Chemistry 112

Buffer plus strong base:

1.0 g KOH is added to the buffer. First change the mass to moles: 1.0 g KOH × 1 mol KOH/(39.1 g + 16.0 g + 1.0 g) = 0.018 mol in 1.00 L

[KOH] = 0.018 M

(The pH of this strong base in the absence of buffer would be pOH = –log(0.018) = 1.74 so pH = 14.00 – 1.74 = 12.26)

NaNO₂(aq) → Na⁺(aq) + NO₂^–(aq)

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

A/B reaction:HNO₂(aq) + KOH(aq) → K⁺(aq) + NO₂^–(aq) + H₂O(l)

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

Initial0.150 0 0.200

A/B reaction–0.018 0 +0.018

Change– x +x +x

Equilibrium0.132 – x x 0.218 + x

Approximate? 0.132/7.2×10^–4 = 180 > 100 Yes.

7.2×10^–4 = (x)(0.218)/(0.132)

x = [H₃O⁺]_e = 4.4×10^–4 M

pH = –log[H₃O⁺] = –log(4.4×10^–4) = 3.36

 

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