Final answer:
To eliminate hard water ions, one must consider the reaction between sodium phosphate and the calcium/magnesium ions in hard water. Without additional information about the concentration of these ions, an exact mass of sodium phosphate cannot be provided. Theoretically, 0.5 moles of sodium phosphate would be needed to eliminate 1.5 moles of hard water ions, assuming a 1:3 molar ratio.
Step-by-step explanation:
To determine what mass of sodium phosphate (Na3PO4) would be required to eliminate hard water ions, we must consider the complete reaction between sodium phosphate and the ions causing water hardness, which are typically calcium (Ca2+) and magnesium (Mg2+) ions. Each formula unit of sodium phosphate can potentially react with three divalent cations, since it provides three sodium ions to exchange with the cations in the hard water.
As the question does not specify the exact concentration of hard water ions, we cannot provide a precise answer here. However, we can discuss the stoichiometry involved in such a reaction. Assuming complete reaction, one mole of sodium phosphate could eliminate three moles of hard water ions, given the 1:3 molar ratio of sodium phosphate to divalent cations.
Therefore, the answer to the question would depend on the exact amount of calcium and magnesium ions in the provided 1.5 L of hard water. In a hypothetical scenario, if there were 1.5 moles of hard water ions in total, then one would require 0.5 moles of sodium phosphate (option A) to completely eliminate them.