Chapter 8, Problem 32P

Interpretation Introduction

(a)

Interpretation:

The number of milliequivalent of ${\text{Na}}^{+}$ ions in $\text{125 mL}$ of Ringer's solution is to be calculated.

Concept introduction:

The concentration of an ion in solution is expressed in equivalents. One equivalent of an ion can be defined as the amount of charge an ion contributes when present in dissolved form. For example, one mole of ${\text{Ca}}^{2+}$ contributes 2 equivalent of ${\text{Ca}}^{2+}$ ions as the charge is +2. Similarly, one mole of ${\text{SO}}_4{}^{2-}$ contributes 2 equivalent of ${\text{SO}}_4{}^{2-}$ ions as the charge is +2.

The concentration of an electrolyte is often expressed in $\text{mEq}/\text{L}$ . The conversion factor to convert milliequivalent to equivalent is as follows:

  $1\text{ mEq}={10}^{-3}\text{ eq}$

To convert one system of unit to another the conversion factors are written such that the unwanted units lie in the denominator and thus are canceled out.

The conversion factor to convert $\text{mL}$ to $\text{L}$ is given as follows:

  $1\text{ mL}={\text{10}}^{-3}\text{ L}$

Interpretation Introduction

(b)

Interpretation:

The concentration of ${\text{Cl}}^{-}$ ions is to be calculated in $\text{mEq}/\text{L}$.

Concept introduction:

The concentration of an ion in solution is expressed in equivalents. One equivalent of an ion can be defined as amount of charge an ion contributes when present in dissolved form. For example, one mole of ${\text{Ca}}^{2+}$ contributes 2 equivalent of ${\text{Ca}}^{2+}$ ions as the charge is +2. Similarly, one mole of ${\text{SO}}_4{}^{2-}$ contributes 2 equivalent of ${\text{SO}}_4{}^{2-}$ ions as the charge is +2.

The concentration of an electrolyte is often expressed in $\text{mEq}/\text{L}$ . The conversion factor to convert milliequivalent to equivalent is as follows:

  $1\text{ mEq}={10}^{-3}\text{ eq}$

To convert one system of unit to another the conversion factors are written such that the unwanted units lie in the denominator and thus are canceled out.

The conversion factor to convert $\text{mL}$ to $\text{L}$ is given as follows:

  $1\text{ mL}={\text{10}}^{-3}\text{ L}$

Any ionic solution contains an equal amount of cationic and anionic charge. For example, a solution of $\text{KI}$ that contains $\text{20 mEq}/\text{L}$ of ${\text{K}}^{\text{+}}$ ions must contain $\text{20 mEq}/\text{L}$ of ${\text{I}}^{-}$ ions.