Multiple Choice
At 25°C, calcium fluoride (CaF2) has a solubility product constant Ksp = 3.5 × 10^-11. What is the solubility of CaF2 in mol/L at this temperature?
18. Aqueous Equilibrium
Solubility Product Constant: Ksp
Struggling with General Chemistry?
Join thousands of students who trust us to help them ace their exams!Watch the first videoMultiple Choice
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
Verified step by step guidance1
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.
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.
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.
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.
Calculate the concentration of Br⁻ ions: [Br⁻] = 2s. This will give you the maximum concentration of Br⁻ ions in the solution.
1:47mWatch next
Master Solubility Product Constant with a bite sized video explanation from Jules
Start learningRelated Videos
Related Practice
Solubility Product Constant: Ksp practice set
Problem sets built by lead tutorsExpert video explanations