Question

A sample of water has the following concentration of ions (and pH = 7.0).

                        cations                                    mg/L      anions            mg/L

                        Ca⁺²      95.0                           HCO₃^-   160.0

                        Mg⁺²     26.0                           SO₄^2-     135.0

                        Na⁺       15.0                           Cl^-         73.0

a) Draw an ion concentration bar (use meq/L for units).

b) What is the total hardness (TH) in mg/L as CaCO₃?

c) What is the carbonate hardness (CH) in mg/L as CaCO₃?

d) What is the non-carbonate hardness (NCH) in mg/L as CaCO₃?

e) What is the alkalinity in mg/L as CaCO₃?

Answer 1

a. Ca2+ = 95 mg/L = 95 mg.L^-1 /20000 mg.eq^-1 = 0.00475 eq/L = 4.75 meq/L

Mg2+ = 26 mg/L = 26 mg.L^-1 /12000 mg.eq^-1 = 0.00216 eq/L = 2.16 meq/L

HCO₃^- = 160 mg/L = 160 mg.L^-1 /61000 mg.eq^-1 = 0.00262 eq/L = 2.62 meq/L

Na⁺ = 15 mg/L = 15 mg.L^-1 /23000 mg.eq^-1 = 0.000652 eq/L = 0.652 meq/L

SO4^2- = 135 mg/L = 135 mg.L^-1 /48000 mg.eq^-1 = 0.00281 eq/L = 2.81 meq/L

Cl- = 73 mg/L = 73 mg.L^-1 /35500 mg.eq^-1 = 0.00205 eq/L = 2.05 meq/L

b. Total hardness is the total concentration of Calcium, Magnesium, and other higher valence metal cations (greater than +1) dissolved in a sample of water.

Total hardness

= (0.00475 + 0.00216) eq/L * 50000 mg/Eq

= 345.5 mg/L as CaCO₃.

c. Carbonate hardness is the measure of hard ions associated with carbonate and bicarbonate anions contained in the water.

CH (for the dissolved Ca2+) = 0.00475 eq/L

= 0.00475 eq/L * 50000 mg/Eq

= 237.5 mg/L as CaCO₃.

d. Noncarbonate hardness is the portion of total hardness in water that is not produced by carbonates, but primarily by sulfate anions.

NCH

= 345.5-237.5

= 108 mg/L as CaCO₃.

e. Alkalinity of water is mainly due to the presence of two forms of the carbonate ions denoted as HCO3(-) and CO3(2-).

Here alkalinity

= 0.00262 eq/L * 50000 mg/Eq

= 131 mg/L as CaCO₃.