Solubility Algebra Question HELP!!!

[NMChick]

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What is the concentration of [Fe3+] in a saturated solution of Fe(OH)3 given that Ksp(Fe(OH)3) = 2.64 x 10^-39?

Solution:
Fe(OH)3--> Fe3+ + 3OH-

Ksp= [Fe3+][OH-]^3
= [x] [3x]^3
=27x^4
2.64 x 10^-39 = 27x^4
(2.64/27) x 10^-39= x^4

I'm stuck here!!! Please explain why the [OH] is raised to the third power in Equilibrium constant AND how the heck do I solve for x when x is raised to the 4th power??

 

[plzNOCarribbean]

What is the concentration of [Fe3+] in a saturated solution of Fe(OH)3 given that Ksp(Fe(OH)3) = 2.64 x 10^-39?

Solution:
Fe(OH)3--> Fe3+ + 3OH-

Ksp= [Fe3+][OH-]^3
= [x] [3x]^3
=27x^4
2.64 x 10^-39 = 27x^4
(2.64/27) x 10^-39= x^4

I'm stuck here!!! Please explain why the [OH] is raised to the third power in Equilibrium constant AND how the heck do I solve for x when x is raised to the 4th power??

yes, your setup is correct. Since Fe (OH)3 is a MX3 salt, the Ksp value should be set to 27 x ^4. Now, if you move the decimal to the right, we have 26.7 x 10^-40. <<< this makes it easy to solve for the molar solubility. Simply divide 26.7/ 27, which is almost 1. and then take the fourth root of 10^-40, which equals 10 ^-10.

So, the answer choice should be around 1 x 10 ^-10.

 

[NMChick]

How do you "take the fourth root of 10^-40 though? This is what is confusing me. Can you explain this a different way? My algebra apparently needs some work.

 

[Temperature101]

What is the concentration of [Fe3+] in a saturated solution of Fe(OH)3 given that Ksp(Fe(OH)3) = 2.64 x 10^-39?

Solution:
Fe(OH)3--> Fe3+ + 3OH-

Ksp= [Fe3+][OH-]^3
= [x] [3x]^3
=27x^4
2.64 x 10^-39 = 27x^4
(2.64/27) x 10^-39= x^4

I'm stuck here!!! Please explain why the [OH] is raised to the third power in Equilibrium constant AND how the heck do I solve for x when x is raised to the 4th power??

You have the dissociation reaction
Fe(OH)3--> Fe3+ + 3OH-
[Fe3+] = x
[3OH-] = 3x
The Ksp is the product of the expression raised to the power of the coefficients of the expression. For instance, if you have Fe2O3, the Ksp expression will be:
[Fe2+]=2x and [3O-]= 3x and the Ksp = [2x]^2*[3x]^3=4x^2*27x^3= 108x^5...you see where I am going.
I dont think mcat would give that kind of heavy calculation.

 

[plzNOCarribbean]

How do you "take the fourth root of 10^-40 though? This is what is confusing me. Can you explain this a different way? My algebra apparently needs some work.

my algebra sucks...but Im pretty sure that when you get a setup like this, whatever exponent that you have should be divisible by a number that it can be cleanly divided by, which is why we moved the decimal to the right 1 place so that we could change 10^-39 to 10 ^-40. That way, we can divide 40 by 4 (since this is the root we will be using), since our exponent in the Ksp was raised to the 4th power.

Thats why if you were given a problem where the Ksp was equal to 4x^3 and you had to se this to say 12 x 10^-8, you would move the decimal 1 pace to the right to get 120 x 10 ^-9, so that you could take the cubed root of 10^9, which would give you 10 ^-3... You are simply dividing

 

[NMChick]

GOT IT!!! Thanks Temperature101 and plzNOCarribbean. That was very helpful!!