Final answer:
The maximum possible concentration of lead in water from lead(II) hydroxide dissolving from the surface of the pipes can be calculated using the solubility product constant (Ksp). The maximum concentration of lead in the water is approximately 1.3 x 10⁻⁶ M.
Step-by-step explanation:
The maximum possible concentration of lead in water from lead(II) hydroxide dissolving from the surface of the pipes can be calculated using the solubility product constant (Ksp) of lead(II) hydroxide.
Since the question provides the Ksp value for lead(II) hydroxide (Ksp = 2.8 x 10⁻¹⁶), we can use this value to calculate the maximum concentration of lead in the water.
To do this, we need to write the balanced equation for the dissolution of lead(II) hydroxide and the Ksp expression:
Pb(OH)₂(s) <=> Pb²⁺(aq) + 2OH⁻(aq)
Ksp = [Pb²⁺][OH⁻]²
Since the concentration of lead(II) hydroxide is equal to the concentration of lead ions, we can assume that the concentration of lead ions is equal to x.
Therefore, Ksp = x * (2x)²
Using the given Ksp value, we can solve for x:
2.8 x 10⁻¹⁶ = x * (2x)²
2.8 x 10⁻¹⁶ = 4x³
x³ = (2.8 x 10⁻¹⁶) / 4
x = ∛((2.8 x 10⁻¹⁶) / 4)
x ≈ 1.3 x 10⁻⁶
Therefore, the maximum possible concentration of lead in the water is approximately 1.3 x 10⁻⁶ M (moles per liter).