Answer
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Step 1:Let's solve this problem step by step, focusing on the solubility of lead(II) chloride (PbCl2).
Step 2:: Understand Solubility Rules
Solubility depends on the interaction between ionic compounds and water. The solubility of an ionic compound is determined by the strength of the ionic bonds and the ability of water molecules to overcome these bonds.
Step 3:: Examine the Ionic Compound
PbCl2 is an ionic compound composed of lead(II) ions ($$Pb^{2+}$$) and chloride ions ($$Cl^{-}$$).
Step 4:: Analyze Solubility Characteristics
- Strong ionic bonds between $$Pb^{2+}$$ and $$Cl^{-}$$ ions
PbCl^2 is considered sparingly soluble due to: - High lattice energy of the crystal - Limited ability of water molecules to overcome these bonds
Step 5:: Solubility Product Constant (K_{sp})
K_{sp} = [Pb^{2+}][Cl^{-}]^{2} = 1.6 \times 10^{-5}$$ at 25°C
The solubility of PbCl^2 is described by its low solubility product constant:
Step 6:: Explanation of Low Solubility
The low $$K_{sp}$$ indicates that very few ions dissociate in water, making PbCl2 essentially insoluble.
Final Answer
PbCl^2 is insoluble in water due to its strong ionic bonds and low solubility product constant, which prevent significant dissociation of ions in water.
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