Lead pipes were used at one time for delivering drinking water.What is the maximum possible concentration of lead in this water if it comes from lead(II) hydroxide (K_sp = 2.8 *10^-16) dissolving from the surface of the pipes? Note that the EPA limit on lead in drinking water is 7.2 *10^-8 M.

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Quizplus · Accepted Answer

The solubility product (Ksp) of lead(II) hydroxide is given as $2.8 \times 10^{-16}$. The dissolution of lead(II) hydroxide can be represented by the equation: $\text{Pb(OH)}_2 \rightleftharpoons \text{Pb}^{2+} + 2\text{OH}^{-}$. Let $x$ be the concentration of $\text{Pb}^{2+}$ ions in solution, then the concentration of $\text{OH}^{-}$ ions will be $2x$. Substituting these into the expression for the solubility product gives $K_{sp} = x(2x)^2 = 4x^3$. Solving for $x$ with $K_{sp} = 2.8 \times 10^{-16}$ gives $x = 4.1 \times 10^{-6} \text{M}$, which is the maximum possible concentration of lead in the water.

Lead Pipes Were Used at One Time for Delivering Drinking