An Introductory Course of Quantitative Chemical Analysis eBook

This eBook from the Gutenberg Project consists of approximately 220 pages of information about An Introductory Course of Quantitative Chemical Analysis.

MnO_{2} + 4HCl —­> MnCl_{2} + Cl_{2} + 2H_{2}O
Cl_{2} + 2KI —­> I_{2} + 2KCl
I_{2} + 2Na_{2}S_{2}O_{3} —­> Na_{2}S_{4}O_{6} + 2NaI.

The chlorine generated by the pyrolusite is passed into a solution of potassium iodide.  The liberated iodine is then determined by titration with sodium thiosulphate, as described on page 78.  This is a direct process, although it involves three steps.]


The titration of iodine against sodium thiosulphate, with starch as an indicator, may perhaps be regarded as the most accurate of volumetric processes.  The thiosulphate solution may be used in both acid and neutral solutions to measure free iodine and the latter may, in turn, serve as a measure of any substance capable of liberating iodine from potassium iodide under suitable conditions for titration, as, for example, in the process outlined in Note 5 on page 74.

The fundamental reaction upon which iodometric processes are based is the following: 

I_{2} + 2 Na_{2}S_{2}O_{3} —­> 2 NaI + Na_{2}S_{4}O_{6}.

This reaction between iodine and sodium thiosulphate, resulting in the formation of the compound Na_{2}S_{4}O_{6}, called sodium tetrathionate, is quantitatively exact, and differs in that respect from the action of chlorine or bromine, which oxidize the thiosulphate, but not quantitatively.


If the formulas of sodium thiosulphate and sodium tetrathionate are written in a manner to show the atoms of oxygen associated with sulphur atoms in each, thus, 2(Na_{2}).S_{2}O_{2} and Na_{2}O.S_{4}O_{5}, it is plain that in the tetrathionate there are five atoms of oxygen associated with sulphur, instead of the four in the two molecules of the thiosulphate taken together.  Although, therefore, the iodine contains no oxygen, the two atoms of iodine have, in effect, brought about the addition of one oxygen atoms to the sulphur atoms.  That is the same thing as saying that 253.84 grams of iodine (I_{2}) are equivalent to 16 grams of oxygen; hence, since 8 grams of oxygen is the basis of normal solutions, 253.84/2 or 126.97 grams of iodine should be contained in one liter of normal iodine solution.  By a similar course of reasoning the conclusion is reached that the normal solution of sodium thiosulphate should contain, per liter, its molecular weight in grams.  As the thiosulphate in crystalline form has the formula Na_{2}S_{2}O_{3}.5H_{2}O, this weight is 248.12 grams.  Tenth-normal or hundredth-normal solutions are generally used.


!Approximate Strength, 0.1 N!

Procedure.—­Weigh out on the rough balances 13 grams of commercial iodine.  Place it in a mortar with 18 grams of potassium iodide and triturate with small portions of water until all is dissolved.  Dilute the solution to 1000 cc. and transfer to a liter bottle and mix thoroughly (Note 1).[1]

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An Introductory Course of Quantitative Chemical Analysis from Project Gutenberg. Public domain.
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