Question

When lead nitrate reacts with potassium iodide then a yellow precipitate of:

• A. $Pb{I}_2$ is formed
• B. $KN{O}_3$ is formed
• C. $Pb(N{O}_3{)}_2$ is formed
• D. $PbI{O}_3$ is formed

Answer 1

The correct option is A $Pb{I}_2$ is formed

The reaction between lead nitrate and potassium iodide is a type of precipitation and double displacement reaction. The lead nitrate $\left[{\text{Pb(NO}}_3{)}_2\right]$ reacts with potassium iodide $\left(\text{Kl}\right)$ causing exchange of ions between the reactants leading to the formation of potassium nitrate $\left({\text{KNO}}_3\right)$ and a yellow precipitate of lead iodide $\left({\text{PbI}}_2\right)$. The reaction involved is given below:

${\text{Pb(NO}}_3{)}_2\text{(aq)}+2\text{KI (aq)}\to {\text{PbI}}_2\text{(s)}(\downarrow )+2{\text{KNO}}_3\text{(aq)}$