Chapter 15, Problem 61QAP

Interpretation Introduction

(a)

Interpretation:

A net ionic equation for the reaction needs to be determined that takes place when 25 ml of 0.500 M iron (II) sulfate is combined with 35.0 ml of 0.332 M barium hydroxide.

Concept introduction:

Net ionic equation is the ionic equation in which reactants are written in the form of ions if they occur as ions in a reaction medium and product form are shown as combination of ions. The charges of ions and each atom in the reaction are balanced

Answer to Problem 61QAP

The net ionic equation is as follows:

${\text{Fe}}^{\text{2+}}\left(\text{aq}\right)+{\text{SO}}_{\text{4}}{}^{2-}\left(\text{aq}\right)+{\text{Ba}}^{\text{2+}}\left(\text{aq}\right)+{\text{2OH}}^{-}\left(\text{aq}\right)\to \text{Fe}{\left(\text{OH}\right)}_{\text{2}}\left(\text{s}\right)+{\text{BaSO}}_{\text{4}}\left(\text{s}\right)$

Explanation of Solution

The equation of reaction of iron (II) sulfate with barium hydroxide is shown as follows:

${\text{FeSO}}_{\text{4}}\left(\text{aq}\right)+\text{Ba}{\left(\text{OH}\right)}_2\left(\text{aq}\right)\to \text{Fe}{\left(\text{OH}\right)}_{\text{2}}\left(\text{s}\right)+{\text{BaSO}}_{\text{4}}\left(\text{s}\right)$

Ionic equation for the above reaction is as follows:

${\text{Fe}}^{\text{2+}}\left(\text{aq}\right)+{\text{SO}}_{\text{4}}{}^{2-}\left(\text{aq}\right)+{\text{Ba}}^{\text{2+}}\left(\text{aq}\right)+{\text{2OH}}^{-}\left(\text{aq}\right)\to \text{Fe}{\left(\text{OH}\right)}_{\text{2}}\left(\text{s}\right)+{\text{BaSO}}_{\text{4}}\left(\text{s}\right)$

As there are no spectator ions that appear on both sides of the arrow, net ionic equation is same as ionic equation.

Hence, net ionic equation is-

${\text{Fe}}^{\text{2+}}\left(\text{aq}\right)+{\text{SO}}_{\text{4}}{}^{2-}\left(\text{aq}\right)+{\text{Ba}}^{\text{2+}}\left(\text{aq}\right)+{\text{2OH}}^{-}\left(\text{aq}\right)\to \text{Fe}{\left(\text{OH}\right)}_{\text{2}}\left(\text{s}\right)+{\text{BaSO}}_{\text{4}}\left(\text{s}\right)$

Interpretation Introduction

(b)

Interpretation:

The mass of the precipitate formed needs to be determined, when $\text{25 ml of }0.\text{5}00\text{M}$ iron (II) sulfate is combined with $\text{35}.0\text{ ml of }0.\text{332 M}$ barium hydroxide.

Concept introduction:

Molarity is one of the ways to express the concentration. Formula to calculate molarity is as follows:

$\text{Molarity=}\frac{\text{Moles}\text{of solute}}{\text{Volume}}$

To calculate moles, the above formula can be rearranged as follows:

$\text{Moles}\text{of solute}=\text{Molarity}\times \text{Volume}$

Answer to Problem 61QAP

The two precipitate formed are $\text{Fe}{\left(\text{OH}\right)}_{\text{2}}\left(\text{s}\right){\text{ and BaSO}}_{\text{4}}\left(\text{s}\right)\downarrow$.

Mass of $\text{Fe}{\left(\text{OH}\right)}_{\text{2}}$ is $1.123\text{g}$

Mass of ${\text{BaSO}}_{\text{4}}$ is $2.707\text{g}$.

