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2 years ago

The maximum solubility of Ag2CO3 in 0.02M Na2CO3 is (Ksp ofAg2003 is 8 x 10-12)

Chemistry

1 answer:

Anna007 [38]2 years ago

3 0

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

<h3>Further explanation</h3>

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

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The maximum solubility of Ag2CO3 in 0.02M Na2CO3 is (Ksp ofAg2003 is 8 x 10-12)​

The maximum solubility of Ag2CO3 in 0.02M Na2CO3 is (Ksp ofAg2003 is 8 x 10-12)