Ask questions and get helpful answers.

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.

  1. 👍
  2. 👎
  3. 👁
  4. ℹ️
  5. 🚩

1 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.

    1. 👍
    2. 👎
    3. ℹ️
    4. 🚩

Answer this Question

Related Questions

Still need help?

You can ask a new question or browse existing questions.