Question

Given a water sample with the following concentrations, calculate the hardness in meq/L and mg/L as CaCO₃

Ca²⁺= 17 mg/L

CO₃^2-= 3.4 mg/L

Answer 1

Ca2+= 17 mg/L

CO32-= 3.4 mg/L

CaCO3 has a molecular weight of 100 g/mol

The Ca2+ cation has a molecular weight of 40 g/mol

Therefore, each milligram of CaCO3 contains 40/100 = 0.4 mg of Ca2+ ions.

The conversion is as follows:

Carbonate Alkalinity as Ca2+ (mg/L) = 0.4 *Carbonate Alkalinity as CaCO3 (mg/L)

17 mg/L = 0.4 * Carbonate Alkalinity as CaCO3 (mg/L)

Carbonate Alkalinity as CaCO3 (mg/L) = 17 mg/L / 0.4

= 42.5 mg/L

Also,

The CO3 2- anion has a molecular weight of 60 g/mol

Therefore, each milligram of CaCO3 contains 60/100 = 0.6 mg of CO3 2-

The conversion is as follows:

Carbonate Alkalinity as CO3 2- (mg/L) = 0.6 *Carbonate Alkalinity as CaCO3 (mg/L)

3.4 mg/L = 0.6 * Carbonate Alkalinity as CaCO3 (mg/L)

Carbonate Alkalinity as CaCO3 (mg/L) = 3.4 mg/L / 0.6

= 5.66 mg/L

Total hardness as CaCO3 = 42.5 + 5.66

= 48.16 mg/L

Ca2+= 17 mg/L = 17 / 20 meq/L

= 0.85 meq/L

Hardness as CaCO3 = 0.85 / 0.4

= 2.125 meq/L

CO32- = 3.4 mg/L = 3.4 / 30

= 0.113 meq/L

Hardness as CaCO3 = 0.113 / 0.6

= 0.188 meq/L

Total hardness as CaCO3 = 2.125 + 0.188

= 2.313 meq/L