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 solution. Suppose that a solution is 0.054 M in calcium chloride and 0.076 M in magnesium nitrate. What mass of sodium phosphate would have to be added to 1.8 L of this solution to completely eliminate the hard water ions? Assume complete reaction.

Answer 1

Ca+2 = 0.054 M

Mg+2 = 0.076 M

Assume

PO4-3 will react with CA+2/Mg2+

3Ca+2 + 2PO4-3 --> Ca3(PO4)2 (s)

3Mg+2 + 2PO4-3 --> Mg3(PO4)2 (s)

now...

ratio is 2/3 mol per each

TotalV = 1.8 L

mol of Ca+2 = MV = 0.054*1.8 = 0.0972 mol of Ca+2

mol of Mg+2 = MV = 0.076*1.8 = 0.1368 mol of Mg+2

Total X+2 ions --> 0.0972 +0.1368 = 0.234 mol

recall that

3 mol of X+2 = 2 mol of PO4-3

0.234mol -- > 2/3*0.234 = 0.156 mol of PO4-3

that is

0.156 mol of Na3PO4 required

mass = mol*Mw = 0.156*163.940671 = 25.574 g required to precipitate all