20.0 mL of 0.500 M HC₂H₃O₂ is titrated with 0.350 M KOH. Determine the pH after 15.0 mL of the base has been added. K_a for HC₂H₃O₂ = 1.8 × 10⁻⁵.

• A. 4.29
• B. 4.59
• C. 4.71
• D. 4.74
• E. 4.79

Answer: (E)

Verified step by step guidance

1. Calculate the initial moles of acetic acid (HC2H3O2) using the formula: \( \text{moles} = \text{volume (L)} \times \text{molarity} \). Convert 20.0 mL to liters and multiply by 0.500 M.
2. Calculate the moles of KOH added using the formula: \( \text{moles} = \text{volume (L)} \times \text{molarity} \). Convert 15.0 mL to liters and multiply by 0.350 M.
3. Determine the moles of acetic acid that react with KOH. Since KOH is a strong base, it will react completely with the acetic acid in a 1:1 ratio. Subtract the moles of KOH from the initial moles of acetic acid to find the moles of acetic acid remaining.
4. Calculate the moles of acetate ion (C2H3O2-) formed, which will be equal to the moles of KOH added, since each mole of KOH reacts to form one mole of acetate ion.
5. Use the Henderson-Hasselbalch equation to find the pH: \( \text{pH} = \text{pK}_a + \log \left( \frac{[\text{C2H3O2}^-]}{[\text{HC2H3O2}]} \right) \). Calculate \( \text{pK}_a \) from \( K_a \) and use the moles of acetate and acetic acid to find their concentrations in the total volume of the solution (initial volume of acetic acid + volume of KOH added).