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)

 

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