Question

Determine which salt CaCO3 or Ag2CO3 is more soluble in water in units of grams per liter?

Answer 1

1)

At equilibrium:

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

   s s

Ksp = [Ca2+][CO32-]

3.36*10^-9=(s)*(s)

3.36*10^-9= 1(s)^2

s = 5.797*10^-5 M

Molar mass of CaCO3,

MM = 1*MM(Ca) + 1*MM(C) + 3*MM(O)

= 1*40.08 + 1*12.01 + 3*16.0

= 100.09 g/mol

Molar mass of CaCO3= 100.09 g/mol

s = 5.797*10^-5 mol/L

To covert it to g/L, multiply it by molar mass

s = 5.797*10^-5 mol/L * 100.09 g/mol

s = 5.802*10^-3 g/L

Solubility of CaCO3 is 5.802*10^-3 g/L

2)

At equilibrium:

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

   2s s

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

8.46*10^-12=(2s)^2*(s)

8.46*10^-12= 4(s)^3

s = 1.284*10^-4 M

Molar mass of Ag2CO3,

MM = 2*MM(Ag) + 1*MM(C) + 3*MM(O)

= 2*107.9 + 1*12.01 + 3*16.0

= 275.81 g/mol

Molar mass of Ag2CO3= 275.81 g/mol

s = 1.284*10^-4 mol/L

To covert it to g/L, multiply it by molar mass

s = 1.284*10^-4 mol/L * 275.81 g/mol

s = 3.54*10^-2 g/L

Solubility of Ag2CO3 is 3.54*10^-2 g/L

Ag2CO3 is more soluble