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.

Would anyone be able to explain to me how to approach this problem? Not quite sure where to start. I just know they are asking for the mass of Na3PO4.

Answer 1

I remember doing a problem like this recently, but I just can't for the life of me remember if I needed the balanced equation or not. It involves precipitation reactions if you're familiar with that.

Answer 2

CaCl2 + MgNO3 + Na3PO4 ---> Ca3PO42 + Mg3PO42 + NaCl + NaNO3. I guess we need to balance that.

Answer 3

Mg(NO3)2*

Answer 4

The molarity of a solution is the # of moles/volume of solution.

Answer 5

That's where I had a problem with. We haven't done equations where we had more than 2 reactants. I would assume that would be the equation too.

Answer 6

I started off by writing out everything that was given in the equation. 2.0L of solution. 5.5x10^(-2) mol CaCl2 = 5.5x10^(-2) mol CaCl2/1L solution and 9.5x10^(-2)Mg(NO3)2=9.5x10^(-2)mols Mg(NO3)2/1L solution

Answer 7

Right. That represents the molarity of the two solutions. Now, we have to find the moles of each compound in 2L of solution. How can we do that?

Answer 8

The only thing I can think of is multiplying 2L of solutions by the molarity to find the moles.

\[2.00L \times \frac{ 5.5x10^{-2}mol CaCl _{2} }{ 1L } = .11 mol CaCl _{2}\]

\[2.00L \times \frac{ 9.5x10^{-2}mol Mg(NO _{3})_{2} }{ 1L } = .19molMg(NO _{3})_{2}\]

Answer 9

Right. Now we have to do stoichiometry on both of these to figure out how many moles of Na3PO4 needed to perform a reaction. Problem is we don't have the balanced equation yet. Can you balance it and go from here?

Answer 10

Honestly I'm stumped. Tried balancing it, but I can't get the Sodium phosphate to balance

Answer 11

I'll give it a try, give me a min.

Answer 12

Hmm, I'm getting stumped too, or maybe it's just too late lol. I know it's not this hard.

Answer 13

Lol it could just be late. I haven't had a problem with the homework until this question. Unfortunately, It's the last problem that is preventing me from completing the homework. I appreciate the help though

Answer 14

3CaCl2 + 2Na3PO4 = Ca3(PO4)2 + 6NaCl
3Mg(NO3)2 + 2Na3PO4 = Mg3(PO4)2 + 6NaNO3

Answer 15

There we go, putting it in 2 separate equations is better, and frankly easier to do the stoichiometry.

Answer 16

Wow I got it. Finally! Thanks! It's basic stoichiometry from there. finding the amount of moles of Sodium Phosphate needed in each reaction. From the moles, I converted to grams and just added the two. I ended up with about 32.788g Sodium Phosphate needed. Thank you for that!

Answer 17

No probs.