Explanation of Solution

When iron sulfate is combined with barium hydroxide, two precipitates are formed. The equation for this gaseous reaction is shown below:

${\text{FeSO}}_{\text{4}}\left(\text{aq}\right)+\text{Ba}{\left(\text{OH}\right)}_2\left(\text{aq}\right)\to \text{Fe}{\left(\text{OH}\right)}_{\text{2}}\left(\text{s}\right)\downarrow +{\text{BaSO}}_{\text{4}}\left(\text{s}\right)\downarrow$

The two precipitate formed are $\text{Fe}{\left(\text{OH}\right)}_{\text{2}}\left(\text{s}\right){\text{ and BaSO}}_{\text{4}}\left(\text{s}\right)\downarrow$

$\begin{array}{c}{\text{Molarity of FeSO}}_4=0.\text{5}00\text{M}\\ \text{Volume}\text{of}{\text{FeSO}}_{\text{4}}=\text{25 ml}\\ =\text{0}\text{.025}\text{L}\end{array}$

$\begin{array}{c}\text{Moles}\text{of}{\text{FeSO}}_4=\text{Molarity}\times \text{Volume}\\ =0.\text{5}00\text{M}\times \text{0}\text{.025}\text{L}\\ =0.0125\text{mol}\end{array}$

$\begin{array}{c}\text{Molarity of Ba}{\left(\text{OH}\right)}_2=0.332\text{M}\\ \text{Volume}\text{of}\text{Ba}{\left(\text{OH}\right)}_2=3\text{5 ml}\\ =\text{0}\text{.035}\text{L}\end{array}$

$\begin{array}{c}\text{Moles}\text{of}\text{Ba}{\left(\text{OH}\right)}_2=\text{Molarity}\times \text{Volume}\\ =0.332\text{M}\times \text{0}\text{.035}\text{L}\\ =0.0116\text{mol}\end{array}$

$\begin{array}{c}\text{Mass}\text{of}\text{Fe}{\left(\text{OH}\right)}_{\text{2}}=0.0125\cancel{\text{mole}{\text{FeSO}}_{\text{4}}}\times \frac{\cancel{1\text{mol}\text{Fe}{\left(\text{OH}\right)}_{\text{2}}}}{1\cancel{\text{mol}{\text{FeSO}}_{\text{4}}}}\times \frac{89.86\text{g}}{\cancel{1\text{mol}\text{Fe}{\left(\text{OH}\right)}_{\text{2}}}}\\ =1.123\text{g}\\ \text{Mass}\text{of}{\text{BaSO}}_{\text{4}}=0.0116\cancel{\text{mol}\text{Ba}{\left(\text{OH}\right)}_{\text{2}}}\times \frac{\cancel{\text{1mol}{\text{BaSO}}_4}}{\text{1}\cancel{\text{mol}\text{Ba}{\left(\text{OH}\right)}_{\text{2}}}}\times \frac{233.38\text{g}}{\cancel{\text{1mol}{\text{BaSO}}_4}}\\ =2.707\text{g}\end{array}$

Hence, mass of

$\text{Fe}{\left(\text{OH}\right)}_{\text{2}}$ ( ppt) is $1.123\text{g}$ and mass of ${\text{BaSO}}_{\text{4}}$ is $2.707\text{g}$.

Interpretation Introduction

(c)

Interpretation:

The equilibrium concentrations of the ions in solutions when $\text{25 mL of }0.\text{5}00\text{M}$ iron (II) sulfate is combined with $\text{35}.0\text{ mL of }0.\text{332 M}$ barium hydroxide needs to be determined.

Concept introduction:

Molarity is one of the ways to express concentration. Formula to calculate molarity is shown as follows:

$\text{Molarity=}\frac{\text{Moles}\text{of solute}}{\text{Volume}}$

To calculate moles, the above formula is rearranged as follows:

$\text{Moles}\text{of solute}=\text{Molarity}\times \text{Volume}$

Limiting reactant in a reaction is that reactant which is completely used to form products.

