[Needaask]

When something dissolves in water say AB A+ +B-  . So when I add in a little AB into a beaker of water. It would dissociate.

However, isn't Ksp a constant? So if a little is added how can the concentrations of A+ and B- reach Ksp? Or is it because Ksp must be used for a saturated solution?

Also, how does the equilibrium tell us when no more AB can be added? Because when I add in more AB, the equilibrium doesn't get shifted anywhere, so why would more AB dissolve or stay as a solid actually? Because even when the solution is saturated, adding solid AB doesn't affect the equilibrium in any way. So I don't get why it would either continue to dissolve or stop dissolving. I mean even when Ksp isn't met, adding more Solid AB wouldn't affect the equilibrium in any way. So why would the equilibrium shift to the right? And when Ksp is met, when more solid AB is added what's stopping it from dissolving further by just using the equilibrium equation?

These are the main problems I face when reading up on the solubility product. Hope you guys can help

[Corribus]

So if a little is added how can the concentrations of A+ and B- reach Ksp?

It won't.  Equilibrium is only reached at the point of saturation.

Also, how does the equilibrium tell us when no more AB can be added? Because when I add in more AB, the equilibrium doesn't get shifted anywhere, so why would more AB dissolve actually?

It won't.  Once you equilibrate at the saturation point, adding more AB won't shift the equilibrium.

[Needaask]

So if a little is added how can the concentrations of A+ and B- reach Ksp?

It won't.  Equilibrium is only reached at the point of saturation.

Also, how does the equilibrium tell us when no more AB can be added? Because when I add in more AB, the equilibrium doesn't get shifted anywhere, so why would more AB dissolve actually?

It won't.  Once you equilibrate at the saturation point, adding more AB won't shift the equilibrium.

But actually why won't it? Because adding in more solid wouldn't shift the equilibrium anywhere. So why should it not precipitate anymore?

Thanks Corribus

[Corribus]

That's a circular post.  Either that or you need to frame your question more clearly.

Let's take a real example rather than speaking in abstractions.

The solubility product of silver chloride in 25 C water is 1.77 × 10⁻¹⁰.

This is for the process AgCl(s) Ag⁺(aq) + Cl^-(aq)

What this practically means is that at saturation the concentrations of both the Ag⁺ and Cl^- ions will be 1.33 x 10^-5 mol/L.  The stoichiometric equivalent of solid AgCl that would give rise to this concentration of dissociated ions in 1 L of water is about 1.9 mg.  If you add less than 1.9 mg of solid AgCl to 1 L of water, it will all dissolve in stoichoimetric ratio, but none of the ionic concentrations will obviously reach the "maximum" level specified by the solubility product equilibrium constant.  If you add more than 1.9 mg of solid AgCl to 1 L, the solution will saturate, meaning 1.33 x 10^-5 mol/L of Ag⁺ and Cl^- ions will form in solution from 1.9 mg of the AgCl.  Any solid AgCl above the 1.9 mg will remain as a solid and settle to the bottom of the flask.  If you add more solid AgCl later, it, too, will settle to the bottom of the flask.

However, this still is an equilibrium constant, so the usual rules apply.  If you add sodium chloride to a saturated solution, this will shift the equilibrium to the left, because now there are more chloride ions available to "react" with the available silver ions.  If you raise the temperature, the equilibrium will shift to the right.  If you change the particle size of the solid to very small (talking nano-regime), this will also impact the equilibrium, although the reason, and the effect, is complicated.

And like all equilibrium situations, the value of the solubility product under any condition is related to thermodynamics - the respective enthalpic and entropic changes (combined as the Gibbs energy change) that occur during dissolution.  Unfortunately predicting solubility of ionic salts from thermodynamic considerations is not always easy due to the number of factors involved, so solubility remains a largely empirical phenomenon.

Here is some more about solubility products:

http://www.chemguide.co.uk/physical/kspmenu.html#top

[Needaask]

That's a circular post.  Either that or you need to frame your question more clearly.

Let's take a real example rather than speaking in abstractions.

The solubility product of silver chloride in 25 C water is 1.77 × 10⁻¹⁰.

This is for the process AgCl(s) Ag⁺(aq) + Cl^-(aq)

What this practically means is that at saturation the concentrations of both the Ag⁺ and Cl^- ions will be 1.33 x 10^-5 mol/L.  The stoichiometric equivalent of solid AgCl that would give rise to this concentration of dissociated ions in 1 L of water is about 1.9 mg.  If you add less than 1.9 mg of solid AgCl to 1 L of water, it will all dissolve in stoichoimetric ratio, but none of the ionic concentrations will obviously reach the "maximum" level specified by the solubility product equilibrium constant.  If you add more than 1.9 mg of solid AgCl to 1 L, the solution will saturate, meaning 1.33 x 10^-5 mol/L of Ag⁺ and Cl^- ions will form in solution from 1.9 mg of the AgCl.  Any solid AgCl above the 1.9 mg will remain as a solid and settle to the bottom of the flask.  If you add more solid AgCl later, it, too, will settle to the bottom of the flask.

