Chapter 15, Problem 20PE

(a)

Interpretation Introduction

Interpretation:

Mass of each ion found in $100\text{ mL}$ of $\text{2}{\text{.25 M FeCl}}_3$ has to be calculated.

Concept Introduction:

Molarity is quantitatively defined as moles of solute in one liter of solution. For example $\text{0}{\text{.070 M AlCl}}_3$ indicates that in $1\text{ L}$ the moles of ${\text{AlCl}}_3$ is $\text{0}\text{.070 mol}$. The expression to evaluate mass is as follows:

  $\text{Mass}=\left(\text{Number of moles}\right)\left(\text{Molar Mass}\right)$

Explanation of Solution

${\text{FeCl}}_3$ can be broken into ${\text{1 mol Fe}}^{3+}$ and $2{\text{ mol Cl}}^{-}$ as follows:

  ${\text{FeCl}}_3\rightleftharpoons {\text{Fe}}^{3+}+3{\text{Cl}}^{-}$

Since ${\text{1 M Fe}}^{3+}$ is furnished by ${\text{1 M FeCl}}_3$ so molarity of ${\text{Fe}}^{3+}$ ion due to $\text{2}{\text{.25 M FeCl}}_3$ is calculated as follows:

  $\begin{array}{c}{\text{Molarity of Fe}}^{3+}=\left(\text{2}{\text{.25 M FeCl}}_3\right)\left(\frac{{\text{1 M Fe}}^{3+}}{{\text{1 M FeCl}}_3}\right)\\ =2.25{\text{ M Fe}}^{3+}\end{array}$

Since $2{\text{ M Cl}}^{-}$ is furnished by ${\text{1 M FeCl}}_3$ so molarity of ${\text{Cl}}^{-}$ ions due to $\text{2}{\text{.25 M FeCl}}_3$ is calculated as follows:

  $\begin{array}{c}{\text{Molarity of Cl}}^{-}=\left(\text{2}{\text{.25 M FeCl}}_3\right)\left(\frac{2{\text{ M Cl}}^{-}}{{\text{1 M FeCl}}_3}\right)\\ =4.5{\text{ M Cl}}^{-}\end{array}$

$2.25{\text{ M Fe}}^{3+}$ indicates that in $1000\text{ mL}$ moles of ${\text{Fe}}^{3+}$ is $\text{2}\text{.25 mol}$ thus moles of ${\text{Fe}}^{3+}$ in $100\text{ mL}$ is calculated as follows:

  $\begin{array}{c}{\text{Moles of Fe}}^{3+}=\left(100\text{ mL}\right)\left(\frac{\text{2}\text{.25 mol}}{1000\text{ mL}}\right)\\ =0.225\text{ mol}\end{array}$

$4.5{\text{ M Cl}}^{-}$ indicates that in $1000\text{ mL}$ the moles of ${\text{Cl}}^{-}$ is $\text{4}\text{.5 mol}$ thus moles of ${\text{Cl}}^{-}$ in $100\text{ mL}$ is calculated as follows:

  $\begin{array}{c}{\text{Moles of Cl}}^{-}=\left(100\text{ mL}\right)\left(\frac{\text{4}\text{.5 mol}}{1000\text{ mL}}\right)\\ =0.45\text{ mol}\end{array}$

The expression to evaluate mass is as follows:

  $\text{Mass}=\left(\text{Number of moles}\right)\left(\text{Molar Mass}\right)$        (1)

Substitute $0.225\text{ mol}$ for number of moles and $\text{55}\text{.845 g/mol}$ for molar mass of ${\text{Fe}}^{3+}$ in equation (1).

  $\begin{array}{c}\text{Mass}=\left(0.225\text{ mol}\right)\left(\text{55}\text{.845 g/mol}\right)\\ =12.5651\text{ g}\end{array}$

Substitute $0.45\text{ mol}$ for number of moles and $\text{35}\text{.453 g/mol}$ for molar mass of ${\text{Cl}}^{-}$ in equation (1).

  $\begin{array}{c}\text{Mass}=\left(0.45\text{ mol}\right)\left(\text{35}\text{.453 g/mol}\right)\\ =15.9538\text{ g}\end{array}$

Thus mass of  ${\text{Fe}}^{3+}$ , ${\text{Cl}}^{-}$ are $12.5651\text{ g}$ and $15.9538\text{ g}$ respectively.

