH --> CH3COONA + H2O
0,2.v 4
0,2.v 0,2.v 0,2.v
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- 4-0,2v 0,2.v
H+=Ka A/G
10^-5=10^-5.4-0,2v/0,2.v
0,2.v=4-0,2V
0,4V=4
V=4/0,4=10 ml