Question

When lead nitrate reacts with potassium iodide, yellow precipitate of

• A. ${PbI}_2$ is formed.
• B. ${KNO}_3$ is formed.
• C. $Pb{\left({NO}_3\right)}_2$ is formed.
• D. ${PbIO}_3$ is formed.

Answer 1

The correct option is A

${\mathrm{PbI}}_2$ is formed.

Explanation:

• In a double displacement reaction, two reactants react by an exchange of ions to form two new compounds.
• When lead nitrate $\left[\mathrm{Pb}{\left({\mathrm{NO}}_3\right)}_2\right]$ reacts with Potassium iodide $\left(\mathrm{KI}\right)$, the formation of Potassium nitrate $\left({\mathrm{KNO}}_3\right)$ along with a yellow precipitate of Lead iodide $\left({\mathrm{PbI}}_2\right)$ takes place. The chemical equation for the reaction is as follows:
• $\mathrm{Pb}{\left({\mathrm{NO}}_3\right)}_2\left(\mathrm{aq}\right)+2\mathrm{KI}\left(\mathrm{aq}\right)\to {\mathrm{PbI}}_2\left(\mathrm{s}\right)+2{\mathrm{KNO}}_3$
• Since, the formation of insoluble solid (precipitate) takes place, thus the reaction is also known as precipitation reaction.

Final Answer:

Hence, option (A) is correct.