Question

1. One way of removing phosphate pollution from lakes that are dying due to excess nitrogen and phosphorous nutrients (a process called eutrophication) is to generate a stable precipitate that settles out to the bottom, where it remains inert.

a) CaHPO4·2H2O has a Ksp of 2.6 x 10-7 . If a stoichiometric amount of Ca2+ were added to Half Moon Lake, how much HPO4 2- (hydrogen phosphate, a primary form in water) would remain dissolved in the lake following treatment? Report this as a mixing ratio, in ppm (w/w).

b) Say that the volume of the lake is 1 x 106 L and that the initial mixing ratio (w/w) of HPO4 2- is 100 ppm. Convert hydrogen phosphate mixing ratios into molar quantities (assuming that the lake density is equal to that of pure water at 15 °C) to calculate the equilibrium molar concentration of aqueous Ca2+ needed to reduce that amount to a final value of 10 ppm.

c) Considering both the amount of CaCl2 required to react stoichiometrically with HPO4 2+, PLUS the amount of excess CaCl2 required to further reduce the phosphate concentration to 10 ppm (w/w), calculate the total mass of CaCl2 required to apply to the lake (in kg).

d) If 100 ppm of F– contamination also existed in Half Moon Lake (very, very unlikely, and a huge environmental problem!), which would precipitate first from addition of CaCl2, calcium fluoride or calcium hydrogen phosphate? (Ksp of CaF2 = 3.9 x 10-11)

Answer 1

Mw of Ca = 40.078 g/mol

Mw of HPO₄ = 95.98 g/mol

Mw of CaCl₂ = 110.98 g/mol

Mw of F = 19 g/mol

a)

CaHPO₄.2H₂O -> Ca²⁺ + HPO₄^2- + 2H2O

                                    S      +      S

Ksp = S² = 2.6 x 10^-7

S = 5.1 x 10^-4 M

Moles of HPO₄^2- remaining dissolved = 95.98 * S

= 95.98 * 5.1 x 10^-4 = 0.0489 g/L

ppm (w/w) = 0.0489 * 1000 = 48.9

b)

Volume of lake, V = 10⁶ L

Mass of lake = 10⁶ * 1000 g/L = 10⁹ g

Initial mixing ratio = 100 ppm

Final mixing ratio = 10 ppm

Initial Mass of HPO₄^2- remaining dissolved = 100 * 10⁹ / 10⁶ g = 10⁵ g

Initial Moles of HPO₄^2- remaining dissolved = 10⁵ / 95.98 = 1.042 x 10³

Initial concentration of HPO₄^2- remaining dissolved =1.042 x 10³ / 10⁶ = 1.042 x 10^-3 M

Final concentration of HPO₄^2- remaining dissolved = 1.042 x 10^-3 M / 10 = 1.042 x 10^-4 M

[Ca2+] [HPO₄2-] = 2.6 x 10^-7

[Ca2+]_initial = 2.6 x 10^-7 / 1.042 x 10^-3 = 2.5 x 10^-4

[Ca2+]_final = 2.6 x 10^-7 / 1.042 x 10^-4 = 2.5 x 10^-3 M

c)

Moles of CaCl2 for 100 ppm mixing volume = [Ca2+]initial * total volume

= 2.5 x 10^-4 * 10⁶ = 2.5 x 10²

Moles of CaCl₂ for 10 ppm mixing volume = [Ca2+]final * total volume

= 2.5 x 10^-3 * 10⁶ = 2.5 x 10³

Mass of CaCl₂ for 10 ppm mixing volume = 2.5 x 10³ * 110.98 g

= 2.77 x 105 g = 277 kg

d)

Moles of F- = 100 / 10⁶ * 10⁹ g/ 19 g/mol = 5.26 x 10³

[F-] = 5.26 x 10³ / 10⁶ = 5.26 x 10^-3 M

[HPO₄^2-] = 1.042 x 10^-3 M

CaF2 : [Ca2+]₁ = Ksp / [F-]² = 1.4 x 10^-6 M

CaHPO4 : [Ca2+]₂ = Ksp / [HPO4^2-] = 2.5 x 10^-4 M

[Ca2+]₁ < [Ca2+]₂ Therefore calcium fluoride will precipitate first.