Answer
The $Na^+$ concentration, in milliequivalents per liter is equal to 55 mEq/L
Work Step by Step
The sum of the equivalents of all the negative ions must be equal to the same of the positive ion(s).
For this solution:
$$mEq \space (Cl^-) + mEq \space (H{PO_4}^{2-}) = mEq \space (Na^+)$$
$$40. \space mEq + 15 \space mEq = mEq \space (Na^+)$$ $$55 \space mEq = mEq \space (Na^+)$$
Therefore, there are 55 mEq of $Na^+$ per liter of solution.
