A 25.0 mL aliquot containing Fe^3+ and Cu^2+ is treated with excess KI (both oxidise iodide, each by one electron per ion); the iodine formed needs 30.0 mL of 0.100 M Na2S2O3. In a second 25.0 mL aliquot, fluoride is added to mask Fe^3+ completely, and then only 12.0 mL of the same thiosulfate is needed. Find the concentration of Fe^3+ in g/L (Fe = 56; two decimal places).

JEE Advanced levelNumerical answerDifficulty 3 of 3eq161

A 25.0 mL aliquot containing $Fe X^{3 +}$ and $Cu X^{2 +}$ is treated with excess KI (both oxidise iodide, each by one electron per ion); the iodine formed needs 30.0 mL of 0.100 M $Na X_{2} S X_{2} O X_{3}$ . In a second 25.0 mL aliquot, fluoride is added to mask $Fe X^{3 +}$ completely, and then only 12.0 mL of the same thiosulfate is needed. Find the concentration of $Fe X^{3 +}$ in g/L (Fe = 56; two decimal places).

Numerical answer type, answer in g/L, tolerance 0.006.

Thiosulfate moles = electrons accepted per ion, with n = 1 for both ions. Total = 3.00 mmol,

Cu X^{2 +}

alone = 1.20 mmol, so

Fe X^{3 +}

= 1.80 mmol in 25.0 mL = 72.0 mmol/L. Mass concentration = 72.0

\times

56 mg/L = 4.03 g/L.