Chapter 4.1, Problem 4.5AFP

Interpretation Introduction

Interpretation:

The amount $\left(\text{mol}\right)$ of each ion in $1.32\text{ L}$ of $0.55M$ sodium phosphate 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 amount of compound in moles when molarity of solution and volume of solution is given is as follows:

$\text{moles of solute}\left(\text{mol}\right)=\left(\text{volume of solution}\left(\text{L}\right)\right)\left(\frac{\text{molarity of solution}\left(\text{mol}\right)}{\text{1}\text{L of solution}}\right)$

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. Sodium phosphate $\left({\text{Na}}_3{\text{PO}}_4\right)$ is an example of an ionic compound and dissociates into ions when dissolved in water. The dissociation reaction of ${\text{Na}}_3{\text{PO}}_4$ is:

${\text{Na}}_3{\text{PO}}_4\left(s\right)\rightleftharpoons 3{\text{Na}}^{+}\left(aq\right)+{\text{PO}}_4^{3-}\left(aq\right)$

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

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