Chapter 15, Problem 19PE

(a)

Interpretation Introduction

Interpretation:

Mass of each ion found in $100\text{ mL}$ of $\text{1}{\text{.25 M CuBr}}_2$ 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

${\text{CuBr}}_2$ can be broken into ${\text{1 mol Cu}}^{2+}$ and $2{\text{ mol Br}}^{-}$ as follows:

  ${\text{CuBr}}_2\rightleftharpoons {\text{Cu}}^{2+}+2{\text{Br}}^{-}$

Since ${\text{1 M Cu}}^{2+}$ is furnished by ${\text{1 M CuBr}}_2$ so molarity of ${\text{Cu}}^{2+}$ ion due to $\text{1}{\text{.25 M CuBr}}_2$ is calculated as follows:

  $\begin{array}{c}{\text{Molarity of Cu}}^{2+}=\left(\text{1}{\text{.25 M CuBr}}_2\right)\left(\frac{{\text{1 M Cu}}^{2+}}{{\text{1 M CuBr}}_2}\right)\\ =1.25{\text{ M Cu}}^{2+}\end{array}$

Since $2{\text{ M Br}}^{-}$ is furnished by ${\text{1 M CuBr}}_2$ so molarity of ${\text{Br}}^{-}$ ions due to $\text{1}{\text{.25 M CuBr}}_2$ is calculated as follows:

  $\begin{array}{c}{\text{Molarity of Br}}^{-}=\left(\text{1}{\text{.25 M CuBr}}_2\right)\left(\frac{2{\text{ M Br}}^{-}}{{\text{1 M CuBr}}_2}\right)\\ =2.5{\text{ M Br}}^{-}\end{array}$

$1.25{\text{ M Cu}}^{2+}$ indicates that in $1000\text{ mL}$ the moles of ${\text{Cu}}^{2+}$ is $\text{2}\text{.5 mol}$ thus moles of ${\text{Cu}}^{2+}$ in $100\text{ mL}$ is calculated as follows:

  $\begin{array}{c}{\text{Moles of Cu}}^{2+}=\left(100\text{ mL}\right)\left(\frac{\text{1}\text{.25 mol}}{1000\text{ mL}}\right)\\ =0.125\text{ mol}\end{array}$

$2.5{\text{ M Br}}^{-}$ indicates that in $1000\text{ mL}$ the moles of ${\text{Br}}^{-}$ is $\text{2}\text{.5 mol}$ thus moles of ${\text{Br}}^{-}$ in

$100\text{ mL}$ is calculated as follows:

  $\begin{array}{c}{\text{Moles of Br}}^{-}=\left(100\text{ mL}\right)\left(\frac{\text{2}\text{.5 mol}}{1000\text{ mL}}\right)\\ =0.25\text{ mol}\end{array}$

The expression to evaluate mass is as follows:

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

Substitute $\text{0}\text{.25 mol}$ for number of moles and $\text{79}\text{.904 g/mol}$ for molar mass of ${\text{Br}}^{-}$ in equation (1).

  $\begin{array}{c}\text{Mass}=\left(\text{25 mol}\right)\left(\text{79}\text{.904 g/mol}\right)\\ =19.976\text{ g}\end{array}$

Substitute $0.125\text{ mol}$ for number of moles and $\text{63}\text{.546 g/mol}$ for molar mass of ${\text{Cu}}^{2+}$ in equation (1).

  $\begin{array}{c}\text{Mass}=\left(0.125\text{ mol}\right)\left(\text{63}\text{.546 g/mol}\right)\\ =7.94\text{ g}\end{array}$

Thus, mass of ${\text{Cu}}^{2+}$ and ${\text{Br}}^{-}$ are $7.94\text{ g}$ and $19.976\text{ g}$ respectively.

(b)

Interpretation Introduction

Interpretation:

Mass of each ion found in $100\text{ mL}$ of $\text{3}{\text{.50 M K}}_3{\text{AsO}}_4$ has to be calculated.

