Discussion
The value of Ka of acetic acid is 1.7 × 10-5 mol/dm3. The pH of a buffer solution prepared by mixing 100 ml of 0.1 M acetic acid with a solution of 100 ml of 0.2 M sodium acetate is
Answer: C
The pH of the buffer is about 5.1.
After mixing, the total volume is 200 ml, so [acid] = 0.01/0.2 = 0.05 M and [acetate] = 0.02/0.2 = 0.10 M. The Henderson-Hasselbalch equation gives
\[ \begin{aligned} pK_{a} &= -\log(1.7 \times 10^{-5}) = 4.77 \\ pH &= pK_{a} + \log\frac{[\text{A}^{-}]}{[\text{HA}]} \\ &= 4.77 + \log 2 = 5.07 \end{aligned} \]
Because both solutions are diluted into the same volume, only the ratio of moles (0.02 to 0.01) matters. 4.5 is the result of inverting the ratio, and 4.1 or 5.5 do not follow from a pKa of 4.77 with a 2:1 salt-to-acid ratio.
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