Question

What mass of silver chloride (m = 143.4) will dissolve in 1.00 l of water? the ksp of agcl is 1.8 × 10–10 ?

Answer 1

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 = $\frac{\text{weight of solute}}{\text{Molecular weight X volume of solution (l)}}$
              ∴  1.34 x 10^-5     = $\frac{\text{weight of AgCl}}{143.4 X 1}$
∴ 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