Ksp: Not Defined for Soluble Salts? Equilibrium Impact

In summary, Ksp is defined for all salts, but it may not be practical to use for highly soluble salts due to thermodynamic effects. Additionally, adding more solid to a saturated solution does not change the concentration or equilibrium, but rather creates more opportunity for precipitation and redissolution at the solid-solution interface according to Le Chatelier's Principle.
  • #1
PFuser1232
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Why is Ksp not defined for soluble salts? Also, when an equilibrium is established between the solid, undissolved salt and the ions in the saturated solution, won't adding more solid shift the equilibrium to the left causing more ions to form?
 
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  • #2
Le Chatelier's Principle:
When a system at equilibrium is subjected to change in concentration, temperature, volume, or pressure, then the equilibrium readjusts itself to counteract the effect of the applied change and a new equilibrium is established.
Of course, when the solution is saturated, adding salt isn't going to change the concentration. The salt just piles up with the rest at the bottom of the beaker. The equilibrium of the solution has not been changed.

If you look at the interface between the solid and solution, you have created more opportunity for both the solution to precipitate and for precipitate to redissolve.
 
  • #3
MohammedRady97 said:
Why is Ksp not defined for soluble salts?

It is perfectly valid for every salt.

Please remember that "soluble" and "insoluble" is not a precise classification. Every salt is in a way soluble - some are better soluble, some are less soluble, and in each case we can describe the solubility with the Ksp.

In practice it doesn't make much sense to use Ksp for highly soluble salts, because of thermodynamic effects which make calculations difficult (high ionic strength of the solution means we have problems evaluating activity coefficients). But the approach is still perfectly valid.
 

What is Ksp and why is it not defined for soluble salts?

Ksp stands for solubility product constant and is a measure of the equilibrium between a solid compound and its dissolved ions. It is not defined for soluble salts because these salts are completely dissolved in water, meaning there is no equilibrium between the solid and the dissolved ions.

How does the lack of a defined Ksp for soluble salts impact equilibrium?

Since there is no Ksp for soluble salts, there is no equilibrium to measure. This means that the concentration of the dissolved ions is not limited by the solubility of the salt, and can continue to increase until it reaches the maximum concentration that water can hold.

Can we still calculate the solubility of a soluble salt without a defined Ksp?

No, we cannot calculate the solubility of a soluble salt without a defined Ksp. Ksp is an essential component in the calculation of solubility, and without it, we cannot determine how much of the salt will dissolve in water.

Are there any exceptions to the lack of a defined Ksp for soluble salts?

In some cases, a soluble salt may have a very low solubility in water, meaning that there is a very small amount of solid that remains in equilibrium with the dissolved ions. In these cases, a Ksp can be determined, but it is not a true measurement of solubility as it is for insoluble salts.

How does the lack of a defined Ksp for soluble salts affect the use of Ksp in predicting precipitate formation?

Since the Ksp for soluble salts is not defined, it cannot be used to predict precipitate formation. Instead, the solubility of the salt must be determined experimentally to determine if a precipitate will form.

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