You are managing a city that has lead in the drinking water service lines. You w

Question

You are managing a city that has lead in the drinking water service lines. You want to use phosphate (PO43-) as a means to control lead corrosion through the formation of lead phosphate (Pb3(PO4)2(s)).

What concentration of phosphate should you have in your distribution lines to ensure that your lead (Pb2+) concentration remains at or below 15 ppb (the EPA’s action level)?

HINTS: Pb3(PO4)2(s) = 3Pb2+ + 2PO43- Ksp = 3*10-44 Recall, when using the Ksp equation, the concentrations are all in M. Recall that M (molar) = moles/Liter The MW of Pb2+ = 207

Explanation / Answer

Ksp = [Pb2+]^3.[PO4^3-]^2 = 3 x 10^-44

[Pb2+] = 15 ppb = 0.015 mg/L = 0.015/1000 x 207 = 2.143 x 10^-6 M

So the [PO4^3-] concentration in the distribution lines should be,

[PO4^3-] = sq.rt.(Ksp/(Pb2+]^3)

               = sq.rt.(3 x 10^-44/(2.143 x 10^-6)^3)

               = 5.52 x 10^-14 M

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