Question

The maximum solubility of Ag2CO3 in 0.02
M Na2CO3 is (Ksp of
Ag2003 is 8 x 10-12)
​

Answer 1

The maximum solubility of Ag₂CO₃ = = 1 x 10⁻⁵

Further explanation

Given

0.02 M Na₂CO₃

Ksp of  Ag₂CO₃ is 8 x 10⁻¹²

Required

The solubility of Ag₂CO₃

Solution

Ag₂CO₃ ⇒ 2Ag⁺ + CO₃²⁻

s                  2s        s

s = solubility

Ksp Ag₂CO₃ = [Ag⁺]² [CO₃²⁻]

Ksp Ag₂CO₃ = (2s)².s

Ksp Ag₂CO₃ = 4s³

In 0.02 M Na₂CO₃ ⇒ 2Na⁺ + CO₃²⁻⇒ [ CO₃²⁻]=0.02, so Ksp Ag₂CO₃ :

8 x 10⁻¹² = [2s]² [0.02]

8 x 10⁻¹² = 4s² [0.02]

4s² = 8 x 10⁻¹² / 2 × 10⁻²

4s² = 4 x 10⁻¹⁰

s² = 1 x 10⁻¹⁰

s = √1 x 10⁻¹⁰

s = 1 x 10⁻⁵