A sample of 0.100 mole of acetic acid ( K _a = 1.8×10^{- 5}) in 50.0 mL of solution is titrated with standard 0.100 M NaOH. What is the pH in the titration flask after addition of 25.0 mL of the base?

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Quizplus · Accepted Answer

At the point of adding 25.0 mL of 0.100 M NaOH to the solution containing 0.100 mole of acetic acid in 50.0 mL, half of the acetic acid has been neutralized by the NaOH. This is the halfway point to the equivalence point in a titration, where the concentration of acetic acid equals the concentration of its conjugate base (acetate ion). At this point, the pH of the solution is equal to the pKa of acetic acid. The pKa can be calculated from the given Ka value (1.8×10^-5) as pKa = -log(Ka) = -log(1.8×10^-5) ≈ 4.74.

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