Question

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

Short answer

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

Step-by-step solution

Given Information.

Potassium loss contains $40mEq/L$

Solution is $1.5L$

To find the moles present in the solution.

Find the equivalent.

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

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

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

Find the moles in K+

Charge of Potassium is $+1$. Therefore,

$1Eqof{K}^{+}=1moleof{K}^{+}$

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

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

Find the moles in Cl-

Charge of Chlorine is $-1$. Therefore,

$1EqofC{l}^{-}=1moleofC{l}^{-}$

Then,

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