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When an analyte that is a reducing agent is titrated directly with a standard iodine solution, the method is called "iodimetry". When an analyte that is an oxidizing agent is added to excess iodide to produce iodine, and the iodine produced is determined by titration with sodium thiosulfate, the method is called "iodometry".
Not usually. Potassium iodide is usually present in a large excess. It is usually the potassium iodate that is the limiting ingredient.
Iodine is much more soluble in ethanol than in water, so it will usually form a homogeneous mixture. Of course, if you add more solid iodine than will dissolve in your quantity of ethanol at the temperature at which you are working, the excess solid will sink to the bottom. In that case, you have a heterogeneous mixture.
If you're asking me to explain how Thiosulfate-Iodine titration works, I'll explain. Usually, this titration is used to calculate the amount of Iodide ions produced in a previous reaction, in order find the concentration of the substance reacted in that reaction. For example, in an attempt to find the percentage of Copper in a coin, the coin is first dissolved in concentrated Nitric acid, where Cu2+ ions are formed. Next, this solution is treated with excess Potassium Iodide solution. The reaction is: 2Cu2+ + 4I- ----> 2CuI + I2 The amount of Iodine liberated is then titrated with a known concentration of Sodium Thiosulfate solution. The reaction is: 2S2O32- + I2 ----> S4O62- + 2I-. Starch is used as indicator for this titration. The color at the end-point is bluish-black. From the volume of Thiosulfate required, the amount of Iodide ions can be calculated(using the second equation). From this, the amount of Copper can be calculated from the first equation. I hope this answers your question.
Long story short, iodine in the solution made earlier through a reducing agent is weakly soluble and prone to loss to the air. Therefore, let's say you are using the titration to determine concentration of the titrant, the concentration would be off from the "true" concentration because you've lost some reactants.
It is used as indicator for the endpoint of a Iodometric (redox) titration: it gives a bluish grey to black color with very minute excess of Iodine-iodide ions (I3-)
When an analyte that is a reducing agent is titrated directly with a standard iodine solution, the method is called "iodimetry". When an analyte that is an oxidizing agent is added to excess iodide to produce iodine, and the iodine produced is determined by titration with sodium thiosulfate, the method is called "iodometry".
Because the iodine is liberated, hence it is called as iodometric titration.
In this titration iodine is liberated ....Added:... from (excess of) iodide by an oxidant. The Iodine is then titrated with thio (di-sodium thio-sulfate) and starch as indicator added just before the expected equivalence point.
Not usually. Potassium iodide is usually present in a large excess. It is usually the potassium iodate that is the limiting ingredient.
iodine solution should be kept in dark because it is highly photosensetive.light accelerates a side reaction in which iodide ions are converted to iodine..
Because starch forms a dark blue colored soluble product with free iodine, so that the appearance of color in the solution from this reaction effectively shows when all the substances in the solution that react more strongly with iodine than does starch have been consumed and the titration is finished.
A thiosulfate titration is mostly carried out to determine the amount of iodine present in the solution. In these reactions, thiosulfate ion acts as the reducing agent. This types titrations are often called as 'iodometric titrations'.
On addition of the KI to your copper (II) solution, you formed Copper (I) iodine solid and produced the tri-iodide ion. It is the tri-iodide ion that you are titrating with the sodium thiosulfate. The tri-iodine ion is what itercalates into the starch molecules to form the dark blue color you are using as an end point in the titration. Some the the tri-iodide ion formed will adsorb to the surface of the solid copper (I) iodine formed. This must be desorbed for a complete titration. The addition of the potassium thiocyanate, displaces the adsorbed tri-iodine ion, and liberates it for titration.
On addition of the KI to your copper (II) solution, you formed Copper (I) iodine solid and produced the tri-iodide ion. It is the tri-iodide ion that you are titrating with the sodium thiosulfate. The tri-iodine ion is what itercalates into the starch molecules to form the dark blue color you are using as an end point in the titration. Some the the tri-iodide ion formed will adsorb to the surface of the solid copper (I) iodine formed. This must be desorbed for a complete titration. The addition of the potassium thiocyanate, displaces the adsorbed tri-iodine ion, and liberates it for titration.
If you are looking at a iodide to iodine redox titration, the solution would turn yellow instead of blue/black. The blue/black color of the iodine-starch complex is very intense and so the end-point is sharper. Without the starch, the endpoint, when the first yellow from the formation of iodine I2, appears, is less sharp and is harder to see.
indirect titration is a process where in the analyte did not react with the titrant, directly,instead..they are connected with the use of iodine.