[aznking1]

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

Here is the part i dont 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 doesnt make sense since the concentration of Ag+ ions did not decrease?

[Sircodekill]

Well, first of all your Ag⁺ precipitated and then your solution went to 2*10^-4M.

You cant just add Ag⁺ ion to the solution, so free ions doesn't exist. You have another equilibrium like

AgNO₃  Ag⁺ + NO₃^-.

Kps gives you the values on a stationary state, when the equilibrium is reached. So your Ag⁺ actually decreased when you added AgNO₃ (for example) trough time, but this is not reflected on this equations.

[Rutherford]

If you add for example 1*10^-5M of AgNO₃, half of the Ag⁺ ions from AgNO3 will precipitate with Cl^- and the other half remains in the solution.

[Borek]

i know it has precipitated but it doesnt make sense since the concentration of Ag+ ions did not decrease?

Concentration did not decrease, but concentration is not the only thing that matters. Amount has changed, because you added some silver to the solution. And initial total amount (silver that was present in the solution plus silver that was added) equals final total amount (silver left in the solution and silver in the precipitate).