Multiple Choice
The Ksp of CuI(s) is 1.1 × 10^-12. Calculate the value of E° for the half-reaction CuI(s) + e^- → Cu(s) + I^-(aq).
18. Aqueous Equilibrium
Solubility Product Constant: Ksp
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The molar solubility of manganese (II) hydroxide, Mn(OH)2, is 3.48 × 10^-3 M. What is the Ksp of Mn(OH)2 at this temperature?
A
2.42 × 10^-6
B
5.67 × 10^-8
C
4.21 × 10^-7
D
1.21 × 10^-5
Verified step by step guidance1
Start by writing the balanced chemical equation for the dissolution of manganese (II) hydroxide in water: Mn(OH)_2(s) ⇌ Mn^{2+}(aq) + 2OH^{-}(aq).
Identify the expression for the solubility product constant (Ksp) for Mn(OH)_2. The Ksp expression is given by: Ksp = [Mn^{2+}][OH^{-}]^2.
Since the molar solubility of Mn(OH)_2 is given as 3.48 × 10^-3 M, this represents the concentration of Mn^{2+} ions in solution at equilibrium.
For every mole of Mn(OH)_2 that dissolves, 2 moles of OH^{-} ions are produced. Therefore, the concentration of OH^{-} ions is 2 × 3.48 × 10^-3 M.
Substitute the concentrations of Mn^{2+} and OH^{-} into the Ksp expression: Ksp = (3.48 × 10^-3) × (2 × 3.48 × 10^-3)^2. Calculate this expression to find the Ksp value.
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