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...
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 was lost in the process?
Answer Details
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