Question

A sample of water has the following concentrations of ions (and pH = 7.0):

                        cations                                    mg/L      anions            mg/L

                        Ca⁺²      40.0                           HCO₃^-   110.0

                        Mg⁺²     10.0                           SO₄^2-     67.2

                        Na⁺       ?                                 Cl^-         11.0

                        K⁺         7.0

Assuming no other constituents are missing, use an anion-cation balance to estimate the concentration of Na⁺ (in mg/L as Na⁺)? Remember that the balance cannot be in mg/L.

Answer 1

(Ca²⁺) = (40 mg/L) x (meq/20 mg) x ( 50 mg CaCO₃/meq) = 100 mg/L as CaCO₃

(Mg²⁺) = (10 mg/L) x (meq/12.2 mg) x (50 mg CaCO₃/meq) = 41 mg/L as CaCO₃

(K⁺) = (7 mg/L) x (meq/39.1 mg) x (50 mg CaCO₃/meq) = 9 mg/L as CaCO₃

(Cl⁻) = (11 mg/L) x (meq/35.5 mg) x (50 mg CaCO₃ /meq) = 15.5 mg/L as CaCO₃

(SO₄^2-) = (67.2 mg/L) x (meq/48 mg) x (50 mg CaCO₃ /meq) = 70 mg/L as CaCO₃

(HCO₃^-) = (110 mg/L) x ( meq/61 mg) x (50 mg CaCO₃ /meq) = 90.2 mg/L as CaCO₃

(Na⁺) = $\sum$ anions − $\sum$ cations = 175.7- 150 = 25.7 mg/L as CaCO₃

Therefore,

Na⁺ = (25.7 mg as CaCO₃/L) x (23 mg/meq) x (meq/ 50 mg CaCO₃) = 11.8 mg/L as CaCO₃