# 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-)# ?

##### 1 Answer

#### 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,*#"Cl"^(-)# , which has a concentration of#"40. mEq/L"# *the hydrogen phosphate anion,*#"HPO"_4^(2-)# ,*which has a concentration of*#"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 *total number of milliequivalents* for the anions will be

#"no. of mEq anions" = "40. mEq" + "15 mEq" = "55 mEq"#

You know that a solution is electrically neutral, so

#color(blue)("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 *for every* liter of solution, which is equivalent to a concentration of

#["Na"^(+)] = color(green)("55 mEq/L")#