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Answer :
Final answer:
The concentration of sodium ions in 4.57 L of a 0.847 M Na3P solution is 2.541 M. This is because each unit of Na3P yields three sodium ions, so the sodium ion concentration triples that of Na3P's molarity.
Explanation:
The question asks about the concentration of sodium ions in a solution of sodium phosphate (Na3P). In a 0.847 M solution of Na3P, there are 0.847 moles of Na3P per liter. Since each formula unit of Na3P contains three sodium ions, the concentration of sodium ions will be three times that of the Na3P.
To find the concentration of sodium ions in moles per liter (molarity), we multiply the concentration of Na3P by 3 (since there are 3 moles of sodium ions for every mole of Na3P):
0.847 M Na3P × 3 = 2.541 M Na+
When we have 4.57 L of this solution, the concentration of sodium ions remains the same because molarity is an intensive property, meaning it does not change with the amount of solution.
Therefore, the concentration of sodium ions in the solution is 2.541 M, which corresponds to option c.
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