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 5.5×10−2M in calcium chloride and 9.5×10−2M in magnesium nitrate.

What mass of sodium phosphate must be added to 2.0L of this solution to completely eliminate the hard water ions? Assume complete reaction.

Answer 1

Ok so, Im going to try to explain this: What we need to do we are going to do this on 3 steps: 1) We need to get the total of moles of calcium + magnesium. 2) Multiply it by 2/3 to get the moles of phosphate 3) from there we need to get the grams of sodium phosphate. So we need those three operations like this:

Moles calcium = 5.5 x 10^-2 moles/liter x 2 liters = 11.0 x 10^-2 moles
Moles magnesium = 9.5 x 10^-2 moles/liter x 2 liters = 19.0 x 10^-2 moles
Total moles of Ca + Mg = 11.0 x 10^-2 + 19.0 x 10^-2 = 30.0 x 10^-2 moles.

Then we include the grams of Na3PO4 that is 163.9408 g/mol and finally you do this operation 163.9408 g/mol x 20.0 x 10^-2 moles = 32.8 grams = 33 grams. And that is how you get it