Open Question
The Ksp of PbBr2 is 6.60×10−6. What is the molar solubility of PbBr2 in pure water?
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
A saturated solution of PbCl2 has [Cl^-] = 2.86×10^-2 M. What is the concentration of Pb^2+? Given that Ksp = 1.17 × 10^-5.
A
1.17 × 10^-5 M
B
5.85 × 10^-3 M
C
8.19 × 10^-3 M
D
2.04 × 10^-2 M
Verified step by step guidance1
Understand the dissolution equilibrium of PbCl2 in water: PbCl2(s) ⇌ Pb^2+(aq) + 2Cl^-(aq). This equation shows that for every mole of PbCl2 that dissolves, one mole of Pb^2+ and two moles of Cl^- are produced.
Given the concentration of Cl^- is 2.86×10^-2 M, use the stoichiometry of the dissolution reaction to determine the concentration of Pb^2+. Since the ratio of Cl^- to Pb^2+ is 2:1, the concentration of Pb^2+ will be half of the concentration of Cl^-.
Calculate the concentration of Pb^2+ using the stoichiometric relationship: [Pb^2+] = [Cl^-] / 2.
Use the solubility product constant (Ksp) expression for PbCl2: Ksp = [Pb^2+][Cl^-]^2. Substitute the known values into this expression to verify the concentration of Pb^2+.
Solve the equation for [Pb^2+] using the given Ksp value: 1.17 × 10^-5 = [Pb^2+](2.86×10^-2)^2. Rearrange and solve for [Pb^2+].
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