sebanyak 0,25 mol natrium asetat dilarutkan ke dalam air sampai 250 mL. jika Ka asam asetat = 10 pangkat min 5. berapakah ph larutan?

M = n/V
M = 0.25 mol/0.25 L
M = 1 M
[H+] = (MxKa)^1/2
[H+] = (1X10^-5)^1/2
[H+] = 3.16 x 10^-3
pH = -log [H+]
pH = -log (3.16 x 10^-3)
pH = 3-log 3.16
pH = 2.5