[haber101]

Calculate the milligrams of metallic ion that can remain at equilibrium in a solution of    Fe(OH)3 having a [OH-] = 1.0 x 10-4 mol/L and Ksp = 6.0 x 10-38.

I have no clue how to start this can someone shed some light

[cliverlong]

Can you write an equilibrium equation for the solution of Fe(OH)₃ ?

http://en.wikipedia.org/wiki/Solubility_equilibrium#Ionic_compounds

What does a K_sp of 10^-38 tell you about the solubility of this material?

[haber101]

Can you write an equilibrium equation for the solution of Fe(OH)₃ ?

http://en.wikipedia.org/wiki/Solubility_equilibrium#Ionic_compounds

What does a K_sp of 10^-38 tell you about the solubility of this material?

that it's not very soluble at all

[cliverlong]

Can you write an equilibrium equation for the solution of Fe(OH)₃ ?

http://en.wikipedia.org/wiki/Solubility_equilibrium#Ionic_compounds

What does a K_sp of 10^-38 tell you about the solubility of this material?

that it's not very soluble at all

OK. So where is the equilibrium equation and the definition of K_sp that supports your assertion about solubility of Fe(OH)₃?

[AWK]

http://www.chemteam.info/Equilibrium/Common-ion-effect.html

[cliverlong]

http://www.chemteam.info/Equilibrium/Common-ion-effect.html

Why do you think common ion effect is relevant when it is not clear from the question what are the reactants/solutes or conditions?

The common ion effect is relevant from this data:

[OH-] = 1.0 x 10-4 mol/L and Ksp = 6.0 x 10^-30

[haber101]

i'm sorry, still not catching on here can someone elaborate more on how to approach this

[haber101]

Ksp= [Fe][OH]^3
6.0x10^-38=[Fe][1.0x10^-4]^3
[Fe]=6.0x10^-26 M

6.0x10^-26 x 1000mL/1L x 1mol/55.85= 1.0x10^-30

whats wrong?

[Borek]

What units is your answer in?

[haber101]

it says mg. so instead 6.0x10^-26 x 1g/1000mg x 1mol/55.85g = 1x10^-30