Question

A Ringer's solution contains the following concentrations $(mEq/L)$of cations: role="math" localid="1652538986943" $N{a}^{+}147,K^{+}4,andC{a}^{2+}4.$If $C{l}^{-}$is the only anion in the solution, what is the $C{l}^{-}$ concentration, in milliequivalents per liter?

Short answer

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

Step-by-step solution

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.

Find the Mole of Na+

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

Transformation$mEq$to $Eq$,

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

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

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

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

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

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,

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

Find the mole in Ca+2

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

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

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,

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

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$0.153mol$

Find the moles to mEq/L.

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