Question

Hydrogen sulfide in waste water can be determined by a coulometric titration with electrically-generated iodine. The products are sulfur and iodide. A 450.0-mL sample of industrial effluent took 36.0 min 52.0 sec for complete oxidation of the hydrogen sulfide, at an average current of 65.0 mA (at a 100% current efficiency). Calculate the hydrogen sulfide concentration in ppm. Use 96,500 as Faraday's Constant. State your answer to one place after the decimal and with units.

Answer 1

According to Faraday’s law ,W = (ECt) / 96500

Where W = mass of metal deposited = ?

           E = Equivalent weight of H₂S = Molar mass / 2 = 34 / 2 = 17

           C = current = 65.0 mA = 65.0x10^-3 A

           t = time taken = 36 min +52 s

                               = (36x60 s) + 52 s

                               = 2212 s

Plug the values we get W = 0.0253 g

Given volume , V = 450 mL = 0.450 L

So concentration of H₂S = 0.0253 g / 0.450 L

                                    = 0.056 g/L

                                    = 56 mg/L

                                   = 56 ppm                Since 1 ppm = 1 mg/L

Therefore the concentration of H2S is 56 ppm