An intravenous solution contains $40mEq/L$of $C{l}^{-}$and $15mEq/L$of$HP{O_4}^{2-}$. If$N{a}^{+}$is the only cation in the solution, what is the$N{a}^{+}$concentration, in milliequivalents per liter?

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In $1L$of solution contains$40mEq/L,15mEq/L$

Converting $mEq$to $Eq$,

$$\begin{gathered} 1\mathrm{Eq}\text{of}{\mathrm{Cl}}^{-}\text{ion}=1000\mathrm{mEq}\text{of}{\mathrm{Cl}}^{-}\text{ion} \\ =40\mathrm{mEq}\text{of}{\mathrm{Cl}}^{-}\text{ion}\times \frac{1\mathrm{Eq}\text{of}{\mathrm{Cl}}^{-}\text{ion}}{1000\mathrm{mEq}\text{of}{\mathrm{Cl}}^{-}\text{ion}} \\ =0.04\mathrm{Eq}{\mathrm{Cl}}^{-}\text{ion} \end{gathered}$$

Then,

$$\begin{gathered} 1\mathrm{Eq}\text{of}{\mathrm{Cl}}^{-}\text{ion}=1\text{mole of}{\mathrm{Cl}}^{-}\text{ion.} \\ 0.04\mathrm{Eq}\text{of}{\mathrm{Cl}}^{-}\text{ion}=0.04\text{mole of}{\mathrm{Cl}}^{-}\text{ion.} \end{gathered}$$

$$\begin{gathered} 1\mathrm{Eq}\text{of}{\mathrm{HPO}}_4^{-2}\text{ion}=1000\mathrm{mEq}\text{of}{\mathrm{HPO}}_4^{-2}\text{ion} \\ =15\mathrm{mEq}\text{of}{\mathrm{HPO}}_4^{-2}\text{ion}\times \frac{1\mathrm{Eq}\text{of}{\mathrm{HPO}}_4^{-2}\text{ion}}{1000\mathrm{mEq}\text{of}{\mathrm{HPO}}_4^{-2}\text{ion}} \\ =0.015\mathrm{Eq}\text{of}{\mathrm{HPO}}_4^{-2}\text{ion} \end{gathered}$$

Therefore, find the mole using the above is,

$$\begin{gathered} 1\mathrm{Eq}\text{of}{\mathrm{HPO}}_4^{2\text{-}}\mathrm{ion}=1\text{mole of}{{\mathrm{HPO}}_4}^{2-}\text{ion.} \\ 0.015\mathrm{Eq}\text{of}{\mathrm{HPO}}_4^{2\text{-}}\mathrm{ion}=0.015\text{mole of}{{\mathrm{HPO}}_4}^{2-}\text{ion.} \end{gathered}$$

Find the total moles present in $1L$of solution,

$$\begin{gathered} \text{Total moles of anions}=0.04\mathrm{mol}+0.015\mathrm{mol} \\ =0.055\mathrm{mol} \end{gathered}$$

Therefore, moles of $N{a}^{+}$ions need $0.055mls/L.$

Converting the moles into $mEq/L.$

The charge of sodium is$+1$. Therefore,

$1\mathrm{Eq}\text{of}{\mathrm{Na}}^{+}\text{ion}=1\text{mole of}{\mathrm{Na}}^{+}\text{ion.}$

$$\begin{gathered} =\frac{0.055\mathrm{mol}\text{of}{\mathrm{Na}}^{+}}{1\mathrm{L}}\times \frac{1\mathrm{Eq}\text{of}{\mathrm{Na}}^{+}}{1\mathrm{mol}\text{of}{\mathrm{Na}}^{+}}\times \frac{1000\mathrm{mEq}}{1\mathrm{Eq}} \\ =55mEq/L \end{gathered}$$