Question

An insoluble salt, MX2, dissociates slightly in water, MX2(s) + M2+ (aq) + 2X-(aq); Ksp = 9.5 x 10-14 at 25 oC. Calculate AG for the dissociation reaction when the concentration of ions are: [M2+] = 6.4 x 10-2 M and {X-} = 9.4 x 10-5 M.

Answer 1

Given that

Ksp = 9.5 *10^-14

First calculate dG° as follows:

dG° = -RT*ln(Ksp)

dG° = -8.314*298*ln( 9.5*10^-14)

dG° =74289.75 J/mol

now...

dG = dG° + RT*ln(Q)

Q= product / reactants

But solid has unity value

Q = [M+2][X-]^2 = (6.4*10^-2)(9.4*10^-5)^2 = 5.66*10^-10

dG = dG° + RT*ln(Q)

dG = 74289.75 + 8.314*298 * ln(5.66*10^-10)

dG = 74289.75 + 8.314*298 * (-21.29)

dG = 74289.75 – 52755.69

dG =21534.06 J/mol

dG = 21.53 kJ/mol