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. A solution is 0.050 M in calcium chloride and 0.085 M in magnesium nitrate. What mass of sodium phosphate would you add to 1.5 L of this solution to completely eliminate the hard water ions.

Answer 1

Answer: The mass of sodium phosphate that would be added is 22.14 grams

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$    .....(1)

• For calcium chloride:

Molarity of calcium chloride solution = 0.050 M

Volume of solution = 1.5 L

Putting values in equation 1, we get:

$0.085M=\frac{\text{Moles of calcium chloride}}{1.5L}\\\\\text{Moles of calcium chloride}=(0.050mol/L\times 1.5L)=0.075mol$

1 mole of calcium chloride contains 1 mole of calcium ions and 2 moles of chloride ions

Moles of calcium ions = 0.075 moles

The chemical equation for the reaction of calcium ions and phosphate ions follows:

$3Ca^{2+}+2PO_4^{3-}\rightarrow Ca_3(PO_4)_2$

By Stoichiometry of the reaction:

3 moles of calcium ions reacts with 2 moles of phosphate ions

So, 0.075 moles of calcium ions will react with = $\frac{2}{3}\times 0.075=0.05mol$ of phosphate ions

• For magnesium nitrate:

Molarity of magnesium nitrate solution = 0.085 M

Volume of solution = 1.5 L

Putting values in equation 1, we get:

$0.085M=\frac{\text{Moles of magnesium nitrate}}{1.5L}\\\\\text{Moles of magnesium nitrate}=(0.085mol/L\times 1.5L)=0.1275mol$

1 mole of magnesium nitrate contains 1 mole of magnesium ions and 2 moles of nitrate ions

Moles of magnesium ions = 0.1275 moles

The chemical equation for the reaction of magnesium ions and phosphate ions follows:

$3Mg^{2+}+2PO_4^{3-}\rightarrow Mg_3(PO_4)_2$

By Stoichiometry of the reaction:

3 moles of magnesium ions reacts with 2 moles of phosphate ions

So, 0.1275 moles of magnesium ions will react with = $\frac{2}{3}\times 0.1275=0.085mol$ of phosphate ions

1 mole of sodium phosphate contains 1 mole of phosphate ions and 2 moles of sodium ions

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Molar mass of sodium phosphate = 164 g/mol

Total moles of phosphate ions = [0.05 + 0.085] = 0.135 moles

Putting values in above equation, we get:

$0.135mol=\frac{\text{Mass of sodium phosphate}}{164g/mol}\\\\\text{Mass of sodium phosphate}=(0.135mol\times 164g/mol)=22.14g$

Hence, the mass of sodium phosphate that would be added is 22.14 grams