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.
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.
To calculate the mass of sodium phosphate needed to completely eliminate the hard water ions, we need to determine the stoichiometry of the reaction between calcium chloride, magnesium nitrate, and sodium phosphate.
The balanced chemical equation for the reaction is:
3 CaCl2 + 2 Na3PO4 -> Ca3(PO4)2 + 6 NaCl
From the equation, we can see that 3 moles of calcium chloride react with 2 moles of sodium phosphate to form 1 mole of calcium phosphate.
First, let's calculate the moles of calcium chloride and magnesium nitrate present in the solution:
Moles of calcium chloride = concentration of calcium chloride x volume of solution
= 5.5 × 10^(-2) mol/L x 1.5 L
= 8.25 × 10^(-2) mol
Moles of magnesium nitrate = concentration of magnesium nitrate x volume of solution
= 9.0 × 10^(-2) mol/L x 1.5 L
= 0.135 mol
Since the stoichiometry of the reaction is 3:2, we need to use the smaller mole value, which is 8.25 × 10^(-2) mol of calcium chloride.
Now we can calculate the moles of sodium phosphate needed using the stoichiometric ratio:
Moles of sodium phosphate = (8.25 × 10^(-2) mol calcium chloride) x (2 mol sodium phosphate/3 mol calcium chloride)
= 5.5 × 10^(-2) mol
Finally, we can calculate the mass of sodium phosphate needed using its molar mass:
Mass of sodium phosphate = Moles of sodium phosphate x molar mass of sodium phosphate
= 5.5 × 10^(-2) mol x (3 x 22.99 g/mol + 2 x 31.00 g/mol + 4 x 16.00 g/mol)
= 5.5 × 10^(-2) mol x 163.94 g/mol
= 9.01 g
Therefore, approximately 9.01 grams of sodium phosphate would have to be added to 1.5 L of the solution to completely eliminate the hard water ions.
To calculate the mass of sodium phosphate needed to completely eliminate the hard water ions, we need to determine the number of moles of calcium and magnesium ions in the solution and then use the stoichiometry of the reaction between these ions and sodium phosphate.
First, let's calculate the number of moles of calcium ions (Ca2+) and magnesium ions (Mg2+) in the solution.
Given:
- Concentration of calcium chloride (CaCl2): 5.5 × 10^−2 M
- Concentration of magnesium nitrate (Mg(NO3)2): 9.0 × 10^−2 M
- Volume of the solution: 1.5 L
To find the number of moles of each ion, we can use the formula:
moles = concentration × volume
For calcium ions (Ca2+):
moles of Ca2+ = 5.5 × 10^−2 M × 1.5 L = 0.0825 moles
For magnesium ions (Mg2+):
moles of Mg2+ = 9.0 × 10^−2 M × 1.5 L = 0.135 moles
Now, let's determine the stoichiometry of the reaction between calcium and magnesium ions with sodium phosphate (Na3PO4). The balanced chemical equation is:
3 Ca2+ + 2 PO4^3- → Ca3(PO4)2 (insoluble precipitates)
From the balanced equation, we can see that 3 moles of calcium ions react with 2 moles of phosphate ions to form 1 mole of calcium phosphate precipitate.
Similarly, 2 moles of magnesium ions react with 3 moles of phosphate ions to form 1 mole of magnesium phosphate precipitate.
Since we want to completely eliminate both calcium and magnesium ions, we need to supply enough phosphate ions (PO4^3-) to react with both ions in a 1:1 ratio.
Hence, the minimum number of moles of sodium phosphate (Na3PO4) needed can be calculated as:
moles of Na3PO4 = min(moles of Ca2+, moles of Mg2+)
moles of Na3PO4 = min(0.0825 moles, 0.135 moles) = 0.0825 moles
Finally, let's calculate the mass of sodium phosphate required using its molar mass:
molar mass of Na3PO4 = (23.0 g/mol × 3) + 31.0 g/mol + (16.0 g/mol × 4) = 163.0 g/mol
mass of Na3PO4 = moles of Na3PO4 × molar mass of Na3PO4
mass of Na3PO4 = 0.0825 moles × 163.0 g/mol ≈ 13.42 grams
Therefore, approximately 13.42 grams of sodium phosphate would need to be added to 1.5 L of this solution to completely eliminate the hard water ions.