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pOH = -logOH- =1,3 pH =14-1,3 = 12,7
b. NaOH + CH3COOH ---> CH3COONa + H2O
m 2,5mmol 3mmol
r 2,5 2,5 2,5 2,5
s - 0,5 2,5
[H+] = Ka x mol asam/mol garam
= 10^-4 x 0,5/2,5
=2x10^-5
pH = 5-Log2
c. Mr NaCH3COO = 70
mol = 0,14gram/70 = 2x10^-3
[NaCH3COO] = mol/v =2x10^-3/0,1 = 0,02M
pH = -log 0,02 = 1,69
d. [H+] =
pH = 5,15