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.
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.
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.
To determine the required pH to start the precipitation of Fe(OH)3 from a 0.380 M solution of FeCl3, we need to use the solubility product constant (Ksp) of Fe(OH)3 and the balanced chemical equation.
The balanced chemical equation for the dissociation of FeCl3 in water is:
FeCl3(aq) ⇌ Fe3+(aq) + 3Cl-(aq)
In solution, Fe3+ ions react with OH- ions to form Fe(OH)3(s):
Fe3+(aq) + 3OH-(aq) ⇌ Fe(OH)3(s)
From the balanced equation, we can see that one Fe3+ ion reacts with three OH- ions to form one molecule of Fe(OH)3.
The solubility product constant (Ksp) expression for Fe(OH)3 is:
Ksp = [Fe3+][OH-]^3
Given that the Ksp of Fe(OH)3 is 4.0e-29 and the concentration of FeCl3 is 0.380 M, we can assume that the concentration of Fe3+ ions is also 0.380 M (since one Fe3+ ion is formed from one FeCl3 molecule).
Let x represent the concentration of OH- ions that react with Fe3+ ions to form Fe(OH)3.
Using the Ksp expression, we can write:
Ksp = (0.380)(x)^3
Now, let's solve for x:
x = (Ksp / 0.380)^(1/3)
x = (4.0e-29 / 0.380)^(1/3)
x ≈ 4.18e-10
The concentration of OH- ions needed to precipitate Fe(OH)3 from the 0.380 M solution of FeCl3 is approximately 4.18e-10 M.
To calculate the pH, we can use the formula:
pOH = -log10[OH-]
pOH = -log10(4.18e-10)
pOH ≈ 9.38
Since pH + pOH = 14, we can calculate the pH:
pH = 14 - pOH
pH = 14 - 9.38
pH ≈ 4.62
Therefore, the required pH to start the precipitation of Fe(OH)3 from a 0.380 M solution of FeCl3 is approximately 4.62.
To find the pH required to start precipitation of Fe(OH)3 from a 0.380 M solution of FeCl3, you need to determine the concentration of Fe3+ ions in solution and compare it with the solubility product constant (Ksp) for Fe(OH)3.
The balanced chemical equation for the dissociation of FeCl3 in water is:
FeCl3(s) -> Fe3+(aq) + 3 Cl-(aq)
Since FeCl3 is a strong electrolyte, it dissociates completely in water. Therefore, the initial concentration of Fe3+ ions in solution is equal to the initial concentration of FeCl3, which is 0.380 M.
Fe(OH)3 dissociates in water according to the following equation:
Fe(OH)3(s) -> Fe3+(aq) + 3 OH-(aq)
The solubility product constant expression for Fe(OH)3 is:
Ksp = [Fe3+][OH-]^3
Given that Ksp = 4.0e-29 and the initial concentration of Fe3+ ions in solution is 0.380 M, the concentration of OH- ions needed to reach the point of precipitation can be calculated.
First, rearrange the Ksp expression to solve for [OH-]:
[OH-]^3 = Ksp / [Fe3+]
[OH-]^3 = (4.0e-29) / (0.380)
[OH-]^3 = 1.0526e-29
Taking the cube root of both sides:
[OH-] = ∛(1.0526e-29)
[OH-] ≈ 1.030e-10 M
Since Fe(OH)3 precipitates when the concentration of OH- ions exceeds the solubility, the pH required to start precipitation can be found by calculating the pOH of the solution and then subtracting it from the value of 14 (pH + pOH = 14).
pOH = -log10[OH-]
pOH = -log10(1.030e-10)
pOH ≈ 10.99
pH = 14 - pOH
pH ≈ 14 - 10.99
pH ≈ 3.01
Therefore, the pH required to start precipitation of Fe(OH)3 from a 0.380 M solution of FeCl3 is approximately 3.01.