Above what Fe2+ concentration will Fe(OH)2 precipitate from a buffer solution that has a pH of 8.74 ?
The 𝐾sp of Fe(OH)2 is 4.87×10−17.
....................Fe(OH)2 ==> Fe^2+ + 2OH^-
Ksp = (Fe^2+)(OH^-)^2 = 4.87E-17
Convert pH = 8.74 to pOH and to (OH^-). Plug OH^- into the Ksp expression and solve for (Fe^2+). Post your work if you get stuck.
To determine the concentration of Fe2+ at which Fe(OH)2 will precipitate from a buffer solution with a pH of 8.74, we need to consider the equilibrium that exists between Fe(OH)2 and its ions in solution.
The chemical equation for the dissociation of Fe(OH)2 is as follows:
Fe(OH)2 ⇌ Fe2+ + 2OH-
Fe(OH)2 will precipitate when the product of the concentrations of Fe2+ and OH- ions exceeds the solubility product constant (Ksp) for Fe(OH)2.
The Ksp expression for Fe(OH)2 is:
Ksp = [Fe2+][OH-]^2
Since Fe(OH)2 is a strong base, OH- concentration can be determined from the pH of the solution using the following equation:
pOH = 14 - pH
OH- = 10^(-pOH)
In this case, the pOH can be calculated as:
pOH = 14 - 8.74
pOH = 5.26
Substituting the calculated pOH value into the equation, we can find the OH- concentration:
OH- = 10^(-5.26)
Now, we can substitute the obtained OH- concentration and the Ksp value into the Ksp expression to find the concentration of Fe2+ at which Fe(OH)2 will precipitate:
Ksp = [Fe2+][OH-]^2
4.87×10^-17 = [Fe2+](10^(-5.26))^2
Let's solve this equation:
4.87×10^-17 = [Fe2+](10^(-10.52))
[Fe2+] = (4.87×10^-17) / (10^(-10.52))
Using scientific notation, we can simplify this:
[Fe2+] = 4.87×10^-17 × 10^(10.52)
[Fe2+] = 4.87 × 10^(-17+10.52)
[Fe2+] = 4.87 × 10^(-6.48)
[Fe2+] = 4.87 × 10^(-6.48) M
Thus, above the concentration of 4.87 × 10^(-6.48) M, Fe(OH)2 will begin to precipitate from the buffer solution with a pH of 8.74.