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Solubility Product

by MohammedRady97
Tags: product, solubility
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MohammedRady97
#1
Jul15-14, 01:08 PM
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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
Jul16-14, 05:44 PM
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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.
Borek
#3
Jul28-14, 04:28 AM
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Quote Quote by MohammedRady97 View Post
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.


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