Answer to Problem 61QAP

The equilibrium concentration of ions are as follows:

$\begin{array}{c}\left[{\text{Fe}}^{\text{2+}}\right]=0.193\text{M}\\ \left[{\text{SO}}_{\text{4}}{}^{2-}\right]=0.193\text{M}\\ \left[{\text{Ba}}^{2+}\right]=0\\ \left[{\text{OH}}^{-}\right]=0\end{array}$

Explanation of Solution

When iron sulfate is combined with barium hydroxide, two precipitates are formed. The equation for this gaseous reaction is as follows:

${\text{FeSO}}_{\text{4}}\left(\text{aq}\right)+\text{Ba}{\left(\text{OH}\right)}_2\left(\text{aq}\right)\to \text{Fe}{\left(\text{OH}\right)}_{\text{2}}\left(\text{s}\right)\downarrow +{\text{BaSO}}_{\text{4}}\left(\text{s}\right)\downarrow$

$\begin{array}{c}{\text{Molarity of FeSO}}_4=0.\text{5}00\text{M}\\ \text{Volume}\text{of}{\text{FeSO}}_{\text{4}}=\text{25 mL}\\ =\text{0}\text{.025}\text{L}\end{array}$

$\begin{array}{c}\text{Moles}\text{of}{\text{FeSO}}_4=\text{Molarity}\times \text{Volume}\\ =0.\text{5}00\text{M}\times \text{0}\text{.025}\text{L}\\ =0.0125\text{mol}\end{array}$

$\begin{array}{c}\text{Molarity of Ba}{\left(\text{OH}\right)}_2=0.332\text{M}\\ \text{Volume}\text{of}\text{Ba}{\left(\text{OH}\right)}_2=3\text{5 mL}\\ =\text{0}\text{.035}\text{L}\end{array}$

The number of moles of $\text{Ba}{\left(\text{OH}\right)}_2$ can be calculated as follows:

$\begin{array}{c}\text{Moles}\text{of}\text{Ba}{\left(\text{OH}\right)}_2=\text{Molarity}\times \text{Volume}\\ =0.332\text{M}\times \text{0}\text{.035}\text{L}\\ =0.0116\text{mol}\end{array}$

Draw ICE table, $\text{Ba}{\left(\text{OH}\right)}_2$ is the limiting reagent as $0.0116\text{mol}$ of $\text{Ba}{\left(\text{OH}\right)}_2$ is used to form product.

$\begin{array}{l}{\text{FeSO}}_{\text{4}}\left(\text{aq}\right)+\text{Ba}{\left(\text{OH}\right)}_2\left(\text{aq}\right)\\ \text{Initial}0.01250.0116\\ \text{Change}0.0116\text{0}\text{.0116}\\ \text{Equilibrium}0.00090\end{array}$

Now,

$\begin{array}{c}\text{Total}\text{volume=0}\text{.025}\text{L+0}\text{.035}\text{L}\\ \text{=0}\text{.060}\text{L}\end{array}$

The reaction is as follows:

${\text{FeSO}}_{\text{4}}\rightleftharpoons {\text{Fe}}^{\text{2+}}+{\text{SO}}_{\text{4}}{}^{2-}$

So $0.0009$ mole ${\text{FeSO}}_{\text{4}}$ produces $0.0009$ mole ${\text{Fe}}^{2+}$ and $0.0009$ mole ${\text{SO}}_{\text{4}}{}^{2-}$.

The concentration of ${\text{Fe}}^{2+}$ will be:

$\begin{array}{c}\left[{\text{Fe}}^{\text{2+}}\right]=\frac{0.0116\text{mol}}{0.060\text{L}}\\ =0.193\text{M}\end{array}$

The concentration of ${\text{SO}}_4{}^{\text{2-}}$ will be:

$\begin{array}{c}\left[{\text{SO}}_4{}^{\text{2-}}\right]=\frac{0.0116\text{mol}}{0.060\text{L}}\\ =0.193\text{M}\end{array}$

As moles of $\text{Ba}{\left(\text{OH}\right)}_2$ in equilibrium is zero so,

$\begin{array}{c}\left[{\text{Ba}}^{2+}\right]=0\\ \left[{\text{OH}}^{-}\right]=0\end{array}$

Hence,

$\begin{array}{c}\left[{\text{Fe}}^{\text{2+}}\right]=0.193\text{M}\\ \left[{\text{SO}}_{\text{4}}{}^{2-}\right]=0.193\text{M}\\ \left[{\text{Ba}}^{2+}\right]=0\\ \left[{\text{OH}}^{-}\right]=0\end{array}$