Analytical Chemistry

Topics: Iodine, Titration, 3rd millennium Pages: 6 (2589 words) Published: April 2, 2013
Name: 1.



Answer the following questions before coming to the laboratory. Give the reason for the following; Iodimetric titrations are usually performed in neutral or mildly alkaline (pH 8) to weakly acid solution.


Iodine solutions are prepared by dissolving I2 in a concentrated solution of potassium iodide.


Iodimetric determination of vitamin C is performed rapidly after preparing vitamin C solution




Redox Titrations are among the most important types of analyses performed in many areas of application, for example, in food analysis, industrial analysis, and pharmaceutical analysis. Tiration of sulfite in wine using iodine is a common example. Alcohol can be determined by reacting with potassium dichromate. Examples in clinical laboratories are rare, since most analyses are for traces, yet these titrations are still extremely useful for standardizing reagents. Iodine is a moderately, strong oxidizing agent that can be used to titrate fairly strong reducing agents. Titrations with I2 are called Iodimetric Methods. In iodimetry , the titrant is I2 and the analyte is a reducing agent. The end point is detected by the appearance of the blue starch-iodine color. Iodimetric titrations are usually performed in neutral or mildly alkaline (pH 8) to weakly acid solutions. If the pH is too alkaline, I2 will disproportionate to hypoiodite and iodide: I2  2OH -  IO-  I-  H 2 O On the other hand, there are three reasons for keeping the solution from becoming strongly acidic. First, the starch used for the detection of the end point is a carbohydrate which tends to hydrolyze or decompose in strongly acidic media, thus effecting the end point. Secondly, the reducing power of several reducing agents is increased in neutral solution. The last reason for avoiding acid solutions is that the I  produced in the reaction tends to be oxidized by dissolved oxygen in acid solution according to the following chemical equation: 4I-  O2  4H   2I2  2H2O

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