50.0 mL of a water sample containing H2S is treated with 25.0 mL of 0.0500 M I2 solution (H2S + I2 -> S + 2H+ + 2I-). The excess iodine is back-titrated with 0.100 M Na2S2O3, needing 8.0 mL. Find the concentration of H2S (molar mass 34) in mg per litre, to the nearest integer.

JEE Advanced levelNumerical answerDifficulty 3 of 3rx210

50.0 mL of a water sample containing $H X_{2} S$ is treated with 25.0 mL of 0.0500 M $I X_{2}$ solution ( $H X_{2} S + I X_{2} S + 2 H X^{+} + 2 I X^{-}$ ). The excess iodine is back-titrated with 0.100 M $Na X_{2} S X_{2} O X_{3}$ , needing 8.0 mL. Find the concentration of $H X_{2} S$ (molar mass 34) in mg per litre, to the nearest integer.

Numerical answer type, answer in mg/L, tolerance 1.

Total

I X_{2}

25.0 \times 0.0500 = 1.25

mmol. Excess

I X_{2}

= (8.0 x 0.100)/2 = 0.400 mmol.

I X_{2}

used = 0.850 mmol = mmol

H X_{2} S

. Mass =

0.850 \times 34 = 28.9

mg in 0.0500 L, so concentration = 578 mg/L.