How many moles of Fe(OH)₂ [K_sp = 1.8 × 10^-15] will dissolve in 1 L of water buffered at pH = 12.00? A) 1.8 × 10^-11 mol B) 1.8 × 10^-9 mol C) 5.0 × 10^-12 mol D) 8.0 × 10^-6 mol E) 4.0 × 10^-8 mol

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To find the solubility of Fe(OH)3 in a solution buffered at pH 12, we use the solubility product constant (Ksp) and the relationship between pH and the concentration of hydroxide ions ([OH-]). Given: - Ksp of Fe(OH)3 = 1.8 × 10^-15- pH = 12First, calculate the concentration of hydroxide ions ([OH-]) using the pH:$$ \text{pOH} = 14 - \text{pH} = 14 - 12 = 2 $$$$ [\text{OH}^-] = 10^{-\text{pOH}} = 10^{-2} = 0.01 \, \text{M} $$The dissolution of Fe(OH)3 can be represented as:$$ \text{Fe(OH)}_3(s) \rightleftharpoons \text{Fe}^{3+}(aq) + 3\text{OH}^-(aq) $$Using the Ksp expression:$$ \text{Ksp} = [\text{Fe}^{3+}][\text{OH}^-]^3 $$Given that the concentration of OH- is much larger than what would be contributed by the dissolution of Fe(OH)3, we can assume [OH-] remains approximately 0.01 M. Solving for [Fe3+] gives:$$ 1.8 \times 10^{-15} = [\text{Fe}^{3+}](0.01)^3 $$$$ [\text{Fe}^{3+}] = \frac{1.8 \times 10^{-15}}{(0.01)^3} $$$$ [\text{Fe}^{3+}] = \frac{1.8 \times 10^{-15}}{10^{-6}} $$$$ [\text{Fe}^{3+}] = 1.8 \times 10^{-9} \, \text{M} $$Therefore, the solubility of Fe(OH)3 in a solution buffered at pH 12 is 1.8 × 10^-9 mol/L, which corresponds to choice B.

How Many Moles of Fe(OH)2 [Ksp = 1

How Many Moles of Fe(OH)₂ [K_sp = 1