What mass of sodium carbonate would be required to remove essentially all of the calcium ion from 750 l of solution containing 43 mg ca2 + per liter?

Ca²⁺ reacts with CO₃²⁻ and forms CaCO₃. The reaction is

CO₃²⁻ (aq) + Ca²⁺ (aq) → CaCO₃ (s)

First, let's find out the moles of Ca²⁺ in 750 L.

1 L has 43 mg of Ca²⁺

Hence, mass of Ca²⁺ in 750 L = 43 mg/L x 750 L

= 32250 mg = 32.25 g

Molar mass of Ca²⁺ = 40.078 g/mol

Hence, moles of Ca²⁺ in 750 L = mass / molar mass

= 32.25 g / 40.078 g/mol

= 0.8047 mol

The stoichiometric ratio between Ca²⁺ and CO₃²⁻ is 1 : 1

Hence, moles of CO₃²⁻ = moles of Ca²⁺

= 0.8047 mol

CO₃²⁻ ions from dissociation of Na₂CO₃. The dissociation of Na₂CO₃ in water is

Na₂CO₃ (s) → 2Na⁺ (aq) + CO₃²⁻ (aq)

The stoichiometric ratio between Na₂CO₃ (s) and CO₃²⁻ (aq) is 1 : 1

Hence, moles of Na₂CO₃ (s) = moles of CO₃²⁻ (aq)

= 0.8047 mol

Molar mass of Na₂CO₃ = 105.9888 g/mol

Hence, mass of Na₂CO₃ = moles x molar mass

= 0.8047 mol x 105.9888 g/mol

= 85.29 g

Hence, mass of Na₂CO₃ needed to react with Ca²⁺ is 85.29 g.