Question

calculate the molar solubility of of Ag2CO3 in a solution that is buffered to a phone of 7.50

Answer 1

Answer – We are given, pH = 7.50

We know

[H⁺] = 10^-pH

         = 10^-7.50

          = 3.16*10^-8 M

We know

Ag₂CO₃ <---> 2Ag⁺ + CO₃^2-

We know the Ksp expression

Ksp = [Ag⁺]² [CO₃^2-]

So, [Ag⁺] = 2 ([CO₃^2-] + [HCO₃^-] + [H₂CO₃]

We know CO₃^2- is the second dissociation of H₂CO₃

HCO₃^- + H₂O <----> H₂CO₃ + OH^-

Kb1 = [H₂CO₃] [OH^-] /[ HCO₃^-] = 2.32*10^-8

[ H₂CO₃] = Kb1.Kb2 /[OH^-]²*[CO₃^2-]

CO₃^2- + H₂O <----> HCO₃^- + OH^-

Kb2 = [HCO₃^-] [OH^-] / [CO₃^2-] = 1.78*10^-4

[HCO₃^-] = Kb2/ [OH^-] *[CO₃^2-]

We also know

Solubility of Ag₂CO₃ = ½ [Ag⁺]

[Ag⁺] = 2 ([CO₃^2-] + [HCO₃^-] + [H₂CO₃]

[Ag⁺] = 2( [CO₃^2-] + Kb2/ [OH^-] *[CO₃^2-] + Kb1.Kb2 /[OH^-]²*[CO₃^2-])

[Ag⁺] = 2 [CO₃^2-] (1+ Kb2/[OH^-] + Kb1.Kb2 /[OH^-]²)

We know

[OH^-] = 1*10^-14 / 3.16*10^-8 M

       = 3.16*10^-7

[Ag⁺] = 2 [CO₃^2-] ( 1.78*10^-4 /(3.16*10^-7) + 2.32*10^-8*1.78*10^-4/(3.16*10^-7)²

[Ag⁺] = 1208.3 [CO₃^2-]

So, [CO₃^2-] = 8.28*10^-4[Ag⁺]

Ksp = [Ag⁺]² [CO₃^2-]

8.1*10^-12 = [Ag⁺]² *8.28*10^-4[Ag⁺]

                  = 8.28*10^-4 x³

So,   8.1*10^-12 /8.28*10^-4 = x³

So, x = 2.14*10^-3 M

So, the molar solubility of of Ag₂CO₃ in a solution that is buffered at pH 7.50 is 2.14*10^-3 M