What is the molar solubility of Fe(OH)3(s) in a solution that is buffered at pH 2.75? The Ksp of Fe(OH)3 is 6.3 × 10-38.
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To find the molar solubility of Fe(OH)3 in a solution buffered at pH 2.75, we need to determine the concentration of Fe3+ ions in the solution.
The first step is to write the balanced equation for the dissociation of Fe(OH)3:
Fe(OH)3(s) ⇌ Fe3+(aq) + 3OH-(aq)
According to the given information, the pH of the solution is 2.75, which tells us the concentration of H+ ions in the solution. Using the equation for the ionization of water, we can find the concentration of OH- ions:
[H+] × [OH-] = 1.0 × 10^-14
Since the solution is buffered, we can assume that the concentration of OH- is equal to the concentration of H+:
[H+]^2 = 1.0 × 10^-14
Taking the square root of both sides:
[H+] = √(1.0 × 10^-14)
[H+] ≈ 1.0 × 10^-7
This tells us that the concentration of OH- ions is also approximately 1.0 × 10^-7 M.
Now, we can substitute this value into the equilibrium equation for Fe(OH)3 to find the concentration of Fe3+ ions:
6.3 × 10^-38 = [Fe3+][OH-]^3
6.3 × 10^-38 = [Fe3+] × (1.0 × 10^-7)^3
6.3 × 10^-38 = [Fe3+] × 1.0 × 10^-21
Therefore, [Fe3+] ≈ 6.3 × 10^-17 M.
So, the molar solubility of Fe(OH)3 in the buffered solution with pH 2.75 is approximately 6.3 × 10^-17 M.