Ionic Equilibrium

When 0.1 mole CuCl is added to saturated solution of AgCl
AgCl (s) ⇌

Ag+

x

+

Cl−

x

CuCl (s) ⇌

Cu+

y

+

Cl−

y

Net Cl− in the solution = (x + y)
Ksp(AgCl) = 1.6 × 10−10 = [Ag+][Cl−] = x (x + y)
x (x + y) = 1.6 × 10−10 (1)
Ksp(CuCl) = 1.0 × 10−6 = [Cu+][Cl−]
1.0 × 10−6 = y (x + y) (2)
Dividing Eqs. (1) and (2), we get

x

y

= 1.6×10−4
y =

104

1.6x

Using this value of y in Eq. (1), we get
x × (x+

104

1.6x

) = 1.6×10−10
x2×

104

1.6

= 1.6×10−10
x2 = (1.6)2×10−14
x = 1.6×10−7