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:

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

Answer to Problem 4.2BFP

$4\text{mol}$ of lithium carbonate gives $8\text{mol}$ of ${\text{Li}}^{+}$ ions and $4\text{mol}$ of ${\text{CO}}_3^{2-}$ ions.

Explanation of Solution

One mole of lithium carbonate $\left({\text{Li}}_{\text{2}}{\text{CO}}_{\text{3}}\right)$ on dissociation produces two moles of ${\text{Li}}^{+}$ ion and one mole of ${\text{CO}}_3^{2-}$ ion.

The expression to calculate the amount of ${\text{Li}}^{+}$ ion in moles is as follows:

$\text{amount}\text{of}{\text{Li}}^{+}\left(\text{mol}\right)=\left({\text{moles of Li}}_{\text{2}}{\text{CO}}_{\text{3}}\left(\text{mol}\right)\right)\left(\frac{{\text{moles of Li}}^{+}\text{ion}\left(\text{mol}\right)}{1{\text{mole of Li}}_{\text{2}}{\text{CO}}_{\text{3}}}\right)$

Substitute $4\text{mol}$ for moles of ${\text{Li}}_{\text{2}}{\text{CO}}_{\text{3}}$ and $\text{2 mol}$ for moles of ${\text{Li}}^{+}$ ion in the above equation as follows:

$\begin{array}{c}\text{amount}\text{of}{\text{Li}}^{+}\left(\text{mol}\right)=\left(\text{4 mol}\right)\left(\frac{\text{2 mol}}{1{\text{mole of Li}}_{\text{2}}{\text{CO}}_{\text{3}}}\right)\\ =8\text{mol}\end{array}$

The expression to calculate the amount of ${\text{CO}}_3^{2-}$ ion in moles is as follows:

$\text{amount}\text{of}{\text{CO}}_3^{2-}\left(\text{mol}\right)=\left({\text{moles of Li}}_{\text{2}}{\text{CO}}_{\text{3}}\left(\text{mol}\right)\right)\left(\frac{{\text{moles of CO}}_3^{2-}\text{ion}\left(\text{mol}\right)}{1{\text{mole of Li}}_{\text{2}}{\text{CO}}_{\text{3}}}\right)$

Substitute $4\text{mol}$ for moles of ${\text{Li}}_{\text{2}}{\text{CO}}_{\text{3}}$ and $\text{1 mol}$ for moles of ${\text{CO}}_3^{2-}$ ion in the above equation as follows:

$\begin{array}{c}\text{amount}\text{of}{\text{CO}}_3^{2-}\left(\text{mol}\right)=\left(\text{4 mol}\right)\left(\frac{\text{1 mol}}{1{\text{mole of Li}}_{\text{2}}{\text{CO}}_{\text{3}}}\right)\\ =4\text{mol}\end{array}$

Conclusion

One mole of lithium carbonate $\left({\text{Li}}_{\text{2}}{\text{CO}}_{\text{3}}\right)$ on dissociation produces two moles of ${\text{Li}}^{+}$ ion and one mole of ${\text{CO}}_3^{2-}$ ion and $4\text{mol}$ of lithium carbonate gives $8\text{mol}$ of ${\text{Li}}^{+}$ ions and $4\text{mol}$ of ${\text{CO}}_3^{2-}$ ions.

(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:

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

Answer to Problem 4.2BFP

$112\text{g}$ of $\text{Iron}\left(\text{III}\right)\text{sulphate}$ gives $0.56\text{mol}$ of ${\text{Fe}}^{3+}$ ion and $0.84\text{mol}$ of ${\text{SO}}_4^{2-}$ ion.

Explanation of Solution

One mole of $\text{iron}\left(\text{III}\right)\text{sulphate}$$\left({\text{Fe}}_{\text{2}}{\left({\text{SO}}_4\right)}_3\right)$ on dissociation produces two moles of ${\text{Fe}}^{3+}$ ion and three moles of ${\text{SO}}_4^{2-}$ ion.

The molecular mass of ${\text{Fe}}_{\text{2}}{\left({\text{SO}}_4\right)}_3$ is $399.88\text{g}$.

