Question

Copper is a toxic metal routinely found in older piping systems. Water treatment plants must report to EPA any values of copper ions in their water that exceed 1.3 mg/L, as it is hazardous above this level. Water running through the pipes contains bicarbonate at a fixed concentration of 95 mg/L (reminder: pK_a2 = 10.35 for carbonate system). Assume that copper is present in pipes as copper (II) carbonate [solid], which can disassociate into water per the equation below: CuCO_3(s) rightarrow Cu^2+ + CO_3^2- K_sp = 1.4 times 10^-10 If copper is naturally present in the water at a concentration of 45 ppb, calculate the pH at which the water at equilibrium with the pipes (e.g. no additional dissolution of copper into water). Ignore Part a. In what pH range will this water be dangerous (Cu^2+ >1.3 mg/L)?

Answer 1

a) Concentration of Cu²⁺ is 45 ppb = 45µg/L = (45µg/L)*(1 mg/1000 µg) = 0.045 mg/L = (0.045 mg/L)*(1 g/1000 mg)*(1 mole/65.546 g) = 7.081*10^-7 mol/L = 7.081*10^-7 M

The dissociation of CuCO₃ follows the equation:

CuCO₃ (s) -------> Cu²⁺ + CO₃^2-

K_sp = [Cu²⁺][CO₃^2-]

====> 1.4*10^-10 = (7.081*10^-7).[CO₃^2-]

====> [CO₃^2-] = (1.4*10^-10)/(7.081*10^-7) = 1.977*10^-4 ≈ 2.0*10^-4

The molar concentration of CO₃^2- in the pipes is 2.0*10^-4 M.

The concentration of bicarbonate is 95 mg/L; the molar concentration of bicarbonate in water is (95 mg/L)*(1 g/1000 mg)*(1 mole/61.0168 g) = 1.5569*10^-3 M ≈ 1.56*10^-3 M.

Consider the dissociation of bicarbonate as

HCO₃^- <======> H⁺ + CO₃^2-

The pH is given by the Henderson-Hasslebach equation as

pH = pK_a2 + log [CO₃^2-]/[HCO₃^-]

===> pH = 10.35 + log (2.0*10^-4)/(1.56*10^-3) = 9.4579 ≈ 9.46

The pH of the water is 9.46 (ans).

b) Find out the molar concentration of CO₃^2- when concentration of Cu²⁺ is 1.3 mg/L.

Molar concentration of Cu²⁺ = (1.3 mg/L)*(1 g/1000 mg)*(1 mole/63.546 g) = 2.0458*10^-5 M.

Given K_sp = 1.4*10^-10,

1.4*10^-10 = (2.0458*10^-5).[CO₃^2-]

===> [CO₃^2-] = (1.4*10^-10)/(2.0458*10^-5) = 6.8433*10^-6

The molar concentration of CO₃^2- is 6.8433*10^-6 M while [HCO₃^-] = 1.56*10^-3 M.

Use the Henderson-Hasslebach equation to find the pH when Cu²⁺ = 1.3 mg/L.

pH = pK_a2 + log [CO₃^2-]/[HCO₃^-] = 10.35 + log (6.8433*10^-6)/(1.56*10^-3) = 7.992 ≈ 7.99

Therefore at all pH < 7.99, the water will be dangerous as Cu²⁺ concentration will exceed the EPA limit of 1.3 mg/L (ans).