Question

The solubility product constant of Ag2CO3 is log Ksp = -11.092. How many milligrams of this substance is dissolved in 100.0mL of saturated solution? (Hint: the dissolution reaction is Ag2CO3⇋Ag +CO32-

Answer 1

log Ksp = -11.092

Ksp   = 10^-11.092

Ksp = 8.09 x 10^-12

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

                                                 2S                  S

Ksp = [Ag+]^2 [CO3-2]

Ksp = (2S)^2 (S)

Ksp = 4 S^3

8.09 x 10^-12   = 4S^3

S = 1.265 x 10^-4 M

solubility = 1.265 x 10^-4 M

solubility = 1.265 x 10^-4 mole / litre

1000 mL ---------------------> 1.265 x 10^-4 mole

100 mL ------------------------> 100 x 1.265 x 10^-4 / 1000 = 1.265 x 10^-5 mol

molar mass of Ag2CO3 = 275.74 g / mol

mass of Ag2CO3 that can dissolve = moles x molar mass

                                                        = 1.265 x 10^-5 x 275.74

                                                         = 3.49 x 10^-3 g

                                                          = 3.49 mg