Q. 9.19 - An Introduction to General, Organic, and Biological Chemistry

Question

An intravenous solution contains $40mEq/L$ of $C{l}^{-}$and $15 mEq/ L$ 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?

Step-by-Step Solution

Answer

Concentration of $N{a}^{+}$ for the solution is $55 mEq/L.$

Step1: Given information

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Step2: Moles of C l -

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

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,

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

Step3: Find moles in H P O 4 2 -

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

Therefore, find the mole using the above is,

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

Step4: Calculate the total moles of anion.

Find the total moles present in $1 L$ 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 $0.055mls/L.$

Step5: Find the moles in m E q

Converting the moles into $mEq/L.$

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

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

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