Final answer:
The solubility product constant (Ksp) for Eu(OH)2 in a solution of 0.065 M KOH is approximately 2.62 × 10⁻⁸, calculated using the provided solubility and the concentration of OH- ions from the dissolved KOH.
Step-by-step explanation:
To calculate the Ksp for Eu(OH)₂, we need to consider its solubility in the presence of 0.065 M KOH. Given that the solubility of Eu(OH)₂ is 6.2 × 10⁻⁶ mol/L, we can establish the dissolution equation as such: Eu(OH)₂(s) → Eu²⁺(aq) + 2 OH⁻(aq).
In pure water, the dissolution would create one Eu²⁺ ion and two OH⁻ ions per formula unit. However, in a KOH solution, the OH⁻ concentration is already elevated due to the presence of KOH, which is a strong base and dissociates completely in water. Therefore, we can assume that the solubility equilibrium is primarily affected by the addition of Eu²⁺ ions.
The Ksp for Eu(OH)₂ can be expressed as:
Ksp = [Eu²⁺][OH⁻]²
Since the solubility of Eu(OH)₂ provides the concentration of Eu²⁺ ions and the concentration of OH⁻ ions is known from KOH, we have:
Ksp = (6.2 × 10⁻⁶)(0.065 ²) = (6.2 × 10⁻⁶)(0.065)(0.065)
Calculating this gives us: Ksp = (6.2 × 10⁻⁶)(0.004225) ≈ 2.62 × 10⁻⁸.
Therefore, the solubility product constant (Ksp) for Eu(OH)₂ in a solution of 0.065 M KOH is approximately 2.62 × 10⁻⁸.