Question

What concentration of the lead ion, Pb 2+ , must be exceeded to precipitate PbBr 2 from a solution that is 1.00×10 −2 M in the bromide ion, Br − ? (Given K sp for lead(II) bromide is 4.67×10 −6.)

A) 1.00×10 −6 M
B) 2.00×10 −6 M
C) 3.00×10 −6 M
D) 4.00×10 −6 M

Answer 1

To precipitate PbBr2 from a solution that is 1.00×10⁻² M in the bromide ion, the concentration of Pb 2+ must exceed 4.00×10⁻² M, according to the given solubility product constant (Ksp) for lead(II) bromide.

Step-by-step explanation:

To determine the concentration of the lead ion (Pb 2+) that must be exceeded to precipitate PbBr2 from a solution that is 1.00×10⁻² M in the bromide ion (Br⁻), we use the given solubility product constant (Ksp) for lead(II) bromide, which is 4.67×10⁻⁶. The Ksp expression for the dissolution of PbBr2 is given by Ksp = [Pb²+][Br⁻]².

We already know the concentration of Br⁻ ions, so we can rearrange the equation to solve for [Pb²+]: 4.67×10⁻⁶ = [Pb²+](1.00×10⁻²)². Solving this gives us [Pb²+] = 4.67×10⁻⁶ / (1.00×10⁻²)² = 4.67×10⁻². Therefore, the correct option is D) 4.00×10⁻² M.