Final answer:
The molar solubility of Pb(OH)₂ in a 0.138-M solution of CH₃NH₂ is 0.0017 M.
Step-by-step explanation:
The molar solubility of Pb(OH)₂ in a 0.138-M solution of CH₃NH₂ can be determined using the concept of solubility product (Ksp).
In this case, we need to consider the dissolution stoichiometry of the compound which shows a 1:1 relation between the molar amounts of the compound and its ions.
Using the given information that [Pb²+] = 0.0017 M, we can conclude that the molar solubility of Pb(OH)₂ in the CH₃NH₂ solution is equal to the concentration of Pb²+ ions, which is 0.0017 M. Therefore, the correct answer is d) 4.0 x 10⁻⁴ M.