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 $\times$ 10^-16) dissolving from the surface of the pipes? Note the U.S. Environmental Protection Agency's limit on lead in drinking water is 7.2 $\times$10^-8 M.

Question

Quizplus · Accepted Answer

The solubility product (Ksp) of lead(II) hydroxide is given as 2.8 x 10^-16. The dissolution of lead(II) hydroxide can be represented by the equation Pb(OH)2(s) ⇌ Pb^2+(aq) + 2OH^-(aq). From the Ksp expression, [Pb^2+][OH^-]^2 = 2.8 x 10^-16. Assuming the concentration of Pb^2+ and OH^- are equal, which is not correct due to the stoichiometry (for every 1 Pb^2+, there are 2 OH^-), but it simplifies the initial estimation. The correct approach considers the stoichiometry, leading to the square root of the Ksp divided by 4 (due to the square of the hydroxide ion concentration in the Ksp expression), which gives the maximum possible concentration of lead in the water, aligning with choice A.

Lead Pipes Were Used at One Time for Delivering Drinking ×\times×

Lead Pipes Were Used at One Time for Delivering Drinking $\times$