Concept Introduction:

Refer to part (a).

Explanation of Solution

${\text{K}}_3{\text{AsO}}_4$ can be broken into ${\text{3 mol K}}^{+}$ and  ${\text{1 mol AsO}}_4{}^{-}$ as follows:

  ${\text{K}}_3{\text{AsO}}_4\rightleftharpoons 3{\text{K}}^{+}+{\text{AsO}}_4{}^{-}$

Since ${\text{3 M K}}^{+}$ is furnished by ${\text{1 M AlCl}}_3$ so molarity of ${\text{K}}^{+}$ ions due to $\text{3}{\text{.50 M K}}_3{\text{AsO}}_4$ is calculated as follows:

  $\begin{array}{c}{\text{Molarity of K}}^{+}=\left(\text{3}{\text{.50 M K}}_3{\text{AsO}}_4\right)\left(\frac{{\text{3 M K}}^{+}}{{\text{1 M K}}_3{\text{AsO}}_4}\right)\\ =10.5{\text{ M K}}^{+}\end{array}$

Since ${\text{1 M AsO}}_4{}^{-}$ is furnished by ${\text{1 M K}}_3{\text{AsO}}_4$ so molarity of ${\text{AsO}}_4{}^{-}$ ion due to $\text{3}{\text{.50 M K}}_3{\text{AsO}}_4$ is calculated as follows:

  $\begin{array}{c}{\text{Molarity of AsO}}_4{}^{-}=\left(\text{3}{\text{.50 M K}}_3{\text{AsO}}_4\right)\left(\frac{{\text{1 M AsO}}_4{}^{-}}{{\text{1 M K}}_3{\text{AsO}}_4}\right)\\ =3.50{\text{ M AsO}}_4{}^{-}\end{array}$

$10.5{\text{ M K}}^{+}$ indicates that in $1000\text{ mL}$ the moles of ${\text{K}}^{+}$ is $\text{10}\text{.5 mol}$ thus moles of ${\text{K}}^{+}$ in $100\text{ mL}$ is calculated as follows:

  $\begin{array}{c}{\text{Moles of K}}^{+}=\left(100\text{ mL}\right)\left(\frac{\text{10}\text{.5 mol}}{1000\text{ mL}}\right)\\ =1.05\text{ mol}\end{array}$

$3.50{\text{ M AsO}}_4{}^{-}$ indicates that in $1000\text{ mL}$ moles of ${\text{AsO}}_4{}^{-}$ is $\text{3}\text{.5 mol}$ thus moles of ${\text{AsO}}_4{}^{-}$ in $100\text{ mL}$ is calculated as follows:

  $\begin{array}{c}{\text{Moles of Br}}^{-}=\left(100\text{ mL}\right)\left(\frac{\text{3}\text{.50 mol}}{1000\text{ mL}}\right)\\ =0.35\text{ mol}\end{array}$

Substitute $\text{1}\text{.05 mol}$ for number of moles and $\text{39}\text{.098 g/mol}$ for molar mass of ${\text{K}}^{+}$ in equation (1).

  $\begin{array}{c}\text{Mass}=\left(\text{1}\text{.05 mol}\right)\left(\text{39}\text{.098 g/mol}\right)\\ =41.053\text{ g}\end{array}$

Substitute $\text{0}\text{.35 mol}$ for number of moles and $\text{138}\text{.919 g/mol}$ for molar mass of ${\text{AsO}}_4{}^{-}$ in equation (1).

  $\begin{array}{c}\text{Mass}=\left(\text{0}\text{.35 mol}\right)\left(\text{138}\text{.919 g/mol}\right)\\ =48.621\text{ g}\end{array}$

Thus, mass of ${\text{K}}^{+}$ and ${\text{AsO}}_4{}^{-}$ are $41.053\text{ g}$ and $48.621\text{ g}$ respectively.

(c)

Interpretation Introduction

Interpretation:

Mass of each ion found in $100\text{ mL}$ of $\text{0}{\text{.75 M NaHCO}}_3$ has to be calculated.

