Chemistry 112

Buffer plus strong acid:

First change the mass to moles: 1.0 g HBr × 1 mol HBr/(1.0 g + 79.9 g) = 0.012 mol in 1.00 L

[HBr] = 0.012 M

(The pH of this strong acid in the absence of buffer would be pH = –log(0.012) = 1.92)

NaNO₂(aq) → Na⁺(aq) + NO₂^–(aq)

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

A/B reaction:HNO₂(aq) ← HBr(aq) + NO₂^–(aq)

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

Initial0.150 0 0.200

A/B reaction+0.012 0 –0.012

Change– x +x +x

Equilibrium0.162 – x x 0.188 + x

Approximate? 0.162/7.2×10^–4 = 230 > 100 Yes.

7.2×10^–4 = (x)(0.188)/(0.162)

x = [H₃O⁺]_e = 6.2×10^–4 M

pH = –log[H₃O⁺] = –log(6.2×10^–4) = 3.21

 

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