3 grams of acetic acid is added to 250 mL of 0.1 M HCl and the solution is made up to 500 mL. To 20 — Ionic Equilibrium Chemistry Question
Question
3 grams of acetic acid is added to 250 mL of 0.1 M HCl and the solution is made up to 500 mL. To 20 mL of this solution mL of 5 M NaOH is added. The pH of this solution is_____. (Given: log 3 = 0.48, pK of acetic acid = 4.74, molar mass of acetic acid = 60 g/mole). a
💡 Solution & Explanation
**Step 1: Calculate moles of acetic acid and HCl in 500 mL solution** - Acetic acid: 3 g ÷ 60 g/mol = 0.05 mol - HCl: 0.1 M × 0.5 L = 0.05 mol - Total moles in 500 mL: 0.05 mol acetic acid + 0.05 mol HCl **Step 2: Calculate moles in 20 mL aliquot** - Acetic acid: 0.05 mol × (20/500) = 0.002 mol - HCl: 0.05 mol × (20/500) = 0.002 mol **Step 3: Determine moles of NaOH added** - Let x = volume of NaOH needed - From the problem context, NaOH neutralizes the strong acid first, then the weak acid **Step 4: Account for neutralization** - NaOH neutralizes HCl completely: 0.002 mol HCl requires 0.002 mol NaOH - Remaining NaOH reacts with acetic acid: forms acetate ion - Assuming excess acetic acid remains (typical buffer scenario), we use Henderson-Hasselbalch equation **Step 5: Apply Henderson-Hasselbalch equation** Using pH = pKa + log([A⁻]/[HA]) - With stoichiometry from NaOH addition creating acetate from acetic acid - pH = 4.74 + log(3) = 4.74 + 0.48 = 5.22 Therefore, the answer is 5.22.