Chapter 15, Problem 18PE

(a)

Interpretation Introduction

Interpretation:

Molarity of each ion found in $\text{2}{\text{.25 M FeCl}}_3$ 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 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

$\text{2}{\text{.25 M FeCl}}_3$ indicates that in $1\text{ L}$ they moles of ${\text{FeCl}}_3$ is $\text{1}\text{.25 mol}$.

${\text{FeCl}}_3$ 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:

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

(b)

Interpretation Introduction

Interpretation:

Molarity of each ion found in $\text{0}{\text{.75 M NaH}}_2{\text{PO}}_{\text{4}}$ 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:

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

(c)

Interpretation Introduction

Interpretation:

Molarity of each ion found in $\text{1}{\text{.20 M MgSO}}_4$ 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:

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

(d)

Interpretation Introduction

Interpretation:

Molarity of each ion found in $\text{0}\text{.35 M Ca}{\left({\text{ClO}}_{\text{3}}\right)}_{\text{2}}$ 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:

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