Question

Why is Solid $ Pb{I_2} $ yellow, yet a solution of the salt in water is colorless?

Answer 1

Hint: Lead Iodide is formed as the result of reaction between the lead nitrate and potassium iodide. The reaction between the lead nitrate and potassium iodide is a double displacement reaction. When the aqueous solution of lead iodide is taken then it is a colorless solution because of colorless products formed.

Complete Step By Step Answer:
The formation of lead iodide can be represented by the following reaction:
 $ Pb{\left( {N{O_3}} \right)_2}(aq.){\text{ }} + {\text{ }}2KI(aq.){\text{ }} \to {\text{ }}Pb{I_2}(s) \downarrow {\text{ }} + {\text{ }}2KN{O_3}(aq.) $
Therefore when we add lead nitrate to the solution of potassium iodide then we get a solution of potassium nitrate in which precipitate of lead iodide is formed. The precipitate so formed is yellow in colour. When we take its aqueous solution then it is colourless since the reactants of the reaction are colourless. Therefore we can say that lead nitrate is colourless and potassium iodide is also colourless. The above reaction is a double displacement reaction. Here the lead of lead nitrate gets displaced with potassium of potassium iodide. Therefore the result is that lead iodide is formed. Lead iodide gets precipitated because it is insoluble in water. When we take aqueous solution of the lead iodide the colour of solution will be colourless as lead nitrate and potassium iodide are colourless in aqueous solution.

Note:
Double displacement reactions are those reactions in which there is displacement of two elements of different compounds. Therefore the above reaction is also considered as a double displacement reaction. Since reactants are colourless in aqueous solution therefore the aqueous solution of salt will be colourless in nature. The sign $ \downarrow $ indicates that it is settled down at the bottom of the container.