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Consider a 0.250 solution of Na₃PO₄. Find the concentration of phosphate ions, mEq/L in

User Niara
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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.

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