Chapter 4, Problem 4.33P

a)

Interpretation Introduction

Interpretation:Balanced total and net equations for reaction of aluminum sulfate with aqueous $\text{NaOH}$ should be determined.

Concept introduction:Three types of equations are used for representation of ionic reactions. These are mentioned below.

1. Molecular equation

2. Total ionic equation

3. Net ionic equation

Molecular equations indicate reactants and products in undissociated forms. In total ionic equations, total ions that are dissociated in reaction are represented whereas in net ionic reactions, only useful ions are written.

Chemical reactions that lead to formation of insoluble salt by combination of two soluble salts with each other are called precipitation reactions. Such an insoluble salt is called precipitate. These reactions are so called due to formation of precipitates in them.

Explanation of Solution

In this reaction, ${\text{Al}}_{\text{2}}{\left({\text{SO}}_4\right)}_3$ reacts with $\text{NaOH}$ and forms ${\text{Na}}_{\text{3}}{\text{SO}}_4$ and $\text{Al}{\left(\text{OH}\right)}_3$ . Since common sulfates are soluble but hydroxides are insoluble, precipitate of $\text{Al}{\left(\text{OH}\right)}_3$ isformed in this reaction. So this reaction is example of precipitation reaction. Chemcal equation for this as follows:

  ${\text{Al}}_{\text{2}}{\left({\text{SO}}_4\right)}_3\left(aq\right)+6\text{NaOH}\left(aq\right)\to 3{\text{Na}}_2{\text{SO}}_4\left(aq\right)+2\text{Al}{\left(\text{OH}\right)}_3\left(s\right)$

Species present in aqueous phases are dissociated into their respective ions to write total ionic equation as follows:

  ${\text{2Al}}^{3+}\left(aq\right)+3{\text{SO}}_4^{2-}\left(aq\right)+6{\text{Na}}^{+}\left(aq\right)+6{\text{OH}}^{-}\left(aq\right)\to \left(\begin{array}{l}6{\text{Na}}^{+}\left(aq\right)+\\ 3{\text{SO}}_4^{2-}\left(aq\right)+2\text{Al}{\left(\text{OH}\right)}_3\left(s\right)\end{array}\right)$

  ${\text{Na}}^{+}$ and ${\text{SO}}_4^{2-}$ are common ions present in the reaction mixture. Therefore net ionic equation for given reaction becomes,

  ${\text{2Al}}^{3+}\left(aq\right)+6{\text{OH}}^{-}\left(aq\right)\to 2\text{Al}{\left(\text{OH}\right)}_3\left(s\right)$

b)

Interpretation Introduction

Interpretation:Mass of precipitate formed by addition of $185.5\text{ mL}$ of $0.533M\text{NaOH}$ to $627\text{ mL}$ of solution that contains $15.8\text{ g}$ of aluminum sulfate per liter is to be determined.

Concept introduction:Quantitative relationships between reactants and products used in chemical reactions are determined by stoichiometry. In such problems, amount of one species is given and that of another species is to be calculated.

Reagent that is fully consumed in any chemical reaction is known as limiting reagent. Yield of product is decided by its presence only. Other species present in any chemical reaction other than limiting reactant are present in excess amount and remain unreacted at end of reaction.

Answer to Problem 4.33P

Mass of $\text{Al}{\left(\text{OH}\right)}_3$ formed by addition of $185.5\text{ mL}$ of $0.533M\text{NaOH}$ is added to $627\text{ mL}$ of a solution that contains $15.8\text{ g}$ of aluminum sulfate per liter is $\text{2}\text{.57 g}$ .

Explanation of Solution

Chemical equation for given reaction is as follows:

  ${\text{Al}}_{\text{2}}{\left({\text{SO}}_4\right)}_3\left(aq\right)+6\text{NaOH}\left(aq\right)\to 3{\text{Na}}_2{\text{SO}}_4\left(aq\right)+2\text{Al}{\left(\text{OH}\right)}_3\left(s\right)$

Formula to calculate moles $\text{NaOH}$ of is as follows:

  $\text{Moles}\text{of}\text{NaOH}=\left(\text{Molarity of NaOH}\right)\left(\text{Volume of}\text{NaOH}\right)$

Molarity of $\text{NaOH}$ is $0.533M$ .

Volume of $\text{NaOH}$ is $185.5\text{mL}$ .

Substitute the values in above equation.

