Asked by abbey
Lead pipes were once used to transport drinking water and are still found in some older homes and city water systems. Suppose the lead concentration and pH of the water determined by lead (II) hydroxide (Ksp = 2.8x10^-16) dissolving to produce a saturated solution. What would the lead concentration (M) and pH of this water be?
I am confused. Does this question actually have anything to do with lead pipes or is it a simple solubility problem?
Pb(OH)2 <--> Pb2+ + 2 OH-
Ksp = [Pb2+]([OH-]^2) = (s)((2s)^2) = 4(s^3)
Solve for s, which is [Pb2+].
Calculate [OH-], which is 2s, and then solve for pOH and lastly pH.
I am confused. Does this question actually have anything to do with lead pipes or is it a simple solubility problem?
Pb(OH)2 <--> Pb2+ + 2 OH-
Ksp = [Pb2+]([OH-]^2) = (s)((2s)^2) = 4(s^3)
Solve for s, which is [Pb2+].
Calculate [OH-], which is 2s, and then solve for pOH and lastly pH.
Answers
Answered by
DrBob222
You have it.
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