Question

What is the molar solubility of Pb(OH)₂ in a 0.138-M solution of CH₃NH₂?

a) 1.0 x 10⁻⁴ M
b) 2.0 x 10⁻⁴ M
c) 3.0 x 10⁻⁴ M
d) 4.0 x 10⁻⁴ M

Answer 1

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.