Question

A 50 L gas stream was passed through a solution containing Cd²⁺ where H₂S (in gas stream) was retained as CdS. The mixture was acidified and treated with 50 ml of 0.004 M I₂. After the reaction \[\ce{S^{2-} + I2 -> S(s) + 2I-}\] was complete, the excess iodine was treated with 15 ml of 0.01 M thiosulphate. The concentration of H₂S in ppm will be ______. Use density of gas stream = 1.7 gm/L.

Options

• 5

• 3

• 6

• 1

Answer 1

A 50 L gas stream was passed through a solution containing Cd²⁺ where H₂S (in gas stream) was retained as CdS. The mixture was acidified and treated with 50 ml of 0.004 M I₂. After the reaction \[\ce{S^{2-} + I2 -> S(s) + 2I-}\] was complete, the excess iodine was treated with 15 ml of 0.01 M thiosulphate. The concentration of H₂S in ppm will be 5.

Explanation:

\[\ce{S^{2-} + I2 -> S(s) + 2I-}\]

\[\ce{I2 + S2O_3^{2-} -> I^- + S4O_6^{2-}}\]

Eqv. of excess I₂ = Eqv. of thiosulphate

= 1 × 15 × 0.01

= 0.15

Millimoles of excess I₂ reacted with thiosulphate = `(0.15/2)`

Millimoles of I₂ reacted with S^2–

= `(50 xx 0.004) - 0.15/2`

= 0.2 - 0.075

= 0.125

ppm of H₂S = `((0.125 xx 10^-3 xx 34 xx 10^6)/(50 xx 1.7))`

ppm of H₂S =  50 = `50/10 = 5`