25 mL of 0.10 M aqueous acetic acid is titrated with 0.10 M NaOH(aq).What is the pH after 15 mL of NaOH(aq)have been added? K_a for acetic acid = 1.8 × 10^-5.

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At the point where 15 mL of 0.10 M NaOH has been added to 25 mL of 0.10 M acetic acid, the solution is not yet at the equivalence point. The amount of NaOH added is not enough to neutralize all the acetic acid. The reaction between acetic acid (CH3COOH) and NaOH is:$$ \text{CH}_3\text{COOH} + \text{NaOH} \rightarrow \text{CH}_3\text{COO}^- + \text{Na}^+ + \text{H}_2\text{O} $$Before the titration, the moles of acetic acid are:$$ \text{Moles of CH}_3\text{COOH} = 0.025 \, \text{L} \times 0.10 \, \text{M} = 0.0025 \, \text{moles} $$After adding 15 mL of 0.10 M NaOH, the moles of NaOH are:$$ \text{Moles of NaOH} = 0.015 \, \text{L} \times 0.10 \, \text{M} = 0.0015 \, \text{moles} $$The reaction consumes equal moles of acetic acid and NaOH, leaving:$$ \text{Moles of CH}_3\text{COOH remaining} = 0.0025 \, \text{moles} - 0.0015 \, \text{moles} = 0.0010 \, \text{moles} $$And produces:$$ \text{Moles of CH}_3\text{COO}^- = 0.0015 \, \text{moles} $$The total volume of the solution is now:$$ \text{Volume} = 25 \, \text{mL} + 15 \, \text{mL} = 40 \, \text{mL} = 0.040 \, \text{L} $$The concentrations of acetic acid and acetate ion are:$$ [\text{CH}_3\text{COOH}] = \frac{0.0010 \, \text{moles}}{0.040 \, \text{L}} = 0.025 \, \text{M} $$$$ [\text{CH}_3\text{COO}^-] = \frac{0.0015 \, \text{moles}}{0.040 \, \text{L}} = 0.0375 \, \text{M} $$Using the Henderson-Hasselbalch equation:$$ \text{pH} = \text{p}K_a + \log \left( \frac{[\text{CH}_3\text{COO}^-]}{[\text{CH}_3\text{COOH}]} \right) $$$$ \text{pH} = -\log(1.8 \times 10^{-5}) + \log \left( \frac{0.0375}{0.025} \right) $$$$ \text{pH} \approx 4.74 + \log(1.5) $$$$ \text{pH} \approx 4.74 + 0.176 $$$$ \text{pH} \approx 4.9 $$Therefore, the pH after adding 15 mL of NaOH is approximately 4.9.

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