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What is the molar concentration of sodium ions (Na⁺) in a 0.350 M Na₃PO₄ solution?

A) 0.350 M
B) 0.700 M
C) 1.05 M
D) 1.40 M

1 Answer

5 votes

Final answer:

The molar concentration of sodium ions in a 0.350 M Na₃PO₄ solution is 1.05 M, since each mole of Na₃PO₄ produces three moles of sodium ions.

Step-by-step explanation:

The molar concentration of sodium ions (Na⁺) in a 0.350 M Na₃PO₄ solution can be calculated based on the stoichiometry of the compound. Each formula unit of Na₃PO₄ dissociates into three sodium ions (Na⁺) and one phosphate ion (PO⁴³⁺) in solution. Therefore, for every 1 mole of Na₃PO₄, there are 3 moles of Na⁺ ions.

To find the concentration of Na⁺ ions, you multiply the molarity of Na₃PO₄ by the number of Na⁺ ions per formula unit:

0.350 M Na₃PO₄ × 3 mol Na⁺/1 mol Na₃PO₄ = 1.05 M Na⁺.

The molar concentration of sodium ions in a 0.350 M Na₃PO₄ solution is 1.05 M, which corresponds to option C.

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