Question

Given:
pH = 3.97
use:
pH = -log [H+]
3.97 = -log [H+]
[H+] = 1.07*10^-4 M

CH3COOH dissociates as:
CH3COOH <-> CH3COO- + H+

So,
Ka = [CH3COO-][H+] / [CH3COOH]
1.8*10^-5 = [CH3COO-] * (1.07*10^-4) / 0.031
[CH3COO-] = 5.21*10^-3 M

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