  $\begin{array}{c}\text{Moles}\text{of}\text{NaOH}=\left(\text{Molarity of NaOH}\right)\left(\text{Volume of}\text{NaOH}\right)\\ =\left(0.533M\right)\left(\text{185}\text{.5 mL}\right)\left(\frac{{10}^3\text{ L}}{1\text{ mL}}\right)\\ =0.0988715\text{ mol}\end{array}$

Since six moles of $\text{NaOH}$ forms two moles of $\text{Al}{\left(\text{OH}\right)}_3$ , moles of $\text{Al}{\left(\text{OH}\right)}_3$ formed in this reaction are calculated as follows:

  $\begin{array}{c}\text{Moles}\text{of}\text{Al}{\left(\text{OH}\right)}_3=\left(0.0988715\text{ mol NaOH}\right)\left(\frac{2\text{mol}\text{Al}{\left(\text{OH}\right)}_3}{6\text{mol}\text{NaOH}}\right)\\ =0.03296\text{ mol}\end{array}$

Formula to calculate mass of $\text{Al}{\left(\text{OH}\right)}_3$ is as follows:

  $\text{Mass of Al}{\left(\text{OH}\right)}_3=\left(\text{Moles of Al}{\left(\text{OH}\right)}_3\right)\left(\text{Molar mass of Al}{\left(\text{OH}\right)}_3\right)$

Moles of $\text{Al}{\left(\text{OH}\right)}_3$ are $0.03296\text{ mol}$ .

Molar mass of $\text{Al}{\left(\text{OH}\right)}_3$ is $78.00\text{ g/mol}$ .

Substitute the values in above equation.

  $\begin{array}{c}\text{Mass of Al}{\left(\text{OH}\right)}_3=\left(\text{Moles of Al}{\left(\text{OH}\right)}_3\right)\left(\text{Molar mass of Al}{\left(\text{OH}\right)}_3\right)\\ =\left(0.03296\text{ mol}\right)\left(78.00\text{ g/mol}\right)\\ =2.57088\text{ g}\\ \approx \text{2}\text{.57 g}\end{array}$

Moles of ${\text{Al}}_{\text{2}}{\left({\text{SO}}_4\right)}_3$ are calculated as follows:

  $\begin{array}{c}{\text{Moles of Al}}_{\text{2}}{\left({\text{SO}}_4\right)}_3=\left(\frac{15.8\text{ g}}{342.14\text{ g/mL}}\right)\left(627\text{ mL}\right)\left(\frac{{10}^{-3}\text{ L}}{1\text{ mL}}\right)\\ =0.02895\text{ mol}\end{array}$

Since one mole of ${\text{Al}}_{\text{2}}{\left({\text{SO}}_4\right)}_3$ forms two moles of $\text{Al}{\left(\text{OH}\right)}_3$ , moles of $\text{Al}{\left(\text{OH}\right)}_3$ formed in this reaction are calculated as follows:

  $\begin{array}{c}\text{Moles}\text{of}\text{Al}{\left(\text{OH}\right)}_3=\left(0.02895{\text{ mol Al}}_{\text{2}}{\left({\text{SO}}_4\right)}_3\right)\left(\frac{2\text{mol}\text{Al}{\left(\text{OH}\right)}_3}{1\text{mol}{\text{Al}}_{\text{2}}{\left({\text{SO}}_4\right)}_3}\right)\\ =0.0579\text{ mol}\end{array}$

Formula to calculate mass of $\text{Al}{\left(\text{OH}\right)}_3$ is as follows:

  $\text{Mass of Al}{\left(\text{OH}\right)}_3=\left(\text{Moles of Al}{\left(\text{OH}\right)}_3\right)\left(\text{Molar mass of Al}{\left(\text{OH}\right)}_3\right)$

Moles of $\text{Al}{\left(\text{OH}\right)}_3$ are $0.0579\text{ mol}$ .

Molar mass of $\text{Al}{\left(\text{OH}\right)}_3$ is $78.00\text{ g/mol}$ .

Substitute the values in above equation.

  $\begin{array}{c}\text{Mass of Al}{\left(\text{OH}\right)}_3=\left(\text{Moles of Al}{\left(\text{OH}\right)}_3\right)\left(\text{Molar mass of Al}{\left(\text{OH}\right)}_3\right)\\ =\left(0.0579\text{ mol}\right)\left(78.00\text{ g/mol}\right)\\ =4.5162\text{ g}\\ \approx \text{4}\text{.52 g}\end{array}$

Since mass of $\text{Al}{\left(\text{OH}\right)}_3$ formed by $\text{NaOH}$ is less than that formed by ${\text{Al}}_{\text{2}}{\left({\text{SO}}_4\right)}_3$ , $\text{NaOH}$ acts as limiting reactant. Therefore mass of $\text{Al}{\left(\text{OH}\right)}_3$ formed is $\text{2}\text{.57 g}$ .