If Ksp for most substances is so negative, then how come we dont see solid in solutions?

[manohman]

So for most substances, the solubility of a substance lies far to the left, being very very close to zero.
This means that for a reaction like:

AB(s) -><----- A+(aq) + B-(aq)

The equilibrium favors the solid much more than the dissolved form with the ions.

If you add enough ions to make the reaction quotient (Q) > Ksp, then the reaction shifts towards the products and you get precipitate (common ion effect).

But if the reaction already lies so far to the left, where the heck is the solid from before we added more ions? Is it dissolved but not dissociated? What's going on?!

---

Members don't see this ad.

 

[Teleologist]

I don't think an equilibrium constant expression is ever going to be negative, considering that concentrations cannot be negative.

 

[chemtopper]

So for most substances, the solubility of a substance lies far to the left, being very negative.
This means that for a reaction like:

AB(s) -><----- A+(aq) + B-(aq)

The equilibrium favors the solid much more than the dissolved form with the ions.

If you add enough ions to make the reaction quotient (Q) > Ksp, then the reaction shifts towards the products and you get precipitate (common ion effect).

But if the reaction already lies so far to the left, where the heck is the solid from before we added more ions? Is it dissolved but not dissociated? What's going on?!

We need some correction :Concentrations are less than one or in 10^negative number but they are never negative.
Now if you have precipitates already in the solution then it means it is already at equilibrium and Ksp = Q
Now if you add more ions in the solution then it becomes Q again because equilibrium is disturbed by adding extra ions .Equilibrium will shift in the direction of solid means more precipitates are formed and new equilibrium will set up.So more solid is formed from the ions which are added from outside.

 

[manohman]

We need some correction :Concentrations are less than one or in 10^negative number but they are never negative.
Now if you have precipitates already in the solution then it means it is already at equilibrium and Ksp = Q
Now if you add more ions in the solution then it becomes Q again because equilibrium is disturbed by adding extra ions .Equilibrium will shift in the direction of solid means more precipitates are formed and new equilibrium will set up.So more solid is formed from the ions which are added from outside.

sorry i meant close to zero! mixed up the 10^-15 power for the constant!

so before you add more solid to shift the equilibrium to the left, there is already some solid sitting there?

 

[chemtopper]

sorry i meant close to zero! mixed up the 10^-15 power for the constant!

so before you add more solid to shift the equilibrium to the left, there is already some solid sitting there?

Yes already solid is there and when you added more ions in the solution above the solid -this results in formation of more solid.

 

[manohman]

Yes already solid is there and when you added more ions in the solution above the solid -this results in formation of more solid.

I see. Sorry I know that seems so obvious when asked like that haha.

What confused me is that we gauge the formation of precipitate by the appearance of a visible solid in solution so it makes it seem like there is no solid in solution. But in reality there is, jut not dissociated into its ions, right?
What exactly does precipitation mean on a physical level then? Besides Q>Ksp, meaning the reaction is pushed to the reactant side. Formation of visible solid?