Chapter 10, Problem 10.145EP

(a)

Interpretation Introduction

Interpretation:

A solution having ${\text{75 mEq Na}}^{\text{+}}{\text{, 25 mEq K}}^{\text{+}}{\text{, 95 mEq Cl}}^{-}{\text{, and 5 mEq NO}}_{\text{3}}{}^{-}$ is whether possible to prepare has to be determined considering charge-balance.

Concept-Introduction:

Charge balance takes place when the sum of concentrations of negative ion and the sum of concentrations of positive ion becomes equal in $\text{mEq/L or Eq/L}$ concentration units.

Answer to Problem 10.145EP

A solution having ${\text{75 mEq Na}}^{\text{+}}{\text{, 25 mEq K}}^{\text{+}}{\text{, 95 mEq Cl}}^{-}{\text{, and 5 mEq NO}}_{\text{3}}{}^{-}$ is possible to prepare.

Explanation of Solution

Given data is shown below:

  $\begin{array}{c}Concentration{\text{ of Na}}^{\text{+}}\text{ = 75 mEq}\\ \\ Concentration{\text{ of K}}^{\text{+}}\text{ = 25 mEq}\\ \\ Concentration{\text{ of NO}}_{\text{3}}{}^{-}\text{ = 5 mEq}\\ \\ Concentration{\text{ of Cl}}^{-}\text{ = 95 mEq}\end{array}$

A charge balance exists between the ions in the given electrolyte solution when the sum of $\text{Eq/L}$ of positive ions is equal to the sum of $\text{Eq/L}$ of negative ions.

Sum of $\text{mEq}$ of negative ions is calculated as shown,

  $\begin{array}{c}\text{mEq of negative ions}={\text{ 95 mEq Cl}}^{-}{\text{+ 5 mEq NO}}_{\text{3}}{}^{-}\\ \\ =\text{ 100 mEq negative ions}\text{.}\end{array}$

Thus, the sum of $\text{mEq}$ of positive ions must be also $\text{100 mEq}\text{.}$

Sum of $\text{mEq}$ of positive ions can be determined as follows,

  $\begin{array}{c}\text{mEq }ofpositiveions{\text{ = 75 mEq Na}}^{\text{+}}+{\text{ 25 mEq K}}^{\text{+}}\\ \\ \text{ = 100 mEq }positive\text{ ions}\end{array}$

Here, the sum of $\text{mEq/L}$ of positive ions is equal to the sum of $\text{mEq/L}$ of negative ions

Therefore,

A solution having ${\text{75 mEq Na}}^{\text{+}}{\text{, 25 mEq K}}^{\text{+}}{\text{, 95 mEq Cl}}^{-}{\text{, and 5 mEq NO}}_{\text{3}}{}^{-}$ is possible to prepare.

(b)

Interpretation Introduction

Interpretation:

A solution having ${\text{73 Eq K}}^{\text{+}}{\text{, 55 Eq Cl}}^{-}{\text{, and 25 Eq C}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}{}^{-}$ is whether possible to prepare has to be determined considering charge-balance.

Concept-Introduction:

Charge balance takes place when the sum of concentrations of negative ion and the sum of concentrations of positive ion becomes equal in $\text{mEq/L or Eq/L}$ concentration units.

Answer to Problem 10.145EP

A solution having ${\text{73 Eq K}}^{\text{+}}{\text{, 55 Eq Cl}}^{-}{\text{, and 25 Eq C}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}{}^{-}$ is not possible to prepare.

Explanation of Solution

Given data is shown below:

  $\begin{array}{c}Conc{\text{ of K}}^{\text{+}}=\text{ 73 Eq }\\ \\ Conc{\text{ of Cl}}^{-}\text{ = 55 Eq}\\ \\ Conc{\text{ of C}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}{}^{-}\text{ = 25 Eq }\end{array}$

A charge balance exists between the ions in the given electrolyte solution when the sum of $\text{Eq/L}$ of positive ions is equal to the sum of $\text{Eq/L}$ of negative ions.

Sum of $\text{Eq}$ of negative ions is calculated as shown,

  $\begin{array}{c}\text{mEq of negative ions}={\text{ 55 Eq Cl}}^{-}{\text{ + 25 Eq C}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}{}^{-}\\ \\ =\text{ 80 Eq negative ions}\text{.}\end{array}$

Thus, the sum of $\text{Eq}$ of positive ions must be also $\text{80 Eq}$ for charge balance to exist.  However, sum of $\text{Eq}$ of positive ions is $\text{73 Eq}\text{.}$

Hence, the sum of $\text{Eq/L}$ of positive ions is not equal to the sum of $\text{Eq/L}$ of negative ions

Therefore,

A solution having ${\text{73 Eq K}}^{\text{+}}{\text{, 55 Eq Cl}}^{-}{\text{, and 25 Eq C}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}{}^{-}$ is not possible to prepare.

(c)

Interpretation Introduction

Interpretation:

A solution having ${\text{750 mEq Na}}^{\text{+}}\text{ and 0}{\text{.750 Eq Cl}}^{-}$ is whether possible to prepare has to be determined considering charge-balance.