(b)

Interpretation Introduction

Interpretation:

Mass of each ion found in $100\text{ mL}$ of $\text{0}{\text{.75 M NaH}}_2{\text{PO}}_{\text{4}}$ has to be calculated.

Concept Introduction:

Refer to part (a).

Explanation of Solution

${\text{NaH}}_2{\text{PO}}_{\text{4}}$ can be broken into ${\text{1 mol Na}}^{+}$ and  ${\text{1 mol H}}_2{\text{PO}}_4^{-}$ as follows:

  ${\text{NaH}}_2{\text{PO}}_{\text{4}}\rightleftharpoons {\text{Na}}^{+}+{\text{H}}_2{\text{PO}}_4^{-}$

Since ${\text{1 M Na}}^{+}$ is furnished by ${\text{1 M NaH}}_2{\text{PO}}_{\text{4}}$ so molarity of ${\text{Na}}^{+}$ ions due to $\text{0}{\text{.75 M NaH}}_2{\text{PO}}_{\text{4}}$ is calculated as follows:

  $\begin{array}{c}{\text{Molarity of Na}}^{+}=\left(\text{0}{\text{.75 M NaH}}_2{\text{PO}}_{\text{4}}\right)\left(\frac{{\text{1 M Na}}^{+}}{{\text{1 M NaH}}_2{\text{PO}}_{\text{4}}}\right)\\ =0.75{\text{ M Na}}^{+}\end{array}$

Since ${\text{1 M H}}_2{\text{PO}}_4^{-}$ is furnished by ${\text{1 M NaH}}_2{\text{PO}}_{\text{4}}$ so molarity of ${\text{H}}_2{\text{PO}}_4^{-}$ ion due to $\text{0}{\text{.75 M NaH}}_2{\text{PO}}_{\text{4}}$ is calculated as follows:

  $\begin{array}{c}{\text{Molarity of H}}_2{\text{PO}}_4^{-}=\left(\text{0}{\text{.75 M NaH}}_2{\text{PO}}_{\text{4}}\right)\left(\frac{{\text{1 M H}}_2{\text{PO}}_4^{-}}{{\text{1 M NaH}}_2{\text{PO}}_{\text{4}}}\right)\\ =0.75{\text{ M H}}_2{\text{PO}}_4^{-}\end{array}$

$0.75{\text{ M Na}}^{+}$ indicates that in $1000\text{ mL}$ the moles of ${\text{Na}}^{+}$ is $\text{0}\text{.75 mol}$ thus moles of ${\text{Na}}^{+}$ in $100\text{ mL}$ is calculated as follows:

  $\begin{array}{c}{\text{Moles of Na}}^{+}=\left(100\text{ mL}\right)\left(\frac{\text{0}\text{.75 mol}}{1000\text{ mL}}\right)\\ =0.075\text{ mol}\end{array}$

$0.75{\text{ M H}}_2{\text{PO}}_4^{-}$ indicates that in $1000\text{ mL}$ moles of ${\text{H}}_2{\text{PO}}_4^{-}$ is $\text{0}\text{.75 mol}$ thus moles of ${\text{H}}_2{\text{PO}}_4^{-}$ in $100\text{ mL}$ is calculated as follows:

  $\begin{array}{c}{\text{Moles of H}}_2{\text{PO}}_4^{-}=\left(100\text{ mL}\right)\left(\frac{\text{0}\text{.75 mol}}{1000\text{ mL}}\right)\\ =0.075\text{ mol}\end{array}$

Substitute $0.075\text{ mol}$ for number of moles and $\text{22}\text{.989 g/mol}$ for molar mass of ${\text{Na}}^{+}$ in equation (1).

  $\begin{array}{c}\text{Mass}=\left(\text{0}\text{.075 mol}\right)\left(\text{22}\text{.989 g/mol}\right)\\ =1.7241\text{ g}\end{array}$

Substitute $0.075\text{ mol}$ for number of moles and $\text{96}\text{.987 g/mol}$ for molar mass of ${\text{H}}_2{\text{PO}}_4^{-}$ in equation (1).

  $\begin{array}{c}\text{Mass}=\left(0.075\text{ mol}\right)\left(\text{96}\text{.987 g/mol}\right)\\ =7.274\text{ g}\end{array}$

Thus masses of ${\text{Na}}^{+}$ and ${\text{H}}_2{\text{PO}}_4^{-}$ is $1.7241\text{ g}$ and $7.274\text{ g}$ respectively.

