Calculating Silver Chloride Solubility at 25℃

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SUMMARY

The solubility of silver chloride (AgCl) at 25℃ is calculated using its solubility product constant (Ksp), which is 1.8x10^-10. The dissociation of AgCl into Ag+ and Cl- ions is represented by the equation AgCl ⇔ Ag+ + Cl-. By applying the Ksp expression, Ksp = [Ag][Cl], and substituting [Ag] = [Cl], the molar solubility of AgCl is determined to be 1.3x10^-5 M.

PREREQUISITES
  • Understanding of solubility product constant (Ksp)
  • Knowledge of chemical equilibrium principles
  • Familiarity with molarity calculations
  • Basic concepts of ionic dissociation
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  • Study the concept of solubility product constants (Ksp) for other salts
  • Learn about the common ion effect on solubility
  • Explore the relationship between temperature and solubility
  • Investigate the calculation of molar solubility for various ionic compounds
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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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