This chemistry question involves key chemical concepts and calculations. The detailed solution below walks through each step, from identifying the reaction type to computing the final answer.
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Lead, silver, and calcium compounds exhibit varying degrees of solubility in water, and their insolubility is primarily governed by lattice energy and hydration energy. Step 1: For lead (Pb) and silver (Ag), many of their common salts, such as chlorides (PbCl₂, AgCl), sulfates (PbSO₄, Ag₂SO₄), and sulfides (PbS, Ag₂S), are insoluble. This is because the lattice energy holding the ions together in their solid crystal structure is greater than the hydration energy released when water molecules surround and stabilize the individual ions. The strong attraction between Pb²⁺/Ag⁺ and Cl⁻/SO₄²⁻/S²⁻ ions outweighs the energy gained from solvation by water. Step 2: For calcium (Ca), its compounds show intermediate solubility. Calcium sulfate (CaSO₄) is only sparingly soluble, meaning a small amount dissolves, but it's often considered insoluble in practical terms. Calcium chloride (CaCl₂) is, however, highly soluble. The solubility of calcium compounds depends on the anion. For CaSO₄, the lattice energy is significant enough to limit solubility, while for CaCl₂, the hydration energy of Ca²⁺ and Cl⁻ ions is much larger, leading to high solubility. Step 3: In general, insolubility occurs when the energy required to break apart the ionic lattice (lattice energy) is greater than the energy released when the ions are surrounded by water molecules (hydration energy). For soluble compounds, the hydration energy is greater than the lattice energy. Send me the next one 📸