(c)

Interpretation Introduction

Interpretation:

Mass of each ion found in $100\text{ mL}$ of $\text{1}{\text{.20 M MgSO}}_4$ has to be calculated.

Concept Introduction:

Refer to part (a).

Explanation of Solution

${\text{MgSO}}_4$ can be broken into ${\text{1 mol Mg}}^{2+}$ and  ${\text{1 mol SO}}_4^{2-}$ as follows:

  ${\text{MgSO}}_4\rightleftharpoons {\text{Mg}}^{2+}+{\text{SO}}_4^{2-}$

Since ${\text{1 M Mg}}^{2+}$ is furnished by ${\text{1 M MgSO}}_4$ so molarity of ${\text{Mg}}^{2+}$ ions due to $\text{1}{\text{.20 M MgSO}}_4$ is calculated as follows:

  $\begin{array}{c}{\text{Molarity of Mg}}^{2+}=\left(\text{1}{\text{.20 M MgSO}}_4\right)\left(\frac{{\text{1 M Mg}}^{2+}}{{\text{1 M MgSO}}_4}\right)\\ =\text{1}{\text{.20 M Mg}}^{2+}\end{array}$

Since ${\text{1 M SO}}_4^{2-}$ is furnished by ${\text{1 M MgSO}}_4$ so molarity of ${\text{SO}}_4^{2-}$ ion due to $\text{1}{\text{.20 M MgSO}}_4$ is calculated as follows:

  $\begin{array}{c}{\text{Molarity of SO}}_4^{2-}=\left(\text{1}{\text{.20 M MgSO}}_4\right)\left(\frac{{\text{1 M SO}}_4^{2-}}{{\text{1 M MgSO}}_4}\right)\\ =\text{1}{\text{.20 M SO}}_4^{2-}\end{array}$

$\text{1}{\text{.20 M Mg}}^{2+}$ indicates that in $1000\text{ mL}$ the moles of ${\text{Mg}}^{2+}$ is $\text{1}\text{.20 mol}$ thus moles of ${\text{Mg}}^{2+}$ in $100\text{ mL}$ is calculated as follows:

  $\begin{array}{c}{\text{Moles of Mg}}^{2+}=\left(100\text{ mL}\right)\left(\frac{\text{1}\text{.20 mol}}{1000\text{ mL}}\right)\\ =0.120\text{ mol}\end{array}$

$\text{1}{\text{.20 M SO}}_4^{2-}$ indicates that in $1000\text{ mL}$ moles of ${\text{SO}}_4^{2-}$ is $\text{1}\text{.20 mol}$ thus moles of ${\text{SO}}_4^{2-}$ in $100\text{ mL}$ is calculated as follows:

  $\begin{array}{c}{\text{Moles of SO}}_4^{2-}=\left(100\text{ mL}\right)\left(\frac{\text{1}\text{.20 mol}}{1000\text{ mL}}\right)\\ =0.120\text{ mol}\end{array}$

Substitute $\text{0}\text{.120 mol}$ for number of moles and $\text{24}\text{.305 g/mol}$ for molar mass of ${\text{Mg}}^{2+}$ in equation (1).

  $\begin{array}{c}\text{Mass}=\left(\text{0}\text{.120 mol}\right)\left(\text{24}\text{.305 g/mol}\right)\\ =2.9166\text{ g}\end{array}$

Substitute $\text{0}\text{.120 mol}$ for number of moles and $\text{96}\text{.06 g/mol}$ for molar mass of ${\text{SO}}_4^{2-}$ in equation (1).

  $\begin{array}{c}\text{Mass}=\left(\text{0}\text{.120 mol}\right)\left(\text{96}\text{.06 g/mol}\right)\\ =11.5272\text{ g}\end{array}$

Thus mass of ${\text{Mg}}^{2+}$, ${\text{SO}}_4^{2-}$ are $2.9166\text{ g}$ and $21.355\text{ g}$ respectively.

(d)

Interpretation Introduction

Interpretation:

Mass of each ion found in $100\text{ mL}$ of $\text{0}\text{.35 M Ca}{\left({\text{ClO}}_{\text{3}}\right)}_{\text{2}}$ has to be calculated.

