. A 500.0 mL buffer solution contains 0.30 M acetic acid (CH3COOH) and 0.20 M sodium acetate (CH3COONa). What will the pH of this solution be after the addition of 20.0 mL of 1.00 M HCl solution

pH = 4.57

Explanation:

pH = pKa + log ([OAc⁺]/[HOAc])

Ka(HOAc) 1.8 x 10⁻⁵ => pKa = -log(1.8 x 10⁻⁵) = 4.74

[OAc⁻] = 0.20M

[HOAc] = 0.30M

pH = 4.74 + log([0.20]/[0.30]) = 4.47 + (-0.17) = 4.57

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