1.0 dm3 of distilled water was used to wash 2.0g of a precipitate of AgCl. If the solubility product of AgCl is 2.0 * 10-10mol2dm-6, what quantity of silver...

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When solid AgCl is mixed with water, it partially dissolves and reaches equilibrium according to the following equation: AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq) The solubility product (Ksp) expression for this equilibrium is: Ksp = [Ag⁺][Cl⁻] Given that the solubility product (Ksp) of AgCl is 2.0 × 10⁻¹⁰ mol² dm⁻⁶, we can use this value and the initial volume of water (1.0 dm³) to calculate the maximum amount of Ag⁺ and Cl⁻ that can dissolve before reaching equilibrium. Let x be the molar solubility of AgCl, which is also equal to the concentration of Ag⁺ and Cl⁻ ions that dissolve. Then, at equilibrium: [Ag⁺] = [Cl⁻] = x Substituting these into the Ksp expression, we have: Ksp = x² Solving for x, we get: x = √Ksp = √(2.0 × 10⁻¹⁰ mol² dm⁻⁶) = 1.414 × 10⁻⁵ mol dm⁻³ This means that at equilibrium, the concentration of Ag⁺ and Cl⁻ ions in the solution is 1.414 × 10⁻⁵ mol dm⁻³. Now, we can use the volume of water and the concentration of Ag⁺ and Cl⁻ ions to calculate the total amount of AgCl that can dissolve: (1.0 dm³) × (1.414 × 10⁻⁵ mol dm⁻³) = 1.414 × 10⁻⁵ mol AgCl This is the maximum amount of AgCl that can dissolve in 1.0 dm³ of water. However, in the given problem, only 2.0 g of AgCl is used, which is equivalent to: (2.0 g) / (143.32 g/mol) = 0.01395 mol AgCl Since this amount is larger than the maximum amount that can dissolve in 1.0 dm³ of water, all of the AgCl will dissolve and none will be lost in the process. Therefore, the answer is option (D) 1.414 × 10⁻⁵ mol dm⁻³