Chemistry 112

This is a typical common ion equilibrium problem:

NaCHO₂(aq) → Na⁺(aq) + CHO₂^–(aq)

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

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

Initial0.050 0 0.050

Change–x +x +x

Equilibrium0.050 – x x 0.050 + x

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

1.8×10^–4 = x(0.050)/0.050

x = 1.8×10^–4 = [H₃O⁺]_e

pH = –log[H₃O⁺]_e = –log(1.8×10^–4) = 3.74

 

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