Confused about solubility product

[mdk2]

I am trying to understand the following question: The molar solubility of Fe(OH)₃ in an aqueous solution was determined to be 4 x 10^-10 M. What is the value of Ksp for Fe(OH)₃ at STP?

I understand that K_sp = [Fe³⁺] [OH^-]³. I know that we are meant to look at the mole ratios of products to reactants, substituting x to get K_sp = (x)(3x)³. And I see that we are meant to then reason that x = 4 x 10^-10 M.

However, this is where I get lost: I thought that the whole point of the K_sp was to convey the relative amount of product vs reactant we have at equilibrium/saturation. If we have a very high K_sp, for example, I thought that it meant that the equilibrium favored the products, signifying high solubility (and low Ksp would favor the reactants, signifying low solubility.) If that is the case, then how can we assume that the concentration of Fe³⁺ is equal to the concentration of Fe(OH)₃ at saturation? 

If the concentration of Fe³⁺ ions, for example, is always the same as the molar solubility of Fe(OH)_3, then I don't see how the K_sp means anything at all. I would have thought that in a situation of very high K_sp, the [Fe³⁺] would be much higher than the molar solubility of Fe(OH)₃ and that in a situation of a very low K_sp, the [Fe³⁺] would be much lower. But this question seems to want us to use simple stoichiometry. 

Am I missing something fundamental regarding the meaning of K_sp and its relationship with concentrations? Thank you so much!

 

 

 

 

[studiot]

14 hours ago, mdk2 said:

I am trying to understand the following question: The molar solubility of Fe(OH)₃ in an aqueous solution was determined to be 4 x 10^-10 M. What is the value of Ksp for Fe(OH)₃ at STP?

I understand that K_sp = [Fe³⁺] [OH^-]³. I know that we are meant to look at the mole ratios of products to reactants, substituting x to get K_sp = (x)(3x)³. And I see that we are meant to then reason that x = 4 x 10^-10 M.

However, this is where I get lost: I thought that the whole point of the K_sp was to convey the relative amount of product vs reactant we have at equilibrium/saturation. If we have a very high K_sp, for example, I thought that it meant that the equilibrium favored the products, signifying high solubility (and low Ksp would favor the reactants, signifying low solubility.) If that is the case, then how can we assume that the concentration of Fe³⁺ is equal to the concentration of Fe(OH)₃ at saturation? 

If the concentration of Fe³⁺ ions, for example, is always the same as the molar solubility of Fe(OH)_3, then I don't see how the K_sp means anything at all. I would have thought that in a situation of very high K_sp, the [Fe³⁺] would be much higher than the molar solubility of Fe(OH)₃ and that in a situation of a very low K_sp, the [Fe³⁺] would be much lower. But this question seems to want us to use simple stoichiometry. 

Am I missing something fundamental regarding the meaning of K_sp and its relationship with concentrations? Thank you so much!

 

 

 

 

Since this is coursework, here are some hints.

But since it is important to understand what you are doing, please ask if there is any part you do not understand.

First it is important to understand the definition of solubility product. So ask if you do not.
Solubility products are different from other constants in that they are not a ratio of concentrations.

When a slightly soluble substance goes into solution, leaving some or most of its substance in physical contact with the solution the undissolved doesn not change in quantity.
The solution is saturated according to the normal reaction equation for solution

[math]Fe{\left( {OH} \right)_3} \Leftrightarrow F{e^{3 + }} + 3{\left( {OH} \right)^ - }[/math]

With an equilibrium constant K_eq given by  (in your case)

[math]{K_{eq}} = \frac{{\left[ {F{e^{3 + }}} \right]{{\left[ {O{H^ - }} \right]}^3}}}{{\left[ {Fe{{\left( {OH} \right)}_3}} \right]}}[/math]

The bottom part is the amount of the substance that has gone into solution.

The concentration of this is therefore the concentration of the ferric hydroxide that is in solution that you are given as 4 x 10^-10M

I have circled this value in pink.

