Question

The Water Treatment Plant and the University Water Treatment Plant remove calcium hardness (Ca2+) by precipitating (making) CaCO3(s). If, after establishing equilibrium with CaCO3(s), the dissolved Ca2+ concentration is 10 mg/L and the pH is 10.5. a. What is the CO32- concentration in mol/L? b. What is the HCO3- concentration in mol/L? c. What is the alkalinity in mg/L as CaCO3?

Answer 1

[Ca²⁺] = 10/ (40*1000) mol/L = 2.5 x 10^-4 mol/L

pH = 10.5

[H+] = 10^-10.5 = 3.16 x 10^-11 mol/L

[OH-] = 10^-14 / 10^-10.5 = 3.16 x 10^-4 mol/L

Solution will contain Ca²⁺, H⁺, HCO₃⁻, CO₃²⁻, OH⁻ ions

Solution must be electrically neutral, we have

2[Ca²⁺] + 2[H⁺] = [HCO₃⁻] + 2[CO₃²⁻] + [OH⁻]   (1)

HCO3− ---> H+ + CO32−

Ka2 = 5.61×10−11 at 25 °C

[CO₃²⁻][H⁺]/[HCO₃⁻] = 5.61×10⁻¹¹

[CO₃²⁻]/[HCO₃⁻] = 5.61×10⁻¹¹ / 3.16 x 10^-11 = 1.775

Using it in equation (1) we get

2[Ca²⁺] + 2[H⁺] = [HCO₃⁻] + 2(1.775[HCO₃⁻]) + [OH⁻]

Or 2[Ca²⁺] + 2[H⁺] = 4.55[HCO₃⁻] + [OH⁻]

b) [HCO₃⁻] = (2 * 2.5 x 10^-4 - 3.16 x 10^-4 )/4.55 = 4.04 x 10^-5 mol/L

a) [CO₃²⁻] = 1.775[HCO₃⁻] = 1.775 * 4.04 x 10^-5 mol/L

= 7.17 x 10^-5 mol/L

c)

Alkalinity is contributed by Ca²⁺, HCO₃⁻ ,HCO₃⁻ and OH^-

1. [Ca²⁺] = 2.5 x 10^-4 mol/L

= 2.5 x 10^-4 mol/L as CaCO₃ = 2.5 x 10^-4 * 100 *1000 = 25 mg/L as CaCO₃

2. [HCO₃⁻] = 4.04 x 10^-5 mol/L

= 4.04 x 10^-5 /2 mol/L as CaCO₃ = 2.02 x 10^-5 mol/L = 2.02 mg/L as CaCO₃

3. [CO₃²⁻] = 7.17 x 10^-5 mol/L

= 7.17 x 10^-5 mol/L as CaCO₃ = 7.17 mg/L as CaCO₃

4. [OH⁻] = 3.16 x 10^-4 mol/L

= 3.16 x 10^-4/2mol/L as CaCO₃ = 3.16 x 10^-4/2*100 *1000 mg/L as CaCO₃ = 15.81 mg/L as CaCO₃

Alkalinity in mg/L as CaCO₃ = 25 + 2.02 + 7.17 + 15.81 mg/L

= 50 mg/L as CaCO₃