A g sample of commercial bleaching powder is suspended in pure water and made up to mL. A mL aliquot — Redox Reactions and Volumetric Analysis Chemistry Question
Question
A $2.0$ g sample of commercial bleaching powder is suspended in pure water and made up to $100$ mL. A $10$ mL aliquot of this suspension is treated with excess $KI$ in an acidic medium. The liberated iodine requires exactly $10$ mL of $0.2 N$ $Na_2S_2O_3$ (hypo) to reach the visual endpoint. What is the percentage of available chlorine by mass in the original bleaching powder?
💡 Solution & Explanation
In the $10$ mL aliquot, milli-equivalents of available $Cl_2$ = meq of $I_2$ = meq of Hypo = $10 \times 0.2 = 2.0$ meq. Because the aliquot is exactly $1/10th$ of the total $100$ mL suspension, the total milli-equivalents of $Cl_2$ available in the full $2.0$ g sample = $2.0 \times 10 = 20$ meq = $0.02$ equivalents. The equivalent weight of $Cl_2$ is $35.5$ (since its molar mass is 71 and n-factor is 2). Mass of available $Cl_2 = 0.02 \times 35.5 = 0.71$ g. Percentage of available $Cl_2 = (0.71 / 2.0) \times 100 = 35.5\%$ .