A solution contains 40. mEq/L of Cl - and 15 mEq/L of HPO4^2-. If Na + is the only cation in the solution, what is the Na + concentration in milliequivalents per liter?

Question

Chemistry

Bradyn Bennett

23 October, 18:19

A solution contains 40. mEq/L of Cl - and 15 mEq/L of HPO4^2-. If Na + is the only cation in the solution, what is the Na + concentration in milliequivalents per liter?

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Kaylyn · Answer 1

The two anions present in solution are:

The chloride anion, Cl -, which has a concentration of 40 mEq/L

The hydrogen phosphate anion, HPO 2 - 4, 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 40. mEq of chloride anions and 15 mEq of hydrogen phosphate anions, which means that the total number of milliequivalents for the anions will be

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

You know that a solution is electrically neutral, so:

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 55 mEq of sodium for every liter of solution, which is equivalent to a concentration of [Na+]=55 mEq/L.