Chapter 4.1, Problem 4.2BFP

(a)

Interpretation Introduction

Interpretation:

Amount of each ion produced when $4\text{mol}$ of lithium carbonate is dissolved in water is to be calculated.

Concept introduction:

A solution is a combination of two parts: solute and solvent. A solute is the substance that is present in small quantity and solvent is the substance in which solute is dissolved. When water acts as a solvent then it is known as an aqueous solution.

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. Lithium carbonate $\left({\text{Li}}_{\text{2}}{\text{CO}}_{\text{3}}\right)$ is an example of an ionic compound and dissociates into ions when dissolved in water. The dissociation reaction of ${\text{Li}}_{\text{2}}{\text{CO}}_{\text{3}}$ is:

${\text{Li}}_{\text{2}}{\text{CO}}_{\text{3}}\left(s\right)\to {\text{2Li}}^{+}\left(aq\right)+{\text{CO}}_3^{2-}\left(aq\right)$

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

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

(b)

Interpretation Introduction

Interpretation:

Amount of each ion produced when $112\text{g}$ of $\text{iron}\left(\text{III}\right)\text{sulphate}$ is dissolved in water is to be calculated.

Concept introduction:

A solution is a combination of two parts: solute and solvent. A Solute is the substance that is present in small quantity and solvent is the substance in which solute is dissolved. When water acts as a solvent then it is known as an aqueous solution.

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. $\text{Iron}\left(\text{III}\right)\text{sulphate}$$\left({\text{Fe}}_{\text{2}}{\left({\text{SO}}_4\right)}_3\right)$ is an example of an ionic compound and dissociate into ions when dissolved in water. The dissociation reaction of ${\text{Fe}}_{\text{2}}{\left({\text{SO}}_4\right)}_3$ is:

${\text{Fe}}_{\text{2}}{\left({\text{SO}}_4\right)}_3\left(s\right)\to {\text{2Fe}}^{3+}\left(aq\right)+3{\text{SO}}_4^{2-}\left(aq\right)$

The expression to calculate the moles of the compound is as follows:

$\text{moles of compound}\left(\text{mol}\right)=\left[\begin{array}{c}\left(\text{given mass of compound}\left(\text{g}\right)\right)\\ \left(\frac{\text{1 mole of compound}}{\text{molecular mass of compound}\left(\text{g}\right)}\right)\end{array}\right]$

(c)

Interpretation Introduction

Interpretation:

Amount of each ion produced when $8.09\times {10}^{22}$ formula unit of aluminium nitrate is dissolved in water is to be calculated.

Concept introduction:

A solution is a combination of two parts: solute and solvent. A solute is the substance that is present in small quantity and solvent is the substance in which solute is dissolved. When water acts as a solvent then it is known as an aqueous solution.

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 nitrate $\left(\text{Al}{\left({\text{NO}}_{\text{3}}\right)}_3\right)$ is an example of an ionic compound and dissociates into ions when dissolved in water. The dissociation reaction of $\text{Al}{\left({\text{NO}}_{\text{3}}\right)}_3$ is:

$\text{Al}{\left({\text{NO}}_{\text{3}}\right)}_3\left(s\right)\to {\text{Al}}^{3+}\left(aq\right)+3{\text{NO}}_3^{-}\left(aq\right)$

A formula unit is used for the ionic compound to represent their empirical formula. The expression to calculate the moles of the compound is as follows:

$\text{moles of compound}\left(\text{mol}\right)=\left[\begin{array}{c}\left(\text{given formula unit of compound}\left(\text{FU}\right)\right)\\ \left(\frac{\text{1 mole of compound}}{\text{6}\text{.022}\times {\text{10}}^{23}\text{FU}}\right)\end{array}\right]$