What pH is required to start precipitation of Fe(OH)3 from a 0.380 M solution of FeCl3? Ksp of Fe(OH)3 = 4.0e-29.
April 3, 2011 2:26pm UTC
Fe(OH)3(s) ==> Fe^+3 + 3OH^-
Ksp = (Fe^+3)(OH^-)^3 = 4.0E-29
(OH^-) = cube root(4E-29/Fe^+3)
(OH^-) = 4.68E-10
Convert that to H^+ and to pH. I get 4.67 for pH.
I have ignored any contribution to the acidity made by the hydrolysis of the FeCl3.
April 3, 2011 8:23pm UTC
Ah, precipitation, the fancy way of saying "the moment when solid particles fall from grace." Now, let me entertain you with some chemistry humor.
To find the pH required to start the precipitation of Fe(OH)3, we need to figure out the concentration of OH- ions. And since we're talking about Fe(OH)3, we'll have 3 OH- ions per formula unit. So, let's get this pH party started!
Using the Ksp expression for Fe(OH)3, which is equal to [Fe3+][OH-]^3, we can say that [OH-] = (Ksp/[Fe3+])^(1/3).
Now, for the punchline! Substituting the given values, we get [OH-] = (4.0e-29/0.380)^(1/3). I could go on with the math, but let's keep it simple: the concentration of OH- ions will be so ridiculously small that it won't significantly impact the pH of the solution.
So, in short, the pH required to start the precipitation of Fe(OH)3 from a 0.380 M solution of FeCl3 is likely to remain pretty much unchanged and won't be influenced by the trace amount of OH- ions.
September 24, 2023 10:11pm UTC