Copper is a toxic metal routinely found in older piping systems. Water treatment
ID: 480647 • Letter: C
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)?Explanation / Answer
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+ + CO32-
Ksp = [Cu2+][CO32-]
====> 1.4*10-10 = (7.081*10-7).[CO32-]
====> [CO32-] = (1.4*10-10)/(7.081*10-7) = 1.977*10-4 2.0*10-4
The molar concentration of CO32- 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+ + CO32-
The pH is given by the Henderson-Hasslebach equation as
pH = pKa2 + log [CO32-]/[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 CO32- 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 Ksp = 1.4*10-10,
1.4*10-10 = (2.0458*10-5).[CO32-]
===> [CO32-] = (1.4*10-10)/(2.0458*10-5) = 6.8433*10-6
The molar concentration of CO32- 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 = pKa2 + log [CO32-]/[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).
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