Q. 9.20 - An Introduction to General, Organic, and Biological Chemistry

Question

A Ringer's solution contains the following concentrations $(mEq/L)$ of cations: $N{a}^{+ }147, K^{+} 4, and C{a}^{2+} 4.$ If $C{l}^{-}$is the only anion in the solution, what is the $C{l}^{-}$ concentration, in milliequivalents per liter?

Step-by-Step Solution

Answer

The Concentration of $C{l}^{-}$ ions in $mEq/L$is $153 mEq/L.$

Step1: Given information

Sodium chloride, potassium chloride, calcium chloride, and sodium bicarbonate are present in solution inside the densities found in bodily secretions. Lactated Ringer's solution is created when sodium lactate is used before sodium bicarbonate.

Step2: Find the Mole of N a +

In $1 L$Ringer's solution contains$147 mEq of K^{2+}, 4 mEq of K^{+}, 4 mEq of C{a}^{2+}$

Transformation$mEq$ to $Eq$,

$1\mathrm{Eq}\text{ of }{\mathrm{Na}}^{+}\text{ion }=1000\mathrm{mEq}\text{ of }{\mathrm{Na}}^{+}\text{ion }$

$=147\mathrm{mEq}\text{ of }{\mathrm{Na}}^{+}\text{ion }\times \frac{1\mathrm{Eq}\text{ of }{\mathrm{Na}}^{+}\text{ion }}{1000\mathrm{mEq}\text{ of }{\mathrm{Na}}^{+}\text{ion }}$

$=0.147 \mathrm{Eq}\text{ of }{\mathrm{Na}}^{+}\text{ion }$

As a result,$0.147 \mathrm{Eq}\text{ of }{\mathrm{Na}}^{+}\text{ion }$ present in $1 L$ of solution.

$$\begin{gathered} 1{\mathrm{Eq}}^{-}{\mathrm{Na}}^{+}\text{ion }=1\text{ mole of }{\mathrm{Na}}^{+}\text{ion. } \\ 0.147{\mathrm{Eq}}^{-}{\mathrm{Na}}^{+}\text{ion }=0.147\text{ mole of }{\mathrm{Na}}^{+}\text{ion. } \end{gathered}$$

Step3: Find the moles in K +

$$\begin{gathered} 1\mathrm{Eq}\text{ of }{\mathrm{K}}^{+}\text{ion }=1000\mathrm{mEq}\text{ of }{\mathrm{K}}^{+}\text{ion } \\ =4\mathrm{mEq}\text{ of }{\mathrm{K}}^{+}\text{ion }\times \frac{1\mathrm{Eq}\text{ of }{\mathrm{K}}^{+}\text{ion }}{1000\mathrm{mEq}\text{ of }{\mathrm{K}}^{+}\text{ion }} \\ =0.004 \mathrm{Eq}\text{ of }{\mathrm{K}}^{+}\text{ion } \end{gathered}$$

Hence,

$$\begin{gathered} 1\mathrm{Eq}\text{ of }{K}^{+}\text{ion }=1\text{ mole of }{K}^{+}\text{ion. } \\ 0.004\mathrm{Eq}\text{ of }{K}^{+}\text{ion }=0.004\text{ mole of }{K}^{+}\text{ion. } \end{gathered}$$

Step4: Find the mole in C a + 2

$1\mathrm{Eq}{\text{ of Ca}}^{+2}\text{ion }=1000 mEq{\text{ of Ca}}^{+2}\text{ion. }$

$$\begin{gathered} =4\mathrm{mEq}\text{ of }{\mathrm{Ca}}^{+2}\text{ ion }\times \frac{1\mathrm{Eq}\text{ of }{\mathrm{Ca}}^{+2}\text{ ion }}{1000\mathrm{mEq}\text{ of }{\mathrm{Ca}}^{+2}\text{ ion }} \\ =0.004 \mathrm{Eq}\text{ of }{\mathrm{Ca}}^{+2}\text{ ion } \end{gathered}$$

Then,

$$\begin{gathered} 2\mathrm{Eq}\text{ of }{\mathrm{Ca}}^{2+}\text{ ion }=1\text{ mole of }{\mathrm{Ca}}^{2+}\text{ ion. } \\ 1\mathrm{Eq}\text{ of }{\mathrm{Ca}}^{2+}\text{ ion }=(1/2)\text{ mole of }{\mathrm{Ca}}^{2+}\text{ ion. } \end{gathered}$$

Hence,

$0.004\mathrm{Eq}\text{ of }{\mathrm{Ca}}^{2+}\text{ ion }=0.002\text{ moles of }{\mathrm{Ca}}^{2+}\text{ ion. }$

Step5: Find the total moles of anion.

Calculate the total moles of anion in $1L$

$$\begin{gathered} \text{ Total moles of cations }=0.147 \mathrm{mol}+0.004 \mathrm{mol}+0.002 \mathrm{mol} \\ =0.153 \mathrm{mol} \end{gathered}$$

The number of moles of $C{l}^{-}$ ions needed to balance cations is $0.153 mol$

Step6: Find the moles to m E q / L .

Charge of chlorine is $-1$. Therefore, $1\mathrm{Eq}\text{ of }{\mathrm{Cl}}^{-}\text{ion }=1\text{ mole of }{\mathrm{Cl}}^{-}\text{ion. }$

$$\begin{gathered} =\frac{0.153 \mathrm{mol}\text{ of }{\mathrm{Cl}}^{-}}{1 \mathrm{L}}\times \frac{1\mathrm{Eq}\text{ of }{\mathrm{Cl}}^{-}}{1 \mathrm{mol}\text{ of }{\mathrm{Cl}}^{-}}\times \frac{1000\mathrm{mEq}}{1\mathrm{Eq}} \\ =153 mEq/L. \end{gathered}$$