A 25.0 mL solution contains CuSO4 and K2Cr2O7. With excess KI in acid, the total iodine liberated needs 36.0 mL of 0.100 M Na2S2O3. A second 25.0 mL portion, which oxidises Fe^2+ only through dichromate (copper(II) does not), consumes 24.0 mL of 0.0500 M Fe^2+. Find the molarity of Cu^2+ in the solution (three decimals).

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JEE Advanced levelNumerical answerDifficulty 3 of 3eq086

A $25.0 mL$ solution contains $CuSO X_{4}$ and $K X_{2} Cr X_{2} O X_{7}$ . With excess $KI$ in acid, the total iodine liberated needs $36.0 mL$ of $0.100 M$ $Na X_{2} S X_{2} O X_{3}$ . A second $25.0 mL$ portion, which oxidises $Fe X^{2 +}$ only through dichromate (copper(II) does not), consumes $24.0 mL$ of $0.0500 M$ $Fe X^{2 +}$ . Find the molarity of $Cu X^{2 +}$ in the solution (three decimals).

Numerical answer type, answer in M, tolerance 0.001.

Answer: 0.096 M

Dichromate = (0.0240 x 0.0500)/6 = 2.00e-04 mol, which needs 6 x 2.00e-04 = 1.20e-03 mol thiosulphate. Total thiosulphate = 0.0360 x 0.100 = 3.60e-03. Copper part = 3.60e-03 - 1.20e-03 = 2.40e-03 mol = mol Cu2+ (1:1). Molarity = 2.40e-03/0.0250 = 0.096 M.

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Answer: 0.096 M

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