Question

Solid Silver chloride has an extremely low solubility in water. Calculate the molar solubility of AgCl in a mixture of AgCl(s) and 0.08 M NaCl(aq). Ksp = 1.8 × 10-10

Answer 1

Solubility of AgCl in solution of NaCl (here common ion is Cl^-)
Solubility product for AgCl can be given as:

                                  AgCl    <====>   Ag⁺ +     Cl^-

Solubility:                    s                         s              s

Ksp = [ Ag⁺][Cl^- ]

Ksp = [s][s]

Ksp = [s]²       ----------------------(1)

As there is common ion Cl^- present (with concentration 0.08 M/L of NaCl), the molar solubility of AgCl will decrease (due to common ion effect) which can be given as sⁱ and at this stage new concentration of [Ag⁺] and [Cl^-] can be given as:

                                AgCl    <====>   Ag⁺ +       Cl^-
Solubility:                    sⁱ                          sⁱ              sⁱ

[Ag⁺] = sⁱ

[Cl^-] = sⁱ + c   = ~ c   (because common ion of Cl^- in AgCl and NaCl in solution)

Now,

Ksp = sⁱ x c     -----------------------(2)

from eq (1) and (2)

sⁱ = Ksp/c

sⁱ = 1.8 x 10^-10/ 0.08

sⁱ = 2.25 x 10^-9 M/L