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In a titration of iodine with sodium thi...

In a titration of iodine with sodium thiosulphate , the formation
of a blue colour on the addition of colourless starch solution
indicates that

A

a blue complex of starch , iodine and sodium thiosulphate has been produced

B

all of the iodine has not been reduced

C

the glassware has not been washed sufficiently

D

all of the iodine has not be oxidized

Text Solution

AI Generated Solution

The correct Answer is:
To solve the question regarding the titration of iodine with sodium thiosulfate and the formation of a blue color upon the addition of starch, we can follow these steps: ### Step-by-Step Solution: 1. **Understand the Reaction**: In the titration, iodine (I2) is being reduced by sodium thiosulfate (Na2S2O3). The reaction can be represented as: \[ I_2 + 2 S_2O_3^{2-} \rightarrow 2 I^- + S_4O_6^{2-} \] Here, iodine is reduced to iodide ions (I^-). 2. **Role of Starch**: Starch is used as an indicator in this titration. When starch is added to a solution containing iodine, it forms a blue complex with iodine (I2). This blue color indicates the presence of free iodine in the solution. 3. **Analyzing the Options**: The question asks what the formation of blue color indicates. We need to determine if all the iodine has reacted or if some remains unreacted. 4. **Conclusion**: Since the blue color indicates the presence of unreacted iodine (I2) in the solution, we conclude that not all iodine has been reduced to iodide ions (I^-). Therefore, the correct answer is that "all of the iodine has not been reduced." ### Final Answer: The formation of blue color on the addition of colorless starch solution indicates that **all of the iodine has not been reduced**. ---
To solve the question regarding the titration of iodine with sodium thiosulfate and the formation of a blue color upon the addition of starch, we can follow these steps: ### Step-by-Step Solution: 1. **Understand the Reaction**: In the titration, iodine (I2) is being reduced by sodium thiosulfate (Na2S2O3). The reaction can be represented as: \[ I_2 + 2 S_2O_3^{2-} \rightarrow 2 I^- + S_4O_6^{2-} \] ...
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