Given that solubility product of AgCl = 1.8 X 10^-10
Dissociation of AgCl can be represented as follows,
AgCl(s) ↔ Ag+(ag) + Cl-(aq)
Let, [Ag+] = [Cl-] = S
∴Ksp = [Ag+][Cl-] = S^2
∴ S = √Ksp = √(1.8 X 10^-10) = 1.34 x 10^-5 mol/dm3
Now, Molarity of solution = [tex] \frac{\text{weight of solute}}{\text{Molecular weight X volume of solution (l)}} [/tex]
∴ 1.34 x 10^-5 = [tex] \frac{\text{weight of AgCl}}{143.4 X 1} [/tex]
∴ Weight of AgCl present in solution = 1.92 X 10^-3 g
Thus, mass of AgCl that will dissolve in 1l water = 1.92 x 10^-3 g
