The average pH of open ocean water is 8.1. What is the maximum value of [Fe3+] in pH 8.1 seawater if the Ksp of Fe(OH)3 is 1.1*10^-36?
Fe(OH)3 ==> Fe^3+ + 3OH^-
pH = 8.1
pH + pOH = pKw = 14
Therefore, pOH = 14-8.1 = ?
pOH = -log(OH^-) and
(OH^-) = ?
Ksp = (Fe^3+)(OH^-)^3
Substitute Ksp and OH^- from above into Ksp expression and solve for Fe^3+.
To determine the maximum value of [Fe3+] in pH 8.1 seawater, we need to find the equilibrium concentration of Fe(OH)3 in the water using the Ksp value.
1. First, write the balanced equation for the dissociation of Fe(OH)3:
Fe(OH)3 ⇌ Fe3+ + 3OH-
2. We can use the Ksp expression to write an equation relating the concentrations of the products and reactants at equilibrium:
Ksp = [Fe3+][OH-]^3
3. Since seawater has a pH of 8.1, it means the concentration of H+ ions is 10^(-8.1) M.
4. To find the concentration of OH- ions, we can use the autoionization of water equation:
Kw = [H+][OH-]
Kw = (10^(-8.1))(OH-)
[OH-] = Kw / (10^(-8.1))
5. The value of Kw (the ion product of water) at room temperature is 1.0 x 10^(-14).
[OH-] = (1.0 x 10^(-14)) / (10^(-8.1))
[OH-] = 1.0 x 10^(-14 + 8.1)
[OH-] = 1.0 x 10^(-5.9)
6. Now, substitute the determined concentration of OH- ions into the Ksp expression:
Ksp = [Fe3+][(1.0 x 10^(-5.9))^3]
1.1 x 10^(-36) = [Fe3+][(1.0 x 10^(-5.9))^3]
7. Calculate the maximum value of [Fe3+] by rearranging the equation:
[Fe3+] = (1.1 x 10^(-36)) / (1.0 x 10^(-5.9))^3
[Fe3+] ≈ 10^(-3.8)
Therefore, the maximum value of [Fe3+] in pH 8.1 seawater is approximately 10^(-3.8) M.