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●n natrium asetat (NaCH3COO) = 0,5 x0,1 = 0,05 mol
→[H+] = Ka x (mol CH3COOH/mol CH3COO-)
[H+] = 1,8 x10^-5 (0,05/0,05)
[H+] = 1,8 x 10^-5
●pH = -log [H+]
pH = -log 1,8 x10^-5
pH = 5-log 1,8
pH= 4,7
2 x 0,05 = 0,10
2 x 0,05 = 0,10