Question

K_sp of AgCl is 1.8 × 10⁻¹⁰. Calculate molar solubility in 1 M AgNO₃.

Answer 1

\[\ce{AgCl_{(s)} ⇌ Ag^+_{( aq)} + Cl^-_{( aq)}}\]

x = solubility of AgCI in 1 M AgNO₃

\[\ce{AgNO3_{(aq)} ⇌ \underset{1 M}{Ag^+_{( aq)}} + \underset{1 M}{NO^-_{3(aq)}}}\]

[Ag⁺] = x + 1 ≃ 1 M ..........(∵ x << 1)

[Cl^–] = x

K_sp = [Ag⁺] [Cl^–]

1.8 × 10⁻¹⁰ = (1) (x)

x = 1.8 × 10⁻¹⁰ M