Chapter 4, Problem 4.47P

Interpretation Introduction

Interpretation:

Molarity of $\text{lead}\left(\text{II}\right)$ ion in the original solution if $38.5\text{ mL}$ of $\text{lead}\left(\text{II}\right)\text{nitrate}$ solution reacts completely with excess sodium iodide solution to yield $0.628\text{ g}$ of precipitate is to be calculated.

Concept introduction:

Precipitation reaction involves the reaction of two soluble ionic compounds to form an insoluble product. The insoluble product is known as a precipitate.

The reason for the precipitation reaction to occur is the formation of a product that is insoluble in nature. The reaction of $\text{lead}\left(\text{II}\right)\text{nitrate}$ and sodium iodide is an example of precipitation reaction.

$\text{Pb}{\left({\text{NO}}_{\text{3}}\right)}_2\left(aq\right)+2\text{NaI}\left(aq\right)\to {\text{PbI}}_2\left(s\right)+2{\text{NaNO}}_{\text{3}}\left(aq\right)$

According to the solubility rules, all the common nitrates are soluble but ${\text{PbI}}_2$ is insoluble in nature.