Chemistry 112

This is a standard weak acid equilibrium problem:

HCHO₂(aq) + H₂O(l) → ← H₃O⁺(aq) + CHO₂^–(aq)

K_a = [H₃O⁺]_e[CHO₂^–]_e                             [HCHO₂]_e = 1.8×10^–4

Initial0.050 0 0

Change–x +x +x

Equilibrium0.050 – x x x

Approximate? 0.050/1.8×10^–4 = 280 Yes

1.8×10^–4 = x²/0.050

x = 3.0×10^–3 = [H₃O⁺]_e

pH = –log[H₃O⁺]_e = –log(3.0×10^–3) = 2.52

 

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