Multiple Choice
C2D3 has a solubility product constant (Ksp) of 9.14 × 10⁻⁹. What is the molar solubility of C2D3 in mol/L?
18. Aqueous Equilibrium
Solubility Product Constant: Ksp
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Calculate the molar solubility of MgF2 in 0.15 M MgCl2 at 25 °C, given that the Ksp of MgF2 is 6.4 x 10^-9.
A
3.2 x 10^-4 M
B
1.6 x 10^-4 M
C
6.4 x 10^-5 M
D
1.0 x 10^-3 M
Verified step by step guidance1
Understand the problem: We need to calculate the molar solubility of MgF2 in a solution that already contains MgCl2. The presence of MgCl2 affects the solubility of MgF2 due to the common ion effect.
Write the dissolution equation for MgF2: MgF2(s) ⇌ Mg²⁺(aq) + 2F⁻(aq). The solubility product constant (Ksp) expression for this equilibrium is Ksp = [Mg²⁺][F⁻]².
Consider the initial concentrations: The solution already contains 0.15 M Mg²⁺ from MgCl2. Let 's' be the molar solubility of MgF2. Therefore, [Mg²⁺] = 0.15 + s and [F⁻] = 2s.
Substitute into the Ksp expression: Ksp = (0.15 + s)(2s)². Given Ksp = 6.4 x 10⁻⁹, set up the equation 6.4 x 10⁻⁹ = (0.15 + s)(4s²).
Assume s is small compared to 0.15 M, simplifying the expression to 6.4 x 10⁻⁹ ≈ 0.15 * 4s². Solve for 's' to find the molar solubility of MgF2.
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