However, this still is an equilibrium constant, so the usual rules apply.  If you add sodium chloride to a saturated solution, this will shift the equilibrium to the left, because now there are more chloride ions available to "react" with the available silver ions.  If you raise the temperature, the equilibrium will shift to the right.  If you change the particle size of the solid to very small (talking nano-regime), this will also impact the equilibrium, although the reason, and the effect, is complicated.

And like all equilibrium situations, the value of the solubility product under any condition is related to thermodynamics - the respective enthalpic and entropic changes (combined as the Gibbs energy change) that occur during dissolution.  Unfortunately predicting solubility of ionic salts from thermodynamic considerations is not always easy due to the number of factors involved, so solubility remains a largely empirical phenomenon.

Here is some more about solubility products:

http://www.chemguide.co.uk/physical/kspmenu.html#top

But actually, the equation doesn't tell us about the Ksp right?

Also, why would the solver chloride continue to dissolve if it isn't saturated? Without the equation I would agree that if it wasn't saturated, Eer AgCl can be dissolved so it would. But when I saw the equation, it doesn't really tell me if more can be dissolved or not actually. Cos AgCl is a solid, so it doesn't push the equilibrium to the right or left.

Thanks again for the help

[Corribus]

K_sp tells us effectively what the maximum amount of dissolved ions will be (for a given temperature).  This amount is determined by the thermodynamics of dissolution.

Adding more of the solute won't shift the equilibrium once equilibrium has been reached.  That is, after the solution is saturated.  Until that point, the system is not at equilibrium, and so adding more of the solid will continue to produce additional dissolved species.

[Needaask]

K_sp tells us effectively what the maximum amount of dissolved ions will be (for a given temperature).  This amount is determined by the thermodynamics of dissolution.

Adding more of the solute won't shift the equilibrium once equilibrium has been reached.  That is, after the solution is saturated.  Until that point, the system is not at equilibrium, and so adding more of the solid will continue to produce additional dissolved species.

But what about the equation? Because from the equation I can't really tell if more is going to dissolve or not.

Thanks

[Corribus]

I'm afraid I don't really understand what you're asking.  Please be specific.

[Needaask]

I'm afraid I don't really understand what you're asking.  Please be specific.

Oops sorry bout that.

I was thinking about AgCl(s) Ag⁺ (aq)+Cl^- (aq). Because I don't really understand why no more AgCl would be reacted once Ksp is reached. Cos when the solid silver chloride is added, there isn't any change to the equilibrium. So if 1) the solution isn't saturated why would the equilibrium shift to the right 2) the solution is already saturated why would it remain as a solid?

The explanation of the Ksp telling us about when it can continue to dissolve makes sense. But when i simply use the chemical equation I can't seem to relate them in anyway actually.

Hope this makes sense thanks!

[Corribus]

So if 1) the solution isn't saturated why would the equilibrium shift to the right

The point is that there is no equilibrium yet if the solution isn't saturated.  So this question doesn't make a whole lot of sense.

Here's an analogy that maybe helps a little bit.  Instead of a solubility, think a conventional reaction A + B C

I think you'll agree that if I put both A and B into a reaction vessel, the two will react until they produce enough C that an equilibrium is established - where the rate of A + B C balances the rate of C A + B.

Suppose now I just put A in.  Can any equilibrium be reached?  It can't.  It'll just sit there until I put some of B in.  Once I do put some B in, the reaction will finally start to go toward equilibrium. 

In short, if it's not possible to reach equilibrium, the reaction will just sit at a non-equilibrium position until the barrier to equilibrium is moved.

2) the solution is already saturated why would it remain as a solid?

Because now the reaction is at equilibrium.  If you add more solid at this point, some more dissociation may occur, but there will almost immediately be a restoring force to return the mixture to equilibrium - which is the point of maximum solubility.

Substances dissolve because it is thermodynamically favorable to do so.  The point of maximum solubility is the point of minimum potential energy of the system.  Dissolve less and the there is more potential energy in the undissolved solute.  Dissolve more and there is more potential energy stored in the solution.  The maximum solubility, defined by the Ksp, is the point where all the forces balance and you have the best compromise between the energy in the solid state and the energy of dissolved ions interacting with each other.