Question

Solubility of AgCl in 0.1 M calcium chloride solution will be (Given: Ksp AgCl = 1.8 x 10-10)​

Answer 1

We have to find the solubility of AgCl in 0.1 M of calcium chloride solution. where Ksp for AgCl is 1.8 × 10¯¹⁰.

solution : let the solubility of AgCl is S.

dissociation reaction of AgCl is..

AgCl ⇔Ag⁺ + Cl¯

S S S

calcium chloride is a strong electrolyte so it completely dissociates.

dissociation reaction of CaCl₂ is ..

CaCl₂⇒Ca²⁺ + 2Cl¯

0 0.1M. 0.2M

So [Ag⁺] = S

[Cl¯] = S + 0.2

so, Ksp = S(S + 0.2)

⇒1.8 × 10¯¹⁰ = S(S + 0.2)

S << 0.2 so, (S + 0.2) ≈ 0.2

⇒1.8 × 10¯¹⁰ = 0.2S

⇒S = 9 × 10¯¹⁰ mol/L

Therefore the solubility of AgCl in 0.1 M CaCl2 solution is 9 × 10¯¹⁰ mol/L