91Ó°ÊÓ

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.

In $1L$Ringer's solution contains$147mEqof{K}^{2+},4mEqof{K}^{+},4mEqofC{a}^{2+}$

Transformation$mEq$to $Eq$,

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

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

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

As a result,localid="1652720092900" $0.147\mathrm{Eq}\text{of}{\mathrm{Na}}^{+}\text{ion}$present in $1L$of solution.

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

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

Hence,

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

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

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

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,

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

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$0.153mol$

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

role="math" localid="1652540717592" $$\begin{gathered} =\frac{0.153\mathrm{mol}\text{of}{\mathrm{Cl}}^{-}}{1\mathrm{L}}×\frac{1\mathrm{Eq}\text{of}{\mathrm{Cl}}^{-}}{1\mathrm{mol}\text{of}{\mathrm{Cl}}^{-}}×\frac{1000\mathrm{mEq}}{1\mathrm{Eq}} \\ =153mEq/L. \end{gathered}$$