Chemistry 112

NaNO₂(aq) → Na⁺(aq) + NO₂^–(aq)

HNO₂(aq) + H₂O(l) → ← H₃O⁺(aq) + NO₂^–(aq)

K_a = [H₃O⁺]_e[NO₂^–]_e                            [HNO₂]_e = 7.2×10^–4

Initial0.150 0 0.200

Change– x +x +x

Equilibrium0.150 – x x 0.200 + x

Approximate? 0.150/7.2×10^–4 = 210 > 100 Yes.

7.2×10^–4 = (x)(0.200)/(0.150)

x = [H₃O⁺]_e = 5.4×10^–4 M

pH = –log[H₃O⁺] = –log(5.4×10^–4) = 3.27

 

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