Calculating Silver Chloride Solubility at 25℃

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The solubility of silver chloride (AgCl) at 25℃ can be calculated using its solubility product constant (Ksp), which is 1.8x10^-10. The dissociation of AgCl into Ag+ and Cl- ions leads to the equation Ksp = [Ag][Cl]. By assuming the concentrations of Ag+ and Cl- are equal, the calculation yields [Ag] = [Cl] = √(1.8x10^-10), resulting in a molar solubility of 1.3x10^-5 M for AgCl. The final conclusion is that the molar solubility of silver chloride at this temperature is 1.3x10^-5 M.
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Homework Statement


Calculate the solubility of silver chloride at 25℃.

Ksp of silver chloride at 25℃=1.8x10^-10

Homework Equations


AgCl⇔Ag+Cl (I believe)
So, Ksp=[Ag][Cl]

The Attempt at a Solution


All I know is the Ksp of silver chloride at that temperature. I have no idea what to do from here. Do I need to know something about Ag or Cl?
 
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I attempted..
Ksp=1.8x10^-10
1.8x10^-10=[Ag][Cl]
[Ag]=[Cl]=√1.8x10^-10
=1.3x10^-5M

So I got the solubility for both elements.. do I just add since AgCl⇔Ag+Cl?
 
Okay, nevermind. 1.3x10^-5M is AgCl's molar solubility..
 
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