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Hard water often contains dissolved Ca^2+ and Mg^2+ ions. One way to soften water is to add phosphates. The phosphate ion forms insoluble precipitates with calcium and magnesium ions, removing them from solution. A solution is 0.055 MM in calcium chloride and 0.075 MM in magnesium nitrate. What mass of sodium phosphate would you add to 1.0 LL of this solution to completely eliminate the hard water ions? Assume complete reaction.

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Answer:

14.2g of sodium phosphate are required

Explanation:

To solve this question we need to find the moles of Ca²⁺ and Mg²⁺ ions. And based on the reactions:

3Ca²⁺ + 2PO₄³⁻ → Ca₃(PO₄)₂ (s)

3Mg²⁺ + 2PO₄³⁻ → Mg₃(PO₄)₂ (s)

we can find the moles of phosphate required to precipitate all these ions and its mass:

Moles Ca²⁺:

1.0L * (0.055mol / L) = 0.055mol

Moles Mg²⁺:

1.0L * (0.075mol / L) = 0.075mol

Total moles = 0.13 moles of ions

Moles of phosphate ion required:

0.13 moles * (2 moles PO₄³⁻ / 3 moles ions) = 0.0867 moles PO₄³⁻

The moles of sodium phosphate (Na₃PO₄) are = 0.0867 moles

The mass is -Molar mass Na₃PO₄: 164g/mol-:

0.0867 moles Na₃PO₄ * (164g / mol) =

14.2g of sodium phosphate are required