# Ksp question...

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#### dental17

##### Full Member
10+ Year Member
The solubility of Fe(OH)3 in aqueous solution was determined to be

4.5 x 10-10 mol/L. what is the value of the Ksp for Fe(OH)3?

- The dissociation reaction is:

Fe(OH)3 (s) ..D.. Fe 3+ (aq) + 3OH- (aq)

For every mole of Fe(OH)3 that dissociate, one mole of Fe 3+ and
three moles of OH- are produced. So,

Ksp = [Fe 3+] [OH-]

[OH-] = 3 [Fe 3+]; [Fe 3+] = 4.5 x 10-10 M

Ksp = [Fe 3+] (3 [Fe 3+])3 = 27[Fe 3+] 4

Ksp = (4.5 x 10-10)[3(4.5 x 10-10 )]3 =27 (4.5 x 10-10 )4

Ksp = 1.1 x 10-36

My question: Why is there a 3 in front the boldfaced [Fe 3+] and why does

[Fe 3+] = 4.5 x 10-10 M (I thought that was the solubility for Fe(OH)2)? I checked out the other thread that specifically asks this question but still don't get it. #### rockclock

##### Full Member
10+ Year Member
The solubility of Fe(OH)3 in aqueous solution was determined to be

4.5 x 10-10 mol/L. what is the value of the Ksp for Fe(OH)3?

- The dissociation reaction is:

Fe(OH)3 (s) ..D.. Fe 3+ (aq) + 3OH- (aq)

For every mole of Fe(OH)3 that dissociate, one mole of Fe 3+ and
three moles of OH- are produced. So,

Ksp = [Fe 3+] [OH-]

[OH-] = 3 [Fe 3+]; [Fe 3+] = 4.5 x 10-10 M

Ksp = [Fe 3+] (3 [Fe 3+])3 = 27[Fe 3+] 4

Ksp = (4.5 x 10-10)[3(4.5 x 10-10 )]3 =27 (4.5 x 10-10 )4

Ksp = 1.1 x 10-36

My question: Why is there a "3" in front the boldfaced [Fe 3+] and why does

[Fe 3+] = 4.5 x 10-10 M (I thought that was the solubility for Fe(OH)2)? I checked out the other thread that specifically asks this question but still don't get it. [OH-] = 3[Fe 3+]

...because for every mole of Fe(OH)3 that decomposes you get one mole of Fe3+ and three moles of OH-, according to that balanced equation. If you start out with no product ([OH-] = 0, [Fe3+] = 0) then you're going to always have a 3:1 ratio between those two as molecules of the reactant decompose.

The "solubility of Fe(OH)3" means that's how much of the reactant decomposed. Again, look at the stoichiometric ratios...1:1:3...they're telling you what that first 1 is, and based on the ratio you can fill in the rest.

#### dental17

##### Full Member
10+ Year Member
Okay, I think I kind of understand the ratio part: so the concentration of Fe3+ is 3 times more than the concentration of OH- at any given point. Right? But then why do we have to raise it to the 3rd power when solving for Ksp (Ksp = [Fe 3+] (3 [Fe 3+])3 = 27[Fe 3+] 4)?

Also does the first 1 in the stoichiometric ratio (1:1:3) you mentioned refer to Fe(OH)3? If yes, then why write it as [Fe3+]= 4.5 x 10-10 M and not [Fe(OH)3]= 4.5 x 10-10 M? Thanks!

[OH-] = 3[Fe 3+]

...because for every mole of Fe(OH)3 that decomposes you get one mole of Fe3+ and three moles of OH-, according to that balanced equation. If you start out with no product ([OH-] = 0, [Fe3+] = 0) then you're going to always have a 3:1 ratio between those two as molecules of the reactant decompose.

The "solubility of Fe(OH)3" means that's how much of the reactant decomposed. Again, look at the stoichiometric ratios...1:1:3...they're telling you what that first 1 is, and based on the ratio you can fill in the rest.

#### rockclock

##### Full Member
10+ Year Member
Okay, I think I kind of understand the ratio part: so the concentration of Fe3+ is 3 times more than the concentration of OH- at any given point. Right? But then why do we have to raise it to the 3rd power when solving for Ksp (Ksp = [Fe 3+] (3 [Fe 3+])3 = 27[Fe 3+] 4)?

Also does the first 1 in the stoichiometric ratio (1:1:3) you mentioned refer to Fe(OH)3? If yes, then why write it as [Fe3+]= 4.5 x 10-10 M and not [Fe(OH)3]= 4.5 x 10-10 M? Thanks!

Wrong. Other way around...[OH-] is three times more than [Fe3+] at any given point. Each Fe(OH)3 splits off into three OH- and one Fe3+.

