Question

If a sample of tap water contains 400 ppm of Ca2+ and 80 ppm of Mg2+, what is the hardness of this water sample in terms of equivalent concentration CaCO3, in mg/L?

Answer 1

1328 mg/L

Explanation

i) Concentration of Ca²⁺ = 400ppm = 400 mg/L

Concentration of Ca²⁺ in mol/L = (0.400g/40.08g/mol)/ L = 0.009980mol/L

Concentration of Mg²⁺ = 80ppm = 80 mg/L

Concentration of Mg²⁺ = (0.080g/24.31g/mol)/ L = 0.003291mol/L

Total hardness in mol/L = 0.009980mol/L+ 0.003291mol/L = 0.01327mol/L

mass of 0.01327moles of CaCO3 = 0.01327mol × 100.09g/mol = 1.328g/L = 1328mg/L

i) The following formula is used to simply calculate total hardness as CaCO3 from mg²⁺ and Ca²⁺ concentrations

[CaCO3] = 2.5[Ca²⁺] + 4.1[Mg²⁺]

[CaCO3]=(2.5 × 400ppm)+ (4.1 × 80ppm)

=1328ppm = 1328 mg/L