Question

When excess of NaOH solution is added to aqueous solution of iodine, the colour of solution becomes:
A) Blue
B) Yellow
C) Colourless
D) Pale green

Answer 1

Hint: Iodine usually changes colour on reaction with various other compounds such as sugars, starch, proteins etc. This property of Iodine is taken advantage of to detect the given type of the compound.

Complete answer:
Initially, the solution containing iodine is violet in colour. As we add excess NaOH, it reacts with iodine to form Sodium Iodide and Sodium hypoiodite. The reaction is accompanied by formation of a water molecule. As a result, the violet colour of the solution becomes colourless.

$${\text{I}}_2\text{ + 2NaOH }\to \text{ NaI + NaIO + }{\text{H}}_2\text{O}$$
$$\text{3}{\text{I}}_2\text{+ 6NaOH }\to \text{ 5NaI + NaI}{\text{O}}_3\text{ + 3}{\text{H}}_2\text{O}$$

Hence, option C is correct.

Additional Information
In the above reaction, we can see that, as the reaction proceeds, Various compounds of Sodium are formed with Iodine. First is, $$\text{NaI}$$ i.e. Sodium iodide. It is an ionic compound. It consists of colourless cubic crystals and since like dissolves like, they are easily soluble in polar solvents. It is mainly used in medicines as a carrier of iodine. Such medicines are mainly used in treatment of thyroid related diseases. ‘Iodised salt’ is basically the addition of a small quantity of Sodium iodide in sodium chloride.

Next compound that we can see is $$\text{NaIO}$$ i.e. Sodium hypoiodite. It is used to check the presence of carbonyl groups in the given unknown substance.
$$\text{NaI}{\text{O}}_3$$ is called Sodium iodate. It is an oxidising agent and is highly inflammable. Majorly it is used in anodising normal salt.

Note: Various products are formed between sodium and iodine as the reaction proceeds. They are formed one after the other, first being $$\text{NaI}$$ i.e. Sodium iodide. If the reaction is stopped at this point, further products will not form.