Question

Which salt, CaCO₃ (Ksp = 2.8 x10-9) or Ag₂CO₃ (Ksp = 8.1 x 10^-12) is more soluble in water in units of moles per liter? How many mole per liter more soluble?

Answer 1

At equilibrium:

CaCO3 <----> Ca2+ + CO32-

   s s

Ksp = [Ca2+][CO32-]

2.8*10^-9=(s)*(s)

2.8*10^-9= 1(s)^2

s = 5.29*10^-5 M or 5.29*10^-5 mol/L

2)

At equilibrium:

Ag2CO3 <----> 2 Ag+ + CO32-

   2s s

Ksp = [Ag+]^2[CO32-]

8.1*10^-12=(2s)^2*(s)

8.1*10^-12= 4(s)³

s = 2.01*10^-4 M or 2.01*10^-4 mol/L

Ag2CO3 is more soluble

excess moles of Ag2CO3 in water= 0.000201-0.0000529= 0.0001481‬= 1.48*10^-4

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