Concept Introduction:

Refer to part (a).

Explanation of Solution

${\text{NaHCO}}_3$ can be broken into ${\text{1 mol Na}}^{+}$ and  ${\text{1 mol HCO}}_3{}^{-}$ as follows:

  ${\text{NaHCO}}_3\rightleftharpoons {\text{Na}}^{+}+{\text{HCO}}_3{}^{-}$

Since ${\text{1 M Na}}^{+}$ is furnished by ${\text{1 M NaHCO}}_3$ so molarity of ${\text{Na}}^{+}$ ions due to $\text{0}{\text{.75 M NaHCO}}_3$ is calculated as follows:

  $\begin{array}{c}{\text{Molarity of Na}}^{+}=\left(\text{0}{\text{.75 M NaHCO}}_3\right)\left(\frac{{\text{1 M Na}}^{+}}{{\text{1 M NaHCO}}_3}\right)\\ =0.75{\text{ M Na}}^{+}\end{array}$

Since ${\text{1 M HCO}}_3{}^{-}$ is furnished by ${\text{1 M NaHCO}}_3$ so molarity of ${\text{HCO}}_3{}^{-}$ ions due to $\text{0}{\text{.75 M NaHCO}}_3$ is calculated as follows:

  $\begin{array}{c}{\text{Molarity of HCO}}_3{}^{-}=\left(\text{0}{\text{.75 M NaHCO}}_3\right)\left(\frac{{\text{1 M HCO}}_3{}^{-}}{{\text{1 M NaHCO}}_3}\right)\\ =0.75{\text{ M HCO}}_3{}^{-}\end{array}$

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

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

$0.75{\text{ M HCO}}_3{}^{-}$ indicates that in $1000\text{ mL}$ moles of ${\text{HCO}}_3{}^{-}$ is $\text{0}\text{.75 mol}$ thus moles of ${\text{HCO}}_3{}^{-}$ in $100\text{ mL}$ is calculated as follows:

  $\begin{array}{c}{\text{Moles of HCO}}_3{}^{-}=\left(100\text{ mL}\right)\left(\frac{\text{0}\text{.75 mol}}{1000\text{ mL}}\right)\\ =0.075\text{ mol}\end{array}$

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).

  $\begin{array}{c}\text{Mass}=\left(\text{0}\text{.075 mol}\right)\left(\text{22}\text{.989 g/mol}\right)\\ =1.724\text{ g}\end{array}$

Substitute $0.075\text{ mol}$ for number of moles and $\text{61}\text{.0168 g/mol}$ for molar mass of ${\text{HCO}}_3{}^{-}$ in equation (1).

  $\begin{array}{c}\text{Mass}=\left(\text{0}\text{.35 mol}\right)\left(\text{61}\text{.0168 g/mol}\right)\\ =21.355\text{ g}\end{array}$

Thus in $100\text{ mL}$ mass of ${\text{Na}}^{+}$ and ${\text{HCO}}_3{}^{-}$ are $1.724\text{ g}$ and $21.355\text{ g}$ respectively.

(d)

Interpretation Introduction

Interpretation:

Mass of each ion found in $100\text{ mL}$ of $\text{0}\text{.65 M }{\left({\text{NH}}_4\right)}_2{\text{SO}}_{\text{4}}$ has to be calculated.

Concept Introduction:

Refer to part (a).

Explanation of Solution

${\left({\text{NH}}_4\right)}_2{\text{SO}}_{\text{4}}$ can be broken into ${\text{2 mol NH}}_4{}^{+}$ and  ${\text{1 mol SO}}_{\text{4}}{}^{2-}$ as follows:

  ${\left({\text{NH}}_4\right)}_2{\text{SO}}_{\text{4}}\rightleftharpoons 2{\text{NH}}_4{}^{+}+{\text{SO}}_{\text{4}}{}^{2-}$