Concept Introduction:

Refer to part (a).

Explanation of Solution

$\text{Ca}{\left({\text{ClO}}_{\text{3}}\right)}_{\text{2}}$ can be broken into ${\text{1 mol Ca}}^{2+}$ and  ${\text{2 mol ClO}}_3^{-}$ as follows:

  $\text{Ca}{\left({\text{ClO}}_{\text{3}}\right)}_{\text{2}}\rightleftharpoons {\text{Ca}}^{2+}+2{\text{ClO}}_3^{-}$

Since ${\text{1 M Ca}}^{2+}$ is furnished by $\text{1 M Ca}{\left({\text{ClO}}_{\text{3}}\right)}_{\text{2}}$ so molarity of ${\text{Ca}}^{2+}$ ions due to $\text{0}\text{.35 M Ca}{\left({\text{ClO}}_{\text{3}}\right)}_{\text{2}}$ is calculated as follows:

  $\begin{array}{c}{\text{Molarity of Ca}}^{2+}=\left(\text{0}\text{.35 M Ca}{\left({\text{ClO}}_{\text{3}}\right)}_{\text{2}}\right)\left(\frac{{\text{1 M Ca}}^{2+}}{\text{1 M Ca}{\left({\text{ClO}}_{\text{3}}\right)}_{\text{2}}}\right)\\ =0.35{\text{ M Ca}}^{2+}\end{array}$

Since ${\text{2 M ClO}}_3^{-}$ is furnished by $\text{1 M Ca}{\left({\text{ClO}}_{\text{3}}\right)}_{\text{2}}$ so molarity of ${\text{ClO}}_3^{-}$ ion due to $\text{0}\text{.35 M Ca}{\left({\text{ClO}}_{\text{3}}\right)}_{\text{2}}$ is calculated as follows:

  $\begin{array}{c}{\text{Molarity of ClO}}_3^{-}=\left(\text{0}\text{.35 M Ca}{\left({\text{ClO}}_{\text{3}}\right)}_{\text{2}}\right)\left(\frac{{\text{2 M ClO}}_3^{-}}{\text{1 M Ca}{\left({\text{ClO}}_{\text{3}}\right)}_{\text{2}}}\right)\\ =0.7{\text{ M ClO}}_3^{-}\end{array}$

$0.35{\text{ M Ca}}^{2+}$ indicates that in $1000\text{ mL}$ the moles of ${\text{Ca}}^{2+}$ is $\text{0}\text{.35 mol}$ thus moles of ${\text{Ca}}^{2+}$ in $100\text{ mL}$ is calculated as follows:

  $\begin{array}{c}{\text{Moles of Ca}}^{2+}=\left(100\text{ mL}\right)\left(\frac{\text{0}\text{.35 mol}}{1000\text{ mL}}\right)\\ =0.035\text{ mol}\end{array}$

$0.7{\text{ M ClO}}_3^{-}$ indicates that in $1000\text{ mL}$ moles of ${\text{ClO}}_3^{-}$ is $\text{0}\text{.7 mol}$ thus moles of ${\text{ClO}}_3^{-}$ in $100\text{ mL}$ is calculated as follows:

  $\begin{array}{c}{\text{Moles of ClO}}_3^{-}=\left(100\text{ mL}\right)\left(\frac{\text{0}\text{.7 mol}}{1000\text{ mL}}\right)\\ =0.07\text{ mol}\end{array}$

Substitute $0.035\text{ mol}$ for number of moles and $\text{40}\text{.078 g/mol}$ for molar mass of ${\text{Ca}}^{2+}$ in equation (1).

  $\begin{array}{c}\text{Mass}=\left(0.035\text{ mol}\right)\left(\text{40}\text{.078 g/mol}\right)\\ =1.40273\text{ g}\end{array}$

Substitute $0.07\text{ mol}$ for number of moles and $\text{99}\text{.45 g/mol}$ for molar mass of ${\text{ClO}}_3^{-}$ in equation (1).

  $\begin{array}{c}\text{Mass}=\left(0.07\text{ mol}\right)\left(\text{99}\text{.45 g/mol}\right)\\ =6.9615\text{ g}\end{array}$

Thus mass of ${\text{Ca}}^{2+}$ and ${\text{ClO}}_3^{-}$ are $1.40273\text{ g}$ and $6.9615\text{ g}$ respectively.