Calculate the pH at 25°C of a solution prepared by adding 50.0 mL of 1.60 M NaOH solution to 0.500 L of 0.0850 M HC5H3O3. Furoic acid (HC5H3O3) has a Ka value of 6.76×10⁻⁴ at 25°C.

• A. pH = 4.25
• B. pH = 6.80
• C. pH = 5.60
• D. pH = 3.45

Answer: (A)

Verified step by step guidance

1. Calculate the moles of NaOH added to the solution using the formula: \( \text{moles of NaOH} = \text{volume (L)} \times \text{molarity (M)} \). Convert 50.0 mL to liters by dividing by 1000.
2. Calculate the moles of furoic acid (HC5H3O3) present in the solution using the formula: \( \text{moles of HC5H3O3} = \text{volume (L)} \times \text{molarity (M)} \).
3. Determine the limiting reactant by comparing the moles of NaOH and HC5H3O3. Since NaOH is a strong base, it will react completely with the furoic acid.
4. Calculate the moles of furoate ion (C5H3O3⁻) formed after the reaction using the stoichiometry of the reaction: \( \text{HC5H3O3} + \text{NaOH} \rightarrow \text{C5H3O3}^- + \text{H2O} \).
5. Use the Henderson-Hasselbalch equation to find the pH of the solution: \( \text{pH} = \text{pKa} + \log \left( \frac{[\text{C5H3O3}^-]}{[\text{HC5H3O3}]} \right) \). Calculate \( \text{pKa} \) from \( \text{Ka} \) using \( \text{pKa} = -\log(\text{Ka}) \).