Question

Calculate the solubility of Ag+ (from AgSCN) in 0.0333 M Mg(ClO4)2 using a) concentrations and b) activities . Compare your results

Answer 1

a)

molar concentrations

ignore Mg(ClO4)2 since it should not affect AgSCN in equilibrium if we ignore activities

so

AgSCN(s) <--> Ag+ + SCN-

Ksp = [Ag+][SCN-]

Ksp for AgSCN = 1.1*10^-12

so... in solution, Ag+ = SCN- due to sotichiometry, let it be "S"

1.1*10^-12 = S*S

S^2 = 1.1*10^-12

S = sqrt(1.1*10^-12) = 0.0000010488 = 1.0488*10^-6 M of AgSCN

b)

using activities

for activity, we require ionic strenght which iuncludes ionic concentrations and charges of all species in solution

so...

first, let us calcualte ionic strenght IS

IS = 1/2*sum(ion charge ^2 * concentration + ... all other species)

Note that Ag+ and SCN- concentrations will be so small compared to Mg+2 and ClO4- so:

Ag+ = 0, SCN- = 0

Mg+2 = 0.0333 M

ClO4- = 2*0.0333 = 0.0666 M

so

IS = 1/2*((2^2)(0.0333 ) + (1^2)(0.0666)) = 0.0999

note that

activities for yAg+ = 0.75 and ySCN- = 0.76

so..

Ksp = A_Ag+ * A_SCN-

Ksp = [Ag+]*yAg+ * [SCN-]*ySCN-

substitute values:

Ksp = [Ag+]*yAg+ * [SCN-]*ySCN-

1.1*10^-12 = [Ag+][SCN-] ( 0.75*0.75)

note that, similar to previous step

Ag+ = SCN- = S

(1.1*10^-12) / (( 0.75*0.75)) = S^2

S = sqrt((1.1*10^-12) / (( 0.75*0.75)) ) = 0.00000139841M

S = 1.39841*10^-6 M

clearly

solubility increases with activity coefficients

1.0488*10^-6 M of AgSCN vs. 1.39841*10^-6 M