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

So

Ksp = [Fe3+][OH-]^3

That's how equilibrium constants work...the coefficients become the exponents. You also know that [OH-] = 3[Fe3+] from above, so

Ksp = [Fe3+][OH-]^3
Ksp = [Fe3+](3[Fe3+])^3 = 27[Fe3+]^4

Now on the "solubility" part...the problems tells us that 4.5e-10 M of Fe(OH)3 reacted. We don't care how much was there initially or what is left because it is a solid, and pure solids and liquids don't show up in the Ksp expression.

4.5e-10 M of Fe(OH)3 reacting...
...produces 4.5e-10 M of Fe3+
...produces 3(4.5e-10 M) of OH- = 1.35e-9 M of OH-

Plug those values into either Ksp equation:

Ksp = [Fe3+][OH-]^3
Ksp = [4.5e-10 M][1.35e-9 M]^3

Ksp = 27[Fe3+]^4
Ksp = 27[4.5e-10 M]^4

Both will get you the same right answer.

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#### dental17

##### Full Member
10+ Year Member
Oops, I meant OH- was 3 times more than Fe3+, sorry for the switch up. Anyway, this makes a lot more sense now. Thank you so much, rockclock!

Wrong. Other way around...[OH-] is three times more than [Fe3+] at any given point. Each Fe(OH)3 splits off into three OH- and one Fe3+.

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

So

Ksp = [Fe3+][OH-]^3

That's how equilibrium constants work...the coefficients become the exponents. You also know that [OH-] = 3[Fe3+] from above, so

Ksp = [Fe3+][OH-]^3
Ksp = [Fe3+](3[Fe3+])^3 = 27[Fe3+]^4

Now on the "solubility" part...the problems tells us that 4.5e-10 M of Fe(OH)3 reacted. We don't care how much was there initially or what is left because it is a solid, and pure solids and liquids don't show up in the Ksp expression.

4.5e-10 M of Fe(OH)3 reacting...
...produces 4.5e-10 M of Fe3+
...produces 3(4.5e-10 M) of OH- = 1.35e-9 M of OH-

Plug those values into either Ksp equation:

Ksp = [Fe3+][OH-]^3
Ksp = [4.5e-10 M][1.35e-9 M]^3

Ksp = 27[Fe3+]^4
Ksp = 27[4.5e-10 M]^4

Both will get you the same right answer.

#### rockclock

##### Full Member
10+ Year Member
Oops, I meant OH- was 3 times more than Fe3+, sorry for the switch up. Anyway, this makes a lot more sense now. Thank you so much, rockclock!

you got it, good luck #### JustwantDDS

##### DrAMG
10+ Year Member
wait a minute- so when it says the molar solubility was determined to be X, thats how much is reacted??? i bolded the part im confused about!
Wrong. Other way around...[OH-] is three times more than [Fe3+] at any given point. Each Fe(OH)3 splits off into three OH- and one Fe3+.

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

So

Ksp = [Fe3+][OH-]^3

That's how equilibrium constants work...the coefficients become the exponents. You also know that [OH-] = 3[Fe3+] from above, so

Ksp = [Fe3+][OH-]^3
Ksp = [Fe3+](3[Fe3+])^3 = 27[Fe3+]^4

Now on the "solubility" part...the problems tells us that 4.5e-10 M of Fe(OH)3 reacted. We don't care how much was there initially or what is left because it is a solid, and pure solids and liquids don't show up in the Ksp expression.

4.5e-10 M of Fe(OH)3 reacting...
...produces 4.5e-10 M of Fe3+
...produces 3(4.5e-10 M) of OH- = 1.35e-9 M of OH-

Plug those values into either Ksp equation:

Ksp = [Fe3+][OH-]^3
Ksp = [4.5e-10 M][1.35e-9 M]^3

Ksp = 27[Fe3+]^4
Ksp = 27[4.5e-10 M]^4

Both will get you the same right answer.

#### rockclock

##### Full Member
10+ Year Member
wait a minute- so when it says the molar solubility was determined to be X, thats how much is reacted??? i bolded the part im confused about!

That's what "solubility" is...the original reactant is a solid i.e. it's not in solution and hasn't yet dissolved into solvent.

Whatever portion of that reactant (here, Fe(OH)3) that is soluble in whatever amount of solvent you have will "react" and dissolve into solution by splitting up into the component ions (here, Fe3+ and OH-).

Based on how much reactant dissolved, you can tell how much of each product (each ion) was produced using the stoichiometric ratio (here, 1:1:3). Then, those values get plugged into the Ksp expression.

Like I said before, you don't care about how much Fe(OH)3 there was to start with or how much remains undissolved because it is a solid and, as such, doesn't appear in the Ksp expression. All you care about is how much reacted...and that's for the purposes of figuring out how much of each product you have in solution.

#### JustwantDDS

##### DrAMG
10+ Year Member
ohhh i get it
sorry i got thrown off by the word react
thanks!!!!