Given a pKa of 4.76 for CH3COOH, what is the pH of a solution prepared by mixing exactly 15.0 mL of 0.400 M CH3COOH with 10.0 mL of 0.400 M KOH?

Here's how to solve this problem, which involves a neutralization reaction between a weak acid (acetic acid) and a strong base (potassium hydroxide): **1. Calculate Moles of Each Reactant:** * Moles of CH₃COOH: (0.015 L) * (0.400 mol/L) = 0.006 mol * Moles of KOH: (0.010 L) * (0.400 mol/L) = 0.004 mol **2. Determine the Limiting Reactant:** * KOH is the limiting reactant since there are fewer moles of it. **3. Neutralization Reaction:** * The reaction between CH₃COOH and KOH is: CH₃COOH + KOH → CH₃COO⁻ + K⁺ + H₂O * Since KOH is limiting, 0.004 mol of CH₃COOH will react with all the KOH, leaving 0.002 mol of CH₃COOH unreacted. This will form 0.004 mol of CH₃COO⁻. **4. Calculate Concentrations After Neutralization:** * Total volume of the solution: 15.0 mL + 10.0 mL = 25.0 mL = 0.025 L * Concentration of CH₃COOH: (0.002 mol) / (0.025 L) = 0.080 M * Concentration of CH₃COO⁻: (0.004 mol) / (0.025 L) = 0.160 M **5. Use the Henderson-Hasselbalch Equation:** Since we have a buffer solution (weak acid and its conjugate base), we use the Henderson-Hasselbalch equation to find the pH: pH =pK_a+\log_{10}\frac{\lbrack A^-\rbrack}{\lbrack HA\rbrack} where: * pKa = 4.76 (given) * [A⁻] = concentration of the conjugate base (CH₃COO⁻) = 0.160 M * [HA] = concentration of the weak acid (CH₃COOH) = 0.080 M pH = \log_{10}\left(\frac{0.160}{0.080}\right) pH = 4.76 + log(2) pH ≈ 4.76 + 0.30 pH ≈ 5.06 **Answer:** The pH of the solution is approximately \boxed{5.06} .