I dont understand how solubility product (Ksp) work?

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The discussion centers on the solubility product constant (Ksp) of AgCl, which is 2 x 10^-10, leading to a saturated solution where the concentrations of Ag+ and Cl- ions are both 1.41 x 10^-5 mol/dm^3. When additional Ag+ ions are introduced, the ionic product exceeds the Ksp, prompting precipitation of AgCl. The confusion arises regarding the decrease in Cl- concentration from 1.41 x 10^-5 to 1 x 10^-6 mol/dm^3, despite the Ag+ concentration remaining high. This decrease in Cl- concentration is necessary to maintain the equilibrium defined by the Ksp, as the precipitation of AgCl consumes Cl- ions. Thus, the concentration of Ag+ does not need to decrease because the system adjusts by reducing Cl- to achieve a new equilibrium.
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Suppose the Ksp value of AgCl is 2 x 10^-10.

Lets say we have a saturated solution of AgCl, i.e concentration of Ag+ ions = concentration of Cl+ions = (2 x 10^-10)^0.5=1.41 x 10^-5

Ok so what i know is that if we add more Ag+ ions in the solution, precipitation of AgCl would occur since ionic product is greater than Ksp.

So let's say we add Ag+ ions such that the concentration of Ag+ ions is now 2x10^-4 moldm^-3.

Here is the part i don't understand.

The remaining chloride concentration would be 2x10^-10 divided by 2x10^-4 mol = 1x10^-6 moldm^-3. My question is where did the chloride ions go? i know it has precipitated but it doesn't make sense since the concentration of Ag+ ions did not decrease?
 
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Why would the concentration of Ag+ need to decrease? You have shown the Cl- concentration decreases from 1.41X10-5 to 1X10-6 so to maintain the Ksp product, Ag+1 must be higher.
 
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