(B) Milli-equivalents of $HCl = 25 \times 0.08 = 2.0 \ mmol$.Milli-equivalents of $NaOH = 25 \times 0.1 = 2.5 \ mmol$.Since $NaOH$ is in excess,the re

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(B) Milli-equivalents of $HCl = 25 \times 0.08 = 2.0 \ mmol$.Milli-equivalents of $NaOH = 25 \times 0.1 = 2.5 \ mmol$.Since $NaOH$ is in excess,the remaining $OH^{-}$ concentration is calculated as:$[OH^{-}] = \frac{2.5 - 2.0}{500 \ mL} = \frac{0.5}{500} = 10^{-3} \ M$.Now,$pOH = -\log[OH^{-}] = -\log(10^{-3}) = 3$.Since $pH + pOH = 14$,we have $pH = 14 - 3 = 11$.