Question

What is the alkalinity, in terms of calcium carbonate, of a water with 80 mg/L bicarbonate and 5 mg/L of carbonate at a pH of 9?

Answer 1

The sample contains 80 mg/L of HCO₃^- ion and 5 mg/L of CO₃^-2 ions.

Gram Equivalent = Molecular weight of compound ÷ Charge on compound

Gram equivalent of HCO₃^- ion = 61/1 = 61  

Gram equivalent of CO₃^-2 ions= 60/2 = 30

Gram equivalent of CaCO₃ = 100/2 = 50

Concentration of ions in terms of CaCO₃ = (concentration of the ion)*(gram equivalent of CaCO₃)/Gram equivalent of the ion.

Then,

Concentration of HCO₃^- ion in terms of CaCO₃ = 80*50/61 = 65.57 mg/L of CaCO₃

Concentration of CO₃^-2 ions in terms of CaCO₃ = 5*50/30 = 8.33 mg/L of CaCO₃

Therefore, total concentration of solutes in the water = 65.57 mg/L of CaCO₃ + 8.33 mg/L of CaCO₃

= 73.90 mg/L of CaCO₃