Chapter 4, Problem 4.48P

Interpretation Introduction

Interpretation:

Molarity of silver ion in the original solution if $25.0\text{ mL}$ of silver nitrate solution reacts with excess potassium chloride solution to yield $0.842\text{ g}$ 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 silver nitrate solution and potassium chloride is an example of precipitation reaction.

${\text{AgNO}}_{\text{3}}\left(aq\right)+\text{KCl}\left(aq\right)\to \text{AgCl}\left(s\right)+{\text{KNO}}_{\text{3}}\left(aq\right)$

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