Q. 9.18 - An Introduction to General, Organic, and Biological Chemistry

Question

An intravenous solution to replace potassium loss contains $40 mEq/L$ each of $K^{+}$ and $C{l}^{-}$. How many moles each of $K^{+}$ and $C{l}^{-}$ are in $1.5 L$ of the solution?

Step-by-Step Solution

Answer

In $1.5 L$of the solution, moles of $K^{+}$and $C{l}^{-}$are $0.06$ and $0.06$ respectively.

Step 1: Given Information.

Potassium loss contains $40 mEq/ L$

Solution is $1.5 L$

To find the moles present in the solution.

Step 2: Find the equivalent.

$1 L$ of intravenous solution contains $40 mEq/ L$

Therefore, convert $mEq$ to $Eq$ as follows:

$$\begin{gathered} 1\mathrm{Eq}=1000\mathrm{mEq} \\ =40 mEq\times \frac{1\mathrm{Eq}}{1000\mathrm{mEq}} \\ =0.04 Eq \end{gathered}$$

Step 3: Find the moles in K +

Charge of Potassium is $+1$. Therefore,

$1 Eq of K^{+}=1 mole of K^{+}$

$$\begin{gathered} =1.5\mathrm{L}\times \frac{0.04\mathrm{Eq}\text{ of }{\mathrm{K}}^{+}}{1\mathrm{L}}\times \frac{1\mathrm{mol}\text{ of }{\mathrm{K}}^{+}}{1\mathrm{Eq}\text{ of }{\mathrm{K}}^{+}} \\ =0.06 mole of K^{+} \end{gathered}$$

Hence, mole of $K^{+}$ ions is $0.06 moles$

Step 4: Find the moles in C l -

Charge of Chlorine is $-1$. Therefore,

$1 Eq of C{l}^{-}=1 mole of C{l}^{-}$

Then,

$$\begin{gathered} =1.5\mathrm{L}\times \frac{0.04\mathrm{Eq}\text{ of }{\mathrm{Cl}}^{-}}{1\mathrm{L}}\times \frac{1\mathrm{mol}\text{ of }{\mathrm{Cl}}^{-}}{1\mathrm{Eq}\text{ of }{\mathrm{Cl}}^{-}} \\ =0.06 moles of C{l}^{-} \end{gathered}$$