To 25.0 mL of an aqueous H2S solution, 50.0 mL of 0.0500 M iodine is added (H2S + I2 -> S + 2H+ + 2I-). The unreacted iodine needs 18.0 mL of 0.100 M Na2S2O3. Taking H = 1 and S = 32, find the concentration of H2S in g/L (two decimals).

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JEE Advanced levelNumerical answerDifficulty 3 of 3eq087

To $25.0 mL$ of an aqueous $H X_{2} S$ solution, $50.0 mL$ of $0.0500 M$ iodine is added ( $H X_{2} S + I X_{2} S + 2 H X^{+} + 2 I X^{-}$ ). The unreacted iodine needs $18.0 mL$ of $0.100 M$ $Na X_{2} S X_{2} O X_{3}$ . Taking H = 1 and S = 32, find the concentration of $H X_{2} S$ in g/L (two decimals).

Numerical answer type, answer in g/L, tolerance 0.02.

Answer: 2.18 g/L

I2 added = 0.0500 x 0.0500 = 2.50e-03 mol. Excess I2 = (0.0180 x 0.100)/2 = 9.00e-04 mol. I2 used by H2S = 1.60e-03 mol = mol H2S in 25.0 mL. Concentration = 1.60e-03/0.0250 = 0.0640 mol/L x 34 = 2.176 g/L.

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To 25.0 mL of an aqueous HX2​S solution, 50.0 mL of 0.0500 M iodine is added (HX2​S+IX2​​S+2HX++2IX−). The unreacted iodine needs 18.0 mL of 0.100 M NaX2​SX2​OX3​. Taking H = 1 and S = 32, find the concentration of HX2​S in g/L (two decimals).

Answer: 2.18 g/L

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