A capsule of vitamin \(\mathrm{C}\), a weak acid, is analyzed by titrating it with \(0.425 M\) sodium hydroxide. It is found that \(6.20 \mathrm{~mL}\) of base is required to react with a capsule weighing \(0.628 \mathrm{~g}\). What is the percentage of vitamin \(\mathrm{C}\) \(\left(\mathrm{C}_{6} \mathrm{H}_{8} \mathrm{O}_{6}\right)\) in the capsule? (One mole of vitamin \(\mathrm{C}\) reacts with one mole of hydroxide ion.)

Answer: To find the percentage of vitamin C in the capsule, follow the steps below: Step 1: Calculate the moles of sodium hydroxide used in the titration. moles of NaOH = molarity × volume = (0.425 M) × (6.20 mL × 0.001 L/mL) = 2.65 × 10^-3 moles Step 2: Calculate the moles of vitamin C. moles of vitamin C = moles of NaOH = 2.65 × 10^-3 moles Step 3: Calculate the mass of vitamin C. molar mass of vitamin C = (6 × 12.01 g/mol) + (8 × 1.01 g/mol) + (6 × 16.00 g/mol) = 176.13 g/mol mass of vitamin C = moles of vitamin C × molar mass of vitamin C = (2.65 × 10^-3 moles) × 176.13 g/mol = 0.466 g Step 4: Calculate the percentage of vitamin C in the capsule. percentage of vitamin C = (mass of vitamin C / mass of capsule) × 100 = (0.466 g / 0.628 g) × 100 = 74.2% The percentage of vitamin C in the capsule is 74.2%.