How many [tex]$mEq$[/tex] of [tex]$Ca^{2+}$[/tex] are present in a solution that also contains the following?

- [tex]$75 \, mEq \, Na^{+}$[/tex]
- [tex]$83 \, mEq \, K^{+}$[/tex]
- [tex]$35 \, mEq \, HCO_3^{-}$[/tex]
- [tex]$153 \, mEq \, Cl^{-}$[/tex]

Input a number only. Use the correct number of significant figures.

[tex]\square \, mEq \, Ca^{2+}[/tex]

To find how many milliequivalents (mEq) of [tex]$ \text{Ca}^{2+} $[/tex] are present in the solution, we first need to ensure the solution is electrically neutral. This means the total positive charges should equal the total negative charges.

Here's the breakdown of the solution:

1. Identify the positive ions and their equivalents:
- [tex]$ \text{Na}^{+} $[/tex]: 75 mEq
- [tex]$ \text{K}^{+} $[/tex]: 83 mEq
- [tex]$ \text{Ca}^{2+} $[/tex] (to be determined): Let's call this [tex]$ \text{mEq}_\text{Ca} $[/tex]. Note that because calcium has a [tex]$2+$[/tex] charge, its contribution to the charge balance is [tex]$2 \times \text{mEq}_\text{Ca}$[/tex].

2. Identify the negative ions and their equivalents:
- [tex]$ \text{HCO}_3^{-} $[/tex]: 35 mEq
- [tex]$ \text{Cl}^{-} $[/tex]: 153 mEq

3. Set up the equation for electrical neutrality:
[tex]\[
\text{Total positive charges} = \text{Total negative charges}
\][/tex]
[tex]\[
75 + 83 + 2 \times \text{mEq}_\text{Ca} = 35 + 153
\][/tex]

4. Solve for [tex]$ \text{mEq}_\text{Ca} $[/tex]:
[tex]\[
75 + 83 + 2 \times \text{mEq}_\text{Ca} = 188
\][/tex]
[tex]\[
158 + 2 \times \text{mEq}_\text{Ca} = 188
\][/tex]
[tex]\[
2 \times \text{mEq}_\text{Ca} = 188 - 158
\][/tex]
[tex]\[
2 \times \text{mEq}_\text{Ca} = 30
\][/tex]
[tex]\[
\text{mEq}_\text{Ca} = \frac{30}{2} = 15
\][/tex]

Therefore, there are 15 mEq of [tex]$ \text{Ca}^{2+} $[/tex] present in the solution.