[Olympiad] The Ksp of Lead(II) iodide (PbI₂) is 7.1 × 10⁻⁹. What is the molar solubility of PbI₂ in pure water?
-
8.4 × 10⁻⁵ M
-
1.2 × 10⁻³ M
-
2.6 × 10⁻³ M
-
1.9 × 10⁻³ M
📝 Detailed explanation
PbI₂(s) ⇌ Pb²⁺ + 2I⁻. Ksp = s × (2s)² = 4s³.
s = ∛(Ksp / 4) = ∛(7.1 × 10⁻⁹ / 4) = ∛(1.775 × 10⁻⁹) ≈ 1.21 × 10⁻³ M.
← Previous
→ Next
Space — Show answer
🎯 Take a test on the topic «Variant #2 of the University Scholarship Examination»
This topic contains 50 questions.
Test your knowledge!
🔐 Log in to take the test
📚 All tests by subject «Bio+Chem»
View all collections and topics in Bio+Chem
Go to subject →
