The solubility of AgCl (s) is 2.5 x 10^-5 M. calculate the ksp
See your post on PbCl2 above.
To calculate the solubility product constant (Ksp) of AgCl, we need to know the molar concentration of the ions formed when AgCl dissociates in water.
The balanced chemical equation for the dissociation of AgCl is:
AgCl (s) ⇌ Ag+ (aq) + Cl- (aq)
Given that the solubility of AgCl is 2.5 x 10^-5 M, we can assume this is the concentration of Ag+ and Cl- ions in the solution.
Since the dissociation results in one mole of Ag+ and one mole of Cl- for every mole of AgCl, we can write the Ksp expression as:
Ksp = [Ag+] * [Cl-]
Substituting the solubility values into the equation, we get:
Ksp = (2.5 x 10^-5 M) * (2.5 x 10^-5 M) = 6.25 x 10^-10 M^2
Therefore, the Ksp of AgCl is 6.25 x 10^-10 M^2.