The solubility products of Fe(OH)2 and Fe(OH)3 are 10^-17 and 10-38 respectively. If the concentrations of Fe2+ and Fe3+ are each 10^-5, at what pH will each hydroxide just begin to precipitate?
Here is how you do the Fe(OH)2. Fe(OH)3 is done the same way.
.......................Fe(OH)2 ==> Fe^2+ + 2OH^-
I.....................solid................0............0
C...................solid.................x...........2x
E...................solid.................x...........2x
Also, the problem tells you that x = (Fe^2+) = 1E-5. Substitute that into the Ksp expression and solve for OH^-, then convert that to pH.
Ksp = 1E-17 = (Fe^2+)(OH^-)^2
Post your work if you get stuck.
To determine the pH at which each hydroxide will begin to precipitate, we need to identify the conditions under which the solubility product of each hydroxide is exceeded.
Let's start with Fe(OH)2. The solubility product expression for Fe(OH)2 is:
Ksp = [Fe2+][OH-]^2
Given that the concentration of Fe2+ is 10^-5, and assuming the concentration of OH- is x (since Fe(OH)2 is a weak base and dissociates into Fe2+ and OH-), we can write:
Ksp = (10^-5)(x)^2
The solubility product constant (Ksp) for Fe(OH)2 is given as 10^-17. Therefore, we have:
10^-17 = (10^-5)(x)^2
Simplifying the equation, we get:
x^2 = 10^-12
Taking the square root of both sides, we find:
x = 10^-6
This concentration of OH- corresponds to the hydroxide precipitating. Since OH- is related to pH by the equation: OH- = 10^-pOH, we can calculate the pOH first and then convert it to pH:
pOH = -log10(10^-6)
pOH = 6
pH = 14 - pOH
pH = 14 - 6
pH = 8
Therefore, Fe(OH)2 will begin to precipitate at pH 8.
Now let's move on to Fe(OH)3. The solubility product expression for Fe(OH)3 is:
Ksp = [Fe3+][OH-]^3
Given that the concentration of Fe3+ is 10^-5, and assuming the concentration of OH- is y (since Fe(OH)3 is a weak base and dissociates into Fe3+ and 3OH-), we can write:
Ksp = (10^-5)(y)^3
The solubility product constant (Ksp) for Fe(OH)3 is given as 10^-38. Therefore, we have:
10^-38 = (10^-5)(y)^3
Simplifying the equation, we get:
(y)^3 = 10^-33
Taking the cube root of both sides, we find:
y = (10^-33)^(1/3)
y = 10^-11
Using the equation OH- = 10^-pOH, we can calculate the pOH and then convert it to pH:
pOH = -log10(10^-11)
pOH = 11
pH = 14 - pOH
pH = 14 - 11
pH = 3
Therefore, Fe(OH)3 will begin to precipitate at pH 3.