25 mL of 0.10 M acetic acid (K_a = 1.8x 10^-5) is titrated with 0.10 M NaOH. At the equivalence point the pH is closest to

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$25 mL$ of $0.10 M$ acetic acid ( $K_{a} = 1.8 \times 1 0^{- 5}$ ) is titrated with $0.10 M$ NaOH. At the equivalence point the pH is closest to

At equivalence the solution is

0.050 M

sodium acetate (volume doubled):

pH = 7 + \frac{1}{2} (p K_{a} + lo g C) = 7 + \frac{1}{2} (4.745 - 1.301) = 8.72

, i.e. basic, which is why phenolphthalein is used.