50 mL of 0.1 M CH COOH is being titrated against 0.1 M NaOH. When 25 mL of NaOH has been added, the — Ionic Equilibrium Chemistry Question

💡 Solution & Explanation

**Step 1: Determine moles of acid and base** - Moles of CH₃COOH = 0.1 M × 50 mL = 5 mmol - Moles of NaOH added = 0.1 M × 25 mL = 2.5 mmol **Step 2: Calculate moles after reaction** CH₃COOH + NaOH → CH₃COONa + H₂O - Moles of CH₃COOH remaining = 5 - 2.5 = 2.5 mmol - Moles of CH₃COONa formed = 2.5 mmol **Step 3: Recognize this is a buffer solution** We have equal moles of weak acid and its conjugate base, so this is a buffer at the half-equivalence point. **Step 4: Apply Henderson-Hasselbalch equation** $$pH = pK_a + \log\frac{[A^-]}{[HA]}$$ $$pH = 4.76 + \log\frac{2.5}{2.5}$$ $$pH = 4.76 + \log(1)$$ $$pH = 4.76 + 0 = 4.76$$ **Step 5: Express in required format** pH = 4.76 = 476 × 10⁻² Therefore, the answer is **476**.