Question

In the United States, table salt is “iodized” and contains NaCl and a small amount of KI. The iodide was added to help prevent thyroid disorders in the population.

The iodide concentration in table salt was measured using the following steps:
1.000 g of table salt was dissolved in 100.0 mL of water. The salt solution was titrated using 4.00 × 10−5 M Ag+ and a Volhard titration (back titration).

15.00 mL of 4.00 × 10−5 M Ag+ solution was added to the 100.0 mL salt solution in excess and the AgI(s) precipitate was removed. Assume that the precipitate was only AgI(s) and all of the I– was precipitated. Then, 50.00 mL of the excess Ag+ solution was titrated with 1.00 × 10−5 M SCN – in the presence of Fe3+. 24.53 mL of 1.00 × 10−5 M SCN– was required to reach the end point of the second titration.

Find the mass percentage (in %) of I– in the original table salt sample.

Answer 1

The millimoles of Ag⁺ used in the titration = 15 mL * 4*10^-5 mmol/mL = 6*10^-4 mmol

The millimoles of SCN- required = (24.53 mL * 10^-5 mmol/mL) = 2.453*10^-6 mmol

i.e. The millimoles of I^- precipitated = 6*10^-4 mmol

i.e. The mass of I^- in the salt solution = 6*10^-4 mmol * 127 g/mol = 0.0762 g

Therefore, the mass percentage of I- in the original sample = (mass of I^-/total mass of salt)*100 = (0.0762/1)*100 = 7.62%