Since ${\text{2 M NH}}_4{}^{+}$ is furnished by $\text{1 M }{\left({\text{NH}}_4\right)}_2{\text{SO}}_{\text{4}}$ so molarity of ${\text{NH}}_4{}^{+}$ ions due to $\text{0}\text{.65 M }{\left({\text{NH}}_4\right)}_2{\text{SO}}_{\text{4}}$ is calculated as follows:

  $\begin{array}{c}{\text{Molarity of NH}}_4{}^{+}=\left(\text{0}\text{.65 M }{\left({\text{NH}}_4\right)}_2{\text{SO}}_{\text{4}}\right)\left(\frac{{\text{2 M NH}}_4{}^{+}}{\text{1 M }{\left({\text{NH}}_4\right)}_2{\text{SO}}_{\text{4}}}\right)\\ =1.3{\text{ M NH}}_4{}^{+}\end{array}$

Since ${\text{1 M SO}}_{\text{4}}{}^{2-}$ is furnished by $\text{1 M }{\left({\text{NH}}_4\right)}_2{\text{SO}}_{\text{4}}$ so molarity of ${\text{SO}}_{\text{4}}{}^{2-}$ ion due to $\text{0}\text{.65 M }{\left({\text{NH}}_4\right)}_2{\text{SO}}_{\text{4}}$ is calculated as follows:

  $\begin{array}{c}{\text{Molarity of SO}}_{\text{4}}{}^{2-}=\left(\text{0}\text{.65 M }{\left({\text{NH}}_4\right)}_2{\text{SO}}_{\text{4}}\right)\left(\frac{{\text{1 M SO}}_{\text{4}}{}^{2-}}{\text{1 M }{\left({\text{NH}}_4\right)}_2{\text{SO}}_{\text{4}}}\right)\\ =0.65{\text{ M SO}}_{\text{4}}{}^{2-}\end{array}$

$1.3{\text{ M NH}}_4{}^{+}$ indicates that in $1000\text{ mL}$ the moles of ${\text{NH}}_4{}^{+}$ is $\text{1}\text{.3 mol}$ thus moles of ${\text{NH}}_4{}^{+}$ in $100\text{ mL}$ is calculated as follows:

  $\begin{array}{c}{\text{Moles of NH}}_4{}^{+}=\left(100\text{ mL}\right)\left(\frac{\text{1}\text{.3 mol}}{1000\text{ mL}}\right)\\ =0.13\text{ mol}\end{array}$

$0.65{\text{ M SO}}_{\text{4}}{}^{2-}$ indicates that in $1000\text{ mL}$ moles of ${\text{SO}}_{\text{4}}{}^{2-}$ is $\text{0}\text{.65 mol}$ thus moles of ${\text{SO}}_{\text{4}}{}^{2-}$ in $100\text{ mL}$ is calculated as follows:

  $\begin{array}{c}{\text{Moles of HCO}}_3{}^{-}=\left(100\text{ mL}\right)\left(\frac{\text{0}\text{.65 mol}}{1000\text{ mL}}\right)\\ =0.065\text{ mol}\end{array}$

Substitute $0.13\text{ mol}$ for number of moles and $\text{18}\text{.039 g/mol}$ for molar mass of ${\text{NH}}_4{}^{+}$ in equation (1).

  $\begin{array}{c}\text{Mass}=\left(0.13\text{ mol}\right)\left(\text{18}\text{.039 g/mol}\right)\\ =2.345\text{ g}\end{array}$

Substitute $0.65\text{ mol}$ for number of moles and $\text{96}\text{.06 g/mol}$ for molar mass of ${\text{SO}}_{\text{4}}{}^{2-}$ in equation (1).

  $\begin{array}{c}\text{Mass}=\left(0.65\text{ mol}\right)\left(\text{96}\text{.06 g/mol}\right)\\ =62.439\text{ g}\end{array}$

Thus mass of ${\text{NH}}_4{}^{+}$ and ${\text{SO}}_{\text{4}}{}^{2-}$ are $2.345\text{ g}$ and $62.439\text{ g}$ respectively.