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 5.0×10−2 m in calcium chloride and 8.5×10−2 m in magnesium nitrate. what mass of sodium phosphate would have to be added to 2.0 l of this solution to completely eliminate the hard water ions? assume complete reaction.

30 g

Explanation:

For CaCl₂

3CaCl₂ + 2Na₃PO₄ ⟶ Ca₃(PO₄)₂ + 6NaCl

Moles of CaCl₂ = c×V = 5.0 × 10⁻² × 2.0 = 0.100 mol

Moles of Na₃PO₄ = 0.100 × 2/3 = 0.0667 mol

Mass of Na₃PO₄ = 0.0667 × 163.94 = 10.9 g

———————————————————

For Mg(NO₃)₂

3Mg(NO₃)₂  + 2Na₃PO₄ ⟶ Mg₃(PO₄)₂ + 6NaCl

Moles of Mg(NO₃)₃ = c×V = 8.5 × 10⁻² × 2.0 = 0.170 mol

Moles of Na₃PO₄ = 0.170 × 2/3 = 0.113 mol

Mass of Na₃PO₄ = 0.113 × 163.94 = 1 g

—————————————————

Total mass of Na₃PO₄ = 10.9 + 18.6 = 29.5 g

To two significant figures, mass = 30 g

in the analysis of systems that involve fluid flow, we frequently encounter the combination of properties u and pv.