Chemistry solubility problem

Question

This reaction represents the dissolution and precipitation of Ag2CO3(s).

Ag2CO3(s)  ⇄  2 Ag+(aq)  +  CO3^(2-)(aq), K = 4.5 ×10^(-12)

If we were to mix 0.00025 mol of solid Ag₂CO₃ in 5.00 L of pure water, how much solid is left when reaction reaches equilibrium?

Answer 1

Ag₂CO₃(s)  ⇄  2 Ag⁺(aq)  +  CO₃^2-(aq) ... K = 4.5 ×10^-12

Ksp = 4.5x10^-12 = [Ag⁺]²[CO₃^2-] =(2x)²(x)

4x³ = 4.5x10^-12

x3 = 1.125x10^-12

x = 1.04x10^-4 M = molar solubility of Ag₂CO₃

Looking at 0.00025 mol Ag₂CO₃ in 5.00 L, we have 0.00025 mol/5.00 L = 5x10_-5 M. This is below the solubility calculated, so all of it should dissolve, and there will be none of the solid Ag₂CO₃ left when the reaction reaches equilibrium.