Chapter 4.1, Problem 4.5BFP

Interpretation Introduction

Interpretation:

The molarity of aluminium ion in a solution that contains $1.25\text{ mol}$ of aluminium sulfate in $875\text{ mL}$ is to be calculated.

Concept introduction:

Molarity $\left(M\right)$ is one of the concentration terms that determine the number of moles of solute present in per liter of solution. Unit of molarity is $\text{mol/L}$.

The expression to calculate the molarity of a solution when moles of solute and volume of solution are given is as follows:

$\text{Molarity of solution}\left(M\right)=\frac{\text{moles of solute}\left(\text{mol}\right)}{\text{volume of solution}\left(\text{L}\right)}$

Ionic compounds are the compounds that are composed of charged ions. They are held together by electrostatic forces. These compounds dissociate into ions when dissolved in water. Aluminium sulfate $\left({\text{Al}}_2{\left({\text{SO}}_4\right)}_3\right)$ is an example of an ionic compound and dissociates into ions when dissolved in water. The dissociation reaction of ${\text{Al}}_2{\left({\text{SO}}_4\right)}_3$ is:

${\text{Al}}_2{\left({\text{SO}}_4\right)}_3\left(s\right)\rightleftharpoons 2{\text{Al}}^{3+}\left(aq\right)+3{\text{SO}}_4^{2-}\left(aq\right)$

The expression to calculate the amount of ions in moles is as follows:

$\text{amount}\text{of}\text{ion}\left(\text{mol}\right)=\left(\text{moles of compound}\left(\text{mol}\right)\right)\left(\frac{\text{moles of ion}\left(\text{mol}\right)}{1\text{mole of compound}}\right)$