Question

The solubility (in mol L⁻¹) of AgCl (K_sp = 1.0 × 10⁻¹⁰) in a 0.1 M KCl solution will be ______.

Options

• 1.0 × 10⁻⁹

• 1.0 × 10⁻¹⁰

• 1.0 × 10⁻⁵

• 1.0 × 10⁻¹¹

Answer 1

The solubility (in mol L⁻¹) of AgCl (K_sp = 1.0 × 10⁻¹⁰) in a 0.1 M KCl solution will be 1.0 × 10⁻⁹.

Explanation:

Let solubility of AgCl = x mole/L

\[\ce{AgCl <=> Ag^+ + Cl^-}\]

i.e., K_sp(AgCl) = x × x

\[\ce{KCl -> K^+ + \underset{0.1}{Cl^-}}\]

[Cl⁻] from KCl = 0.1 m

Total [Cl⁻] in solution = x + 0.1

K_sp(AgCl) = [Ag⁺] [Cl⁻] = x (x + 0.1)

1.0 × 10⁻¹⁰ = x (x + 0.1)

1.0 × 10⁻¹⁰ = x² + 0.1x

1.0 × 10⁻¹⁰ = 0.1x ....(as x² << 1)

x = 1.0 × 10⁻⁹ mol/L