# How many milliequivalents of sodium cations are present per liter of a solution that contains "40. mEq/L" of chloride anions, "Cl"^(-), and "15 mEq/L" of hydrogen phosphate anions, "HPO"_4^(2-) ?

Dec 5, 2015

$\text{55 mEq/L}$

#### Explanation:

The idea here is that you need to use the fact that a solution is electrically neutral to determine how many milliequivalents of sodium cations are needed to balance the milliequivalents of the two anions.

The two anions present in solution are

• the chloride anion, ${\text{Cl}}^{-}$, which has a concentration of $\text{40. mEq/L}$
• the hydrogen phosphate anion, ${\text{HPO}}_{4}^{2 -}$, which has a concentration of $\text{15 mEq/L}$

As you know, an equivalent for an ion is calculated by multiplying the number of moles of that ion by its valence.

In this case, every liter of solution will contain $\text{40. mEq}$ of chloride anions and $\text{15 mEq}$ of hydrogen phosphate anions, which means that the total number of milliequivalents for the anions will be

$\text{no. of mEq anions" = "40. mEq" + "15 mEq" = "55 mEq}$

You know that a solution is electrically neutral, so

$\textcolor{b l u e}{\text{no. of mEq anions " = " no. of mEq cations}}$

Because sodium is the only cation present in this solution, its number of milliequivalents must be equal to the total number of milliequivalents for the anions.

This means that the solution will contain $\text{55 mEq}$ of sodium for every liter of solution, which is equivalent to a concentration of

$\left[\text{Na"^(+)] = color(green)("55 mEq/L}\right)$