How many grams of should be added in 500 ml of 2 M acetic acid solution to get a buffer solution of — Ionic Equilibrium Chemistry Question

💡 Solution & Explanation

Initial moles of CH3COOH in 500 mL of 2 M solution = 2 M × 0.5 L = 1.0 mol. For maximum buffer capacity, [CH3COOH] = [CH3COO-]. Thus, exactly half of the acetic acid must be neutralized by $NaOH$: CH3COOH + $NaOH$ -> CH3COONa + $H_2O$. Therefore, moles of $NaOH$ needed = 0.5 mol. Molar mass of $NaOH$ = 40 g/mol. Mass of $NaOH$ = 0.5 mol × 40 g/mol = 20.0 g.