24.5k views
5 votes
Consider a 0.250 solution of Na₃PO₄. Find the concentration of phosphate ions, mEq/L in

User Niara
by
7.9k points

1 Answer

3 votes

Final answer:

The concentration of phosphate ions in a 0.250 M solution of sodium phosphate is 0.750 mEq/L, calculated by multiplying the molarity by the valence of the phosphate ion.

Step-by-step explanation:

To find the concentration of phosphate ions (PO43-) in a 0.250 M solution of sodium phosphate (Na3PO4), we need to consider the chemical formula of sodium phosphate and how it dissociates in water. Sodium phosphate dissociates into three sodium ions (Na+) and one phosphate ion (PO43-) in solution. Therefore, a 0.250 M solution of sodium phosphate will produce 0.250 M of phosphate ions as well, because each molecule of sodium phosphate produces one phosphate ion upon dissociation.

To convert this concentration to milliequivalents per liter (mEq/L), we can use the equation mEq/L = Molarity (M) × valence. The valence of phosphate ions is 3, as indicated by the charge on the ion (3-). Thus, the concentration in mEq/L would be:

0.250 M × 3 = 0.750 mEq/L

This is the concentration of phosphate ions in milliequivalents per liter for the given solution of sodium phosphate.

User Zunino
by
8.4k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.