Concept-Introduction:

Charge balance takes place when the sum of concentrations of negative ion and the sum of concentrations of positive ion becomes equal in $\text{mEq/L or Eq/L}$ concentration units.

Unit Conversion:

  $1{\text{ Eq = 10}}^3\text{ mEq}$

Answer to Problem 10.145EP

A solution having ${\text{750 mEq Na}}^{\text{+}}\text{ and 0}{\text{.750 Eq Cl}}^{-}$ is possible to prepare.

Explanation of Solution

Given data is shown below:

  $\begin{array}{c}Concentration{\text{ of Na}}^{\text{+}}\text{ = 750 mEq }\\ \\ Concentration{\text{ of Cl}}^{-}\text{ = 0}\text{.750 Eq }\end{array}$

A charge balance exists between the ions in the given electrolyte solution when the sum of $\text{Eq/L}$ of positive ions is equal to the sum of $\text{Eq/L}$ of negative ions.

$Eq$ of positive and negative ions follows as,

  $\begin{array}{c}Eq{\text{ of positive ions = 750 mEq Na}}^{+}\\ \\ \text{= 750 mEq positive ions}\\ \\ Eq\text{ of negative ions = 0}\text{.750 Eq }C{l}^{-}\\ \\ \text{ = }\left(\text{0}\text{.750 Eq}\right)\frac{\left({10}^3\text{ mEq}\right)}{1\text{ Eq}}\\ \\ \text{= 750 mEq negative ions}\end{array}$

Here, the sum of $\text{mEq/L}$ of positive ions is equal to the sum of $\text{mEq/L}$ of negative ions

Therefore,

A solution having ${\text{750 mEq Na}}^{\text{+}}\text{ and 0}{\text{.750 Eq Cl}}^{-}$ is possible to prepare.

(d)

Interpretation Introduction

Interpretation:

A solution having $\text{0}{\text{.025 mole Na}}^{\text{+}}\text{, 0}{\text{.025 mole Ca}}^{\text{2+}}\text{ and 0}{\text{.075 mole Cl}}^{-}$ is whether possible to prepare has to be determined considering charge-balance.

Concept-Introduction:

Charge balance takes place when the sum of concentrations of negative ion and the sum of concentrations of positive ion becomes equal in $\text{mEq/L or Eq/L}$ concentration units.

Unit Conversion:

  $1{\text{ Eq = 10}}^3\text{ mEq}$

Answer to Problem 10.145EP

A solution having $\text{0}{\text{.025 mole Na}}^{\text{+}}\text{, 0}{\text{.025 mole Ca}}^{\text{2+}}\text{ and 0}{\text{.075 mole Cl}}^{-}$ is possible to prepare.

Explanation of Solution

Given data is shown below:

  $\begin{array}{l}\text{No}{\text{. of moles of Na}}^{\text{+}}\text{ = 0}\text{.025 mole}\\ \\ \text{No}{\text{. of moles of Ca}}^{\text{2+}}\text{ = 0}\text{.025 mole}\\ \\ \text{No}{\text{. of moles of Cl}}^{-}\text{ = 0}\text{.075 mole }\end{array}$

A charge balance exists between the ions in the given electrolyte solution when the sum of $\text{Eq/L}$ of positive ions is equal to the sum of $\text{Eq/L}$ of negative ions.

Calculate the equivalence of each ions:

$\begin{array}{c}{\text{Charge of Na}}^{\text{+}}=\text{ 1+}\\ \\ \text{Hence,1 mole = 1 equivalence}\\ \\ \text{0}{\text{.025 mole Na}}^{\text{+}}\text{ = 0}{\text{.025 Eq Na}}^{\text{+}}\\ \\ {\text{Charge of Ca}}^{\text{2+}}=\text{ 1+}\\ \\ \text{Hence,1 mole = 2 equivalence}\\ \\ \text{0}{\text{.025 mole Na}}^{\text{+}}\text{ = }\left(2\times \text{0}{\text{.025 mole Ca}}^{\text{2+}}\right)\text{ = 0}{\text{.050 Eq Ca}}^{\text{2+}}\\ \\ {\text{Charge of Cl}}^{-}=\text{ 1}-\\ \\ \text{Hence,1 mole = 1 equivalence}\\ \\ \text{0}{\text{.075 mole Cl}}^{-}\text{ = 0}{\text{.075 Eq Cl}}^{-}\end{array}$

Sum of $Eq$ of positive and sum of $Eq$ of negative ions follows as,

$\begin{array}{c}SumofEq\text{ of positive ions = 0}{\text{.025 Eq Na}}^{\text{+}}\text{ + 0}{\text{.050 Eq Ca}}^{\text{2+}}\\ \\ \text{= 0}\text{.075 Eq positive ions}\\ \\ Eq\text{ of negative ions = 0}{\text{.075 Eq Cl}}^{-}\\ \\ Therefore,\\ \\ Eq\text{ of positive ions = }Eq\text{ of negative ions }\end{array}$

Therefore,

A solution having $\text{0}{\text{.025 mole Na}}^{\text{+}}\text{, 0}{\text{.025 mole Ca}}^{\text{2+}}\text{ and 0}{\text{.075 mole Cl}}^{-}$ is possible to prepare.