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
C + 273.15 = temp. in Kelvins.
25° C would be 298.15 k.
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