Given the following data: Volume of HA used = 9.19 mL, Volume of NaOH at equivalence point = 8.42 mL, pH at equivalence point = 8.66, and Concentration of NaOH = 0.1471 M, what is the pKa of the acid HA?

• A. pKa = 6.50
• B. pKa = 7.00
• C. pKa = 4.33
• D. pKa = 8.66

Answer: (D)

Verified step by step guidance

1. Start by understanding that at the equivalence point in a titration, the amount of acid (HA) is stoichiometrically equal to the amount of base (NaOH) added. This means that the moles of HA initially present are equal to the moles of NaOH added.
2. Calculate the moles of NaOH used at the equivalence point using the formula: \( \text{moles of NaOH} = \text{Volume of NaOH} \times \text{Concentration of NaOH} \). Convert the volume from mL to L by dividing by 1000.
3. Since the moles of HA are equal to the moles of NaOH at the equivalence point, use the moles of NaOH calculated to find the initial concentration of HA. Use the formula: \( \text{Concentration of HA} = \frac{\text{moles of HA}}{\text{Volume of HA}} \). Again, convert the volume from mL to L.
4. Recognize that at the equivalence point, the solution contains the conjugate base of HA, and the pH is determined by the hydrolysis of this conjugate base. The pH at the equivalence point is given as 8.66.
5. Use the Henderson-Hasselbalch equation to find the pKa of HA: \( \text{pH} = \text{pKa} + \log \left( \frac{[A^-]}{[HA]} \right) \). At the equivalence point, \([A^-] = [HA]\), so \( \log \left( \frac{[A^-]}{[HA]} \right) = 0 \). Therefore, \( \text{pH} = \text{pKa} \). Thus, the pKa of HA is equal to the pH at the equivalence point, which is 8.66.