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?
chemistry - DrBob222, Friday, December 7, 2012 at 10:14pm
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+.
I cant seem to get the right answer. I have no idea if I'm doing it right. So when you subtract 8.1 from 14, you get 5.9. Is that the value for OH that you plug into the pOH equation?
No, it isn't OH^-. That's the pOH. Then
pOH = -log(OH^-) and calculate OH from that, then substitute into Ksp expression and solve for Fe&3+
To find the maximum value of [Fe3+] in pH 8.1 seawater, we need to calculate the concentration of OH^- in the water.
Given:
pH = 8.1
pOH = 14 - pH = 14 - 8.1 = 5.9
Now, we know that pOH is the negative logarithm of the concentration of OH^- ions. To find the concentration of OH^-, we need to take the inverse logarithm (antilogarithm) of pOH.
OH^- = 10^(-pOH) = 10^(-5.9)
Now, we can substitute the value of OH^- into the Ksp expression for Fe(OH)3:
Ksp = [Fe^3+][OH^-]^3
Substituting the known values:
1.1 * 10^(-36) = [Fe^3+] * (10^(-5.9))^3
Simplifying:
1.1 * 10^(-36) = [Fe^3+] * 10^(-5.9 * 3)
1.1 * 10^(-36) = [Fe^3+] * 10^(-17.7)
1.1 * 10^(-36) = [Fe^3+] * (1 / 10^17.7)
1.1 * 10^(-36) = [Fe^3+] / 10^17.7
To solve for [Fe^3+], multiply both sides by 10^17.7:
1.1 * 10^(-36) * 10^17.7 = [Fe^3+]
1.1 * 10^(17.7 - 36) = [Fe^3+]
1.1 * 10^(-18.3) = [Fe^3+]
Therefore, the maximum value of [Fe^3+] in pH 8.1 seawater is 1.1 * 10^(-18.3).