Question

Hard water often contains dissolved Ca2+ and Mg2+ ions. One way to soften water is to add phosphates. The phosphate ion forms insoluble precipitates with calcium and magnesium ions, removing them from the solution. Suppose that a solution is 5.5×10−2 M in calcium chloride and 9.0×10−2 M in magnesium nitrate.
Part A
What mass of sodium phosphate would have to be added to 1.5 L of this solution to completely eliminate the hard water ions? Assume complete reaction.

Answer 1

5.5×10−2 M in calcium chloride and 9.0×10−2 M in magnesium nitrate.
so 0.055 M Ca^2+ + 0.090 M in Mg^2+
millimoles Ca^2+ M in 1.5 L = mL x M = 1500 mL x 0.055 = 82.5 or 0.0825 moles.
millimoles Mg^2+ = 1500 mL x 0.090 = 135 or 0.135 moles.
3Mg^2+ + 2Na3PO4 ==> Mg3(PO4)2 6Na^+
mols Na3PO4 = mols Mg^2+ + (2 mols Na3PO4/3 mols Mg^2+) = 0.0825 x 2/3 = ?. Then g Na3PO4 = mols Na3PO4 x molar mass Na3PO4 = ?
3Ca^2+ + 2Na3PO4 = Ca3(PO4)4 + 6Na^+
mols Na3PO4 = 0.135 x 2/3 = ?
g Na3PO4 = mols x molar mass = ?
Then total g Na3PO4 = g to ppt Mg + g to ppt Ca.
Post your work if you get stuck.