Given the solubility product constant (Ksp) for PbBr₂ is 6.3 x 10⁻⁶, calculate the maximum concentration of Br⁻ ions that can exist in 0.5 L of 0.1 M PbBr₂ solution.

• A. 0.2 M
• B. 0.1 M
• C. 0.4 M
• D. 0.3 M

Answer: (D)

Verified step by step guidance

1. Understand the dissolution of PbBr₂ in water: PbBr₂(s) ⇌ Pb²⁺(aq) + 2Br⁻(aq). This equation shows that one mole of PbBr₂ produces one mole of Pb²⁺ ions and two moles of Br⁻ ions.
2. Write the expression for the solubility product constant (Ksp): Ksp = [Pb²⁺][Br⁻]². Here, [Pb²⁺] is the concentration of lead ions and [Br⁻] is the concentration of bromide ions at equilibrium.
3. Let 's' be the solubility of PbBr₂ in mol/L. At equilibrium, [Pb²⁺] = s and [Br⁻] = 2s. Substitute these into the Ksp expression: Ksp = (s)(2s)² = 4s³.
4. Given Ksp = 6.3 x 10⁻⁶, set up the equation: 4s³ = 6.3 x 10⁻⁶. Solve for 's' to find the solubility of PbBr₂.
5. Calculate the concentration of Br⁻ ions: [Br⁻] = 2s. This will give you the maximum concentration of Br⁻ ions in the solution.