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 0.049 M in calcium chloride and 0.094 M in magnesium nitrate. What mass of sodium phosphate would have to be added to 1.4 L of this solution to completely eliminate the hard water ions? Assume complete reaction. Enter a numerical answer in terms of grams to 1 decimal place

Answer 1

Consider the dissociations of calcium chloride, CaCl₂ and magnesium nitrate, Mg(NO₃)₂ in water.

CaCl₂ (aq) -------> Ca²⁺ (aq) + 2 Cl^- (aq)

Mg(NO₃)₂ (aq) ------> Mg²⁺ (aq) + 2 NO₃^- (aq)

Due to the nature of the dissociations. [Ca²⁺] = 0.049 M and [Mg²⁺] = 0.094 M.

We have 1.4 L of the mixed solution. Find out the moles of each ion.

Moles Ca²⁺ = (1.4 L)*(0.049 mol/L) [1 M = 1 mol/L] = 0.0686 mole.

Moles Mg²⁺ = (1.4 L)*(0.094 mol/L) = 0.1316 mole.

Next, write down the balanced chemical equations for the reactions of sodium phosphate, Na₃PO₄ with both the ions.

3 Ca²⁺ (aq) + 2 Na₃PO₄ (aq) --------> Ca₃(PO₄)₂ (aq) + 6 Na⁺ (aq)

3 Mg²⁺ (aq) + 2 Na₃PO₄ (aq) --------> Mg₃(PO₄)₂ (aq) + 6 Na⁺ (aq)

As per the stoichiometric equations above,

3 moles Ca²⁺ = 2 moles Na₃PO₄.

Therefore, 0.0686 mole Ca²⁺ = (0.0686 mole Ca²⁺)*(2 mole Na₃PO₄/3 mole Ca²⁺) = 0.04573 mole Na₃PO₄.

Again, 3 moles Mg²⁺ = 2 moles Na₃PO₄.

Therefore, 0.1316 mole Mg²⁺ = (0.1316 mole Mg²⁺)*(2 mole Na₃PO₄/3 mole Mg²⁺) = 0.08773 mole Na₃PO₄.

The total moles of Na₃PO₄ required for complete precipitation of the two metal ions is (0.04573 + 0.08773) mole = 0.13346 mole.

The molar mass of Na₃PO₄ is (3*22.989 + 1*30.974 + 4*15.9994) g/mol = 163.9386 g/mol.

The mass of sodium phosphate required for complete precipitation = (0.13346 mole)*(163.9386 g/mol) = 21.8792 g ≈ 21.9 g (correct to one decimal place, ans).