Question

The sulfur in a 5.00 g sample of steel was evolved as H₂S and then collected in a solution of CdCl₂ to produce the precipitate CdS. The CdS was then titrated with excess I₂ and the remaining I₂ was then back titrated with 4.82 ml of 0.0510 M sodium thiosulfate. If the concentration of the I₂ is 0.0600 M and a total of 10.0 ml is added, calculate the %S in the steel.

The relevant reactions are:

CdS + I₂ -> S + Cd²⁺ + 2I^-

I₂ + 2S₂O₃^- -> S₂O₆^2- +2I^-

Answer 1

Total millimoles of I2 = Consumed by CdS    + Consumed By Hypo (sodium thiosulfate)

CdS + I₂ ---------> S + Cd²⁺ + 2I^-

I₂ + 2S₂O₃^--- -------------------> S₂O₆^2-     +        2I^-

Millimoles of I2 consumed by Hypo (sodium thiosulfate) = 1/2(Millimoles of sodium thiosulfate)

                                                                                   = 1/2(4.82x0.0510)

                                                                                    = 0.12291 millmoles

Total Millimoles of I2 added = M x V = 0.06x10 = 0.6 millmioles

Millimoles of I2 consumed by CdS = total added I2 - Consumed by sodium thiosulfate

                                                 = 0.6 - 0.12291 = 0.47709

Millimoles of I2 = Millimoles of Cds   = 0.47709 = Millimoles of S

So mass of S = Molar mass x Millimoles = 0.47709x32 = 15.266 milligram = 0.015266 gram

% of S in steel = 100x(mass of S)/mass of steel sample = 100x0.015266/5 =0.30 %