The expression to calculate the moles of ${\text{Fe}}_{\text{2}}{\left({\text{SO}}_4\right)}_3$ in moles is as follows:

${\text{moles of Fe}}_{\text{2}}{\left({\text{SO}}_4\right)}_3\left(\text{mol}\right)=\left[\begin{array}{c}\left({\text{given mass of Fe}}_{\text{2}}{\left({\text{SO}}_4\right)}_3\left(\text{g}\right)\right)\\ \left(\frac{{\text{1 mole of Fe}}_{\text{2}}{\left({\text{SO}}_4\right)}_3}{{\text{molecular mass of Fe}}_{\text{2}}{\left({\text{SO}}_4\right)}_3\left(\text{g}\right)}\right)\end{array}\right]$

Substitute $112\text{g}$ for the given mass of ${\text{Fe}}_{\text{2}}{\left({\text{SO}}_4\right)}_3$ and $399.88\text{g}$ for the molecular mass of ${\text{Fe}}_{\text{2}}{\left({\text{SO}}_4\right)}_3$ ion in the above equation as follows:

$\begin{array}{c}{\text{moles of Fe}}_{\text{2}}{\left({\text{SO}}_4\right)}_3\left(\text{mol}\right)=\left(112\text{g}\right)\left(\frac{\text{1 mole of Mg}{\left({\text{C}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}\right)}_2}{399.88\text{g}\text{g}}\right)\\ =0.2801\text{mol}\\ \approx \text{0}\text{.28}\text{mol}\end{array}$

Hence, $112\text{g}$ of ${\text{Fe}}_{\text{2}}{\left({\text{SO}}_4\right)}_3$ on dissociation produces double of moles of ${\text{Fe}}_{\text{2}}{\left({\text{SO}}_4\right)}_3$ that is $0.56\text{mol}$ of ${\text{Fe}}^{3+}$ ion and thrice of moles of ${\text{Fe}}_{\text{2}}{\left({\text{SO}}_4\right)}_3$ that is $0.84\text{mol}$ of ${\text{SO}}_4^{2-}$ ion.

Conclusion

$112\text{g}$ of $\text{Iron}\left(\text{III}\right)\text{sulphate}$ gives $0.56\text{mol}$ of ${\text{Fe}}^{3+}$ ion and $0.84\text{mol}$ of ${\text{SO}}_4^{2-}$ ion.

(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:

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

Answer to Problem 4.2BFP

$8.09\times {10}^{22}$ formula unit of aluminium nitrate gives $0.134\text{mol}$ of ${\text{Al}}^{3+}$ ion and $0.403\text{mol}$ of ${\text{NO}}_3^{-}$ ion.

Explanation of Solution

One mole of aluminium nitrate $\left(\text{Al}{\left({\text{NO}}_{\text{3}}\right)}_3\right)$ on dissociation produces one mole of ${\text{Al}}^{3+}$ ion and three mole of ${\text{NO}}_3^{-}$ ion.

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

$\text{moles of Al}{\left({\text{NO}}_{\text{3}}\right)}_3\left(\text{mol}\right)=\left[\begin{array}{c}\left(\text{given formula unit of Al}{\left({\text{NO}}_{\text{3}}\right)}_3\left(\text{FU}\right)\right)\\ \left(\frac{\text{1 mole of Al}{\left({\text{NO}}_{\text{3}}\right)}_3}{\text{6}\text{.022}\times {\text{10}}^{23}\text{FU}}\right)\end{array}\right]$

Substitute $8.09\times {10}^{22}\text{FU}$ formula unit for given formula unit of $\text{Al}{\left({\text{NO}}_{\text{3}}\right)}_3$ in the above equation as follows:

$\begin{array}{c}\text{moles of Al}{\left({\text{NO}}_{\text{3}}\right)}_3\left(\text{mol}\right)=\left(8.09\times {10}^{22}\text{FU}\right)\left(\frac{\text{1 mole of Al}{\left({\text{NO}}_{\text{3}}\right)}_3}{\text{6}\text{.022}\times {\text{10}}^{23}\text{FU}}\right)\\ =0.1343\text{mol}\\ \approx 0.134\text{mol}\end{array}$

Hence, $8.09\times {10}^{22}\text{FU}$ of $\text{Al}{\left({\text{NO}}_{\text{3}}\right)}_3$ on dissociation produces $0.134\text{mol}$ of ${\text{Al}}^{3+}$ ion and thrice of that number of ${\text{NO}}_3^{-}$ ion that is $0.403\text{mol}$.

Conclusion

$8.09\times {10}^{22}\text{FU}$ of $\text{Al}{\left({\text{NO}}_{\text{3}}\right)}_3$ on dissociation produces $0.134\text{mol}$ of ${\text{Al}}^{3+}$ ion and thrice of that number of ${\text{NO}}_3^{-}$ ion that is $0.403\text{mol}$.