Calculate the concentration of Fe (3 ) (which is saturated in Fe3 ). pH=7.41 and Ksp of Fe(OH)3 = 4.0x10^-4M

Question

i tried finding the pOH and then the concentration of OH and substituting that into the equation and solving for only Fe but that doesn't seem to work. but i realized that the X that you solve for in Fe is the same X as in OH (where in this case it would be (3x)^3)

Answer 1

First calculate the OH- concentration.

pH = 7.41

[H+] = 3.89*10^-8M

10^-14 = [H+] *[OH-]

10^-14 = [OH-] *(3.89*10^-8M)

[OH-] = 2.57*10^-7M



What I did was set up a reaction table. They are real simple, so don't over look there importance.

................ Fe(OH)3(s) ..... <--> ...... Fe3+(aq) ....... + ...... OH-(aq)

Initial............... - ............................... 0 ..................... 2.57*10^-7M

Reacting.......... - ............................. +x ........................... +3x

Equilibrium....... - .............................. x .................. 2.57*10^-7 + 3x

Ksp = [Fe3+][OH-]

[OH-] = 2.57*10^-7+3x  [Fe3+] = x

Substitute the values into the equilibium expression.

4.0*10^-4 = (x)(2.57*10^-7 + 3x)^3

Foil and then simplify.

4.0*10^-4 = 27x^4 +6.940*10^-6x^3 +5.946*10^13x^2 +1.698*10^-20x  -4.0*10^-4 = 0

To solve this crazy @ss equation you need a graphing calculator. By deductive reasoning you can find the correct x value. I found my x to be.

x = .06204

x = [Fe3+] = .06204M

Answer 2

when the pH or pOH is given we don't do an "ICE" to determine the [OH-], because they gave it to us, & there is no need for any 3x's

when the pH = 7.41, ....since pH +pOH = 14,....the pOH = 6.59

when the pOH = 6.59 , ... the [OH-] = 2.57e-7 Molar

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Fe(OH)3 --> Fe+3  & 3OH-

K = [Fe+3]  OH-]^3

4.0e-4 = [Fe+3]  [2.57e-7]^3

Fe+3 = 4.0e-4 / 1.70 e-20

Fe+3 = 2.35e16 you are right, that's a pretty rediculous answer

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more than likely , it  is because almost all of the references that I checked have the Ksp closer to 2.7e-39...

I am going to re-do the calculations using the Ksp from the CRC Handbook of Chemistry & Physics; 2.64 e-39

2.64 e-39 = [Fe+3]  [2.57e-7]^3

Fe+3 = 2.64 e-39 / 1.70 e-20

Fe+3 = 1.55e-19

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alternatively what if your constant given was meant to be 4.0e-40, well then:

4.0e-40 = [Fe+3]  [2.57e-7]^3

Fe+3 = 4.0e-40 / 1.70 e-20

Fe+3 = 12.35e-20

whatever