 

So to move on we multiply both sides of this equation by the concentration of the ferric hydroxide to get the solubility product, K_sp.

[math]{K_{sp}} = \left[ {Fe{{\left( {OH} \right)}_3}} \right]{K_{eq}} = \left[ {F{e^{3 + }}} \right]{\left[ {O{H^ - }} \right]^3}[/math]

Substituting values and forming the other equations of equilibrium should enable you to complete this question.

Come back and let us know how you get on.

 

 

 

 

[Strange]

!

Moderator Note

Moved to Homework Help

 

[hypervalent_iodine]

5 hours ago, studiot said:

Since this is coursework, here are some hints.

But since it is important to understand what you are doing, please ask if there is any part you do not understand.

First it is important to understand the definition of solubility product. So ask if you do not.
Solubility products are different from other constants in that they are not a ratio of concentrations.

When a slightly soluble substance goes into solution, leaving some or most of its substance in physical contact with the solution the undissolved doesn not change in quantity.
The solution is saturated according to the normal reaction equation for solution

$Fe{\left( {OH} \right)_3} \Leftrightarrow F{e^{3 + }} + 3{\left( {OH} \right)^ - }$

With an equilibrium constant K_eq given by  (in your case)

${K_{eq}} = \frac{{\left[ {F{e^{3 + }}} \right]{{\left[ {O{H^ - }} \right]}^3}}}{{\left[ {Fe{{\left( {OH} \right)}_3}} \right]}}$

The bottom part is the amount of the substance that has gone into solution.

The concentration of this is therefore the concentration of the ferric hydroxide that is in solution that you are given as 4 x 10^-10M

I have circled this value in pink.

 

So to move on we multiply both sides of this equation by the concentration of the ferric hydroxide to get the solubility product, K_sp.

${K_{sp}} = \left[ {Fe{{\left( {OH} \right)}_3}} \right]{K_{eq}} = \left[ {F{e^{3 + }}} \right]{\left[ {O{H^ - }} \right]^3}$

Substituting values and forming the other equations of equilibrium should enable you to complete this question.

Come back and let us know how you get on.

 

 

 

 

I can see what you’re maybe getting at here, but it is confusing. The Keq would not be written like that as the equilibrium is not dependant on the concentration of the solid salt. The expression for Ksp is not actually different from how you would write other equilibrium expressions, since you typically don’t include terms for solid or liquid components in any expression for Keq. This is the same in, for example, expressions for Ka, where water is not included. The reasoning is that these components are normally in such large amounts relative to other components, that minor changes to their concentration has no effect on the overall equilibrium. This holds true for Ksp since we assume at a minimum that the solution is at saturation point; adding more salt has no effect on the concentration of ions in solution, and therefore no effect on the equilibrium. 

 

19 hours ago, mdk2 said:

If that is the case, then how can we assume that the concentration of Fe³⁺ is equal to the concentration of Fe(OH)₃ at saturation? 

The answer to this is simply stoichiometry. If x number of moles of a salt has dissolved, and the ratio of it to one of its ions is 1:1, then x number of moles of that ion is in solution. Note of course that this concentration is not the same as the amount of salt that might have been added to solution, just how much at equilibrium is actually in solution. I think you understand most of the concepts here, so I don’t think you’re necessarily missing anything. Ksp and molar solubility are closely linked - if one is low, so is the other. We might use the numbers for different things or to work out other problems. Practically, molar solubility is a more useful and easily relatable number if you just want to know how soluble something is. However, if you want to know the ion concentration in more complicated scenarios (I’m talking common ion effect) or if you want to say predict whether something will precipitate in complex solutions, you would need to use Ksp. Perhaps have a read of this: https://www.chem.purdue.edu/gchelp/howtosolveit/Equilibrium/Solubility_Products.htm

Edited April 26, 2020 by hypervalent_iodine
Clarification in the first paragraph