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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.

Verified Answer :

Dejanras

Answer is: mass of sodium phosphate must be added is 32.79 grams.

Balanced chemical reaction 1:

3CaCl₂(aq) + 2Na₃PO₄(aq) → Ca₃(PO₄)₃(s) + 6NaCl(aq).

Balanced chemical reaction 2:

3Mg(NO₃)₂(aq) + 2Na₃PO₄(aq) → Mg₃(PO₄)₃(s) + 6NaNO₃(aq).

c(CaCl₂) = 0.055 M; molarity of calcium chloride.

V(CaCl₂) = 2 L; volume of calcium chloride solution.

n(CaCl₂) = c(CaCl₂) · V(CaCl₂).

n(CaCl₂) = 0.055 mol/L · 2 L.

n(CaCl₂) = 0.11 mol; amount of calcium chloride.

From balanced chemical reaction 1: n(CaCl₂) : n(Na₃PO₄) = 3 : 2.

n₁(Na₃PO₄) = 2 · 0.11 mol ÷ 3.

n₁(Na₃PO₄) = 0.073 mol; amount of sodium phosphate in reaction 1.

2) c(Mg(NO₃)₂) = 0.095 M; amount of magnesium nitrate.

V(Mg(NO₃)₂) = 2 L; volume of magnesium nitrate solution.

n(Mg(NO₃)₂) = c(Mg(NO₃)₂) · V(Mg(NO₃)₂).

n(Mg(NO₃)₂) = 0.095 mol/L · 2 L.

n(Mg(NO₃)₂) = 0.19 mol; amount of magnesium nitrate.

From balanced chemical reaction 2: n(Mg(NO₃)₂) : n(Na₃PO₄) = 3 : 2.

n₂(Na₃PO₄) = 2 · 0.19 mol ÷ 3.

n₂(Na₃PO₄) = 0.126mol; amount of sodium phosphate in reaction 2.

3) n(Na₃PO₄) = n₁(Na₃PO₄) + n₂(Na₃PO₄).

n(Na₃PO₄) = 0.073 mol + 0.126 mol.

n(Na₃PO₄) = 0.2 mol; amount of sodium phosphate.

M(Na₃PO₄) = 3Ar(Na) + Ar(P) + 4Ar(O) · g/mol.

m(Na₃PO₄) = n(Na₃PO₄) · M(Na₃PO₄).

m(Na₃PO₄) = 0.2 mol · 163.94 g/mol.

n(Na₃PO₄) = 32.788 g; mass of sodium phosphate.

Hard water has a high content of dissolved calcium and magnesium salts. 32.79 gm of sodium phosphate is needed to be added to 2.0 L of the solution.

What is mass?

Mass is the multiplication product of moles and the molar mass of the compound.

The balanced chemical reactions are shown as,

\(\rm 3CaCl_{2}(aq) + 2Na_{3}PO_{4}(aq) \rightarrow Ca_{3}(PO_{4})_{3}(s) + 6NaCl(aq)\;\; (reaction\; 1)\)

And,

\(\rm 3Mg(NO_{3})_{2}(aq) + 2Na_{3}PO_{4}(aq) \rightarrow Mg_{3}(PO_{4})_{3}(s) + 6NaNO_{3}(aq) \;\; (reaction\; 2)\)

Given,

Molarity of calcium chloride (c) = 0.055 M

Volume of  calcium chloride (V) = 2.0 L

Moles of calcium chloride are calculated as:

\(\begin{aligned} \rm moles (n) &= \rm molarity \times volume\\\\ &= 0.055 \times 2.0 \\\\&= 0.11 \;\rm mol\end{aligned}\)

From reaction 1 moles of calcium chloride: moles of trisodium phosphate = 3:2.

Moles of trisodium phosphate will be 0.073 moles.

In reaction 2 the moles of magnesium nitrate is calculated as:

Given,

Molarity of magnesium nitrate (c) = 0.095 M

Volume of magnesium nitrate (V) = 2.0 L

\(\begin{aligned}\rm moles (n) &= \rm molarity \times volume \\\\&= 0.095\times 2.0 \\\\&= 0.19 \;\rm mol\end{aligned}\)

From reaction 2 moles of magnesium nitrate: moles of trisodium phosphate = 3:2.

Moles of trisodium phosphate will be 0.126 moles.

Total moles of sodium phosphate are: 0.073 mol + 0.126 mol = 0.2 mol.

Mass of sodium phosphate is calculated as:

\(\begin{aligned}\rm mass &= \rm moles \times molar\; mass\\\\&= 0.2 \times 163.94\\\\&= 32.78\;\rm g\end{aligned}\)

Therefore, 32.78 g mass must be added to the solution.

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