Question concerning solubility equilibriums for salts

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In this discussion, the focus is on the solubility product constant (Ksp) of a salt with low solubility and the implications of its concentration in solution. The key point is that in the expression for Ksp, the concentration of the undissolved salt ([AB]) is often treated as constant and can be considered equal to 1 because it represents the activity of the solid phase rather than its concentration in solution. This means that when a weakly soluble salt is added to water, the concentration of the dissolved ions ([A+] and [B-]) is what matters for calculating Ksp, while the undissolved salt remains in a separate phase and does not contribute to the concentration in the solution. This concept is illustrated through the analogy of a large crystal of salt acting as a wall, where the concentration of the undissolved substance in the solution is effectively zero, but the ions can still enter the solution from the solid phase.
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Hi. Let's say we have a salt AB with a very low Ksp. A bunch of it gets thrown into the water and mixed.

So:

K=[A+][B-]/[AB].

Ksp=[A+][B-]

Why is it OK to remove [AB], then? ie why does [AB] equal 1? I thought [AB] was concentration of the substance in the water, but it is the substance's activity in the water, according to wikipedia?

Please explain this so a high-schooler can understand.
 
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[AB] would be concentration of the substance, but the substance is not dissolved.

Imagine you have a single large crystal of a weakly soluble salt, that is a wall of the vessel in which you are dissolving the substance. Concentration of the undissociated substance in the solution is zero, however, it is in the contact with water, so the ions can can be leached into the solution. When you have smaller crystals situation is identical, they form a separate phase.
 
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