The iodide content of a solution was determined by titration with cerium (IV) sulphate in the presence of HCl, in which I- is converted to ICl. A 250 ml sample of the solution required 20 ml 0.05 N ce4+ solution. What is the iodide concentration in the original solution?
Answers
Answer:potassium iodate
IO3- + 6HCl + 4e- ---> ICl + 5Cl- + 3H2O
In this case, iodide ions serve as the reducing agent thus:
I- ---> I+ + 2e-
Hence the overall reaction is as shown below, and the end point may be readily detected by the disappearance of free iodine and its replacement by the pale yellow colour of iodine monochloride.
IO3- + 2I- + 6HCl ---> 3ICl + 3Cl- + 3H2O
Method
(Conc. HCL should be handled in the fume hood. Avoid contact with skin!)
Explanation:potassium iodate, when used as an oxidising agent was itself reduced to elemental iodine. However, if the reagent is used in the presence of an excess of concentrated hydrochloric acid, the reduction product is not iodine but iodine monochloride, where iodine is formally in the +1 oxidation state.
IO3- + 6HCl + 4e- ---> ICl + 5Cl- + 3H2O
In this case, iodide ions serve as the reducing agent thus:
I- ---> I+ + 2e-
Hence the overall reaction is as shown below, and the end point may be readily detected by the disappearance of free iodine and its replacement by the pale yellow colour of iodine monochloride.
IO3- + 2I- + 6HCl ---> 3ICl + 3Cl- + 3H2O
Method
(Conc. HCL should be handled in the fume hood. Avoid contact with skin!)