Chapter 15, Problem 17PE

(a)

Interpretation Introduction

Interpretation:

Molarity of each ion found in $\text{1}{\text{.25 M CuBr}}_2$ 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 $1\text{ L}$ volume moles of ${\text{AlCl}}_3$ is $\text{0}\text{.070 mol}$. The formula to evaluate volume from molarity is given as follows:

  $M=\frac{n}{V}$

Here,

$V$ represents volume.

$M$ represents molarity.

$n$ represents number of moles.

Explanation of Solution

$\text{1}{\text{.25 M CuBr}}_2$ indicates that in $1\text{ L}$ they moles of ${\text{CuBr}}_2$ is $\text{1}\text{.25 mol}$.

${\text{CuBr}}_2$ 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:

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

Hence, molariry of ${\text{ Cu}}^{2+}$ and ${\text{Br}}^{-}$ is $1.25\text{ M}$ and $2.5\text{ M}$ respectively.

(b)

Interpretation Introduction

Interpretation:

Molarity of each ion found in $\text{3}{\text{.50 M K}}_3{\text{AsO}}_4$ 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:

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

Hence, molariry of ${\text{K}}^{+}$ and ${\text{AsO}}_4{}^{-}$ is $10.5\text{ M}$ and $3.50\text{ M}$ respectively.

(c)

Interpretation Introduction

Interpretation:

Molarity of each ion found in $\text{0}{\text{.75 M NaHCO}}_3$ 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:

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

Hence, molariry of ${\text{Na}}^{+}$ and ${\text{HCO}}_3{}^{-}$ is $0.75\text{ M}$$10.5\text{ M}$ and $0.75\text{ M}$ respectively.

(d)

Interpretation Introduction

Interpretation:

Molarity of each ion found in $\text{0}\text{.65 M }{\left({\text{NH}}_4\right)}_2{\text{SO}}_{\text{4}}$ 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:

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

Hence, molariry of ${\text{NH}}_4{}^{+}$ and ${\text{SO}}_{\text{4}}{}^{2-}$ is $0.75\text{ M}$$1.3\text{ M}$ and $0.65\text{ M}$ respectively.