Question

Calculate the concentration of Ag+ in an aqueous solution containing 0.500M NaNO3 in equilibrium with solid AgI (ion strength and activity coefficients should be considered).

The answer is 1.5*10^-8 but please show me how to get there. Thank you

Answer 1

Recall that ionic strength considers all ions in solution, and its charges. It is typically used to calculate the ionic activity of other ions. The stronger the electrolytes, the more ionic strength they will have.

The formula:

I.S. = 1/2*sum( Ci * Zi^2)

Where

I.S. = ionic strength, M (also miu / μ ) used

Ci = concentration of ion “i”

Zi = Charge of ion “i”

The exercise:

IS = 1/2*( 0.5(1)2 + (0.5)(1) = 0.5

KIO4(s) = Ag+(aq) + I-(aq)

Ksp = Activity of Ag+ * Activity of I-

Ksp = Y_Ag+ *[ Ag+ ] * Y_I- * [-]

8.52*10^-17= Y_Ag+ *[ Ag+ ] * Y_I- * [-]

Recall that:

-log(γ) = 0.51*(Zi^2)*sqrt(I.S.) / ( 1 + (α * sqrt(I.S)/305))

Where

γi = activity coefficient for species “i”

αi = theoretical diameter in pm (10^-12 m)

Zi = Charge of ion

I.S. = ionic Strength (usually used as μ as well)

If we wanted only γ

γ = 10^-(0.51*(Zi^2)*sqrt(I.S.) / ( 1 + (α * sqrt(I.S)/305)))

alpha Ag = 250

alpha I- = 300

γ = 10^-(0.51*(1^2)*sqrt(0.5) / ( 1 + (250 * sqrt(0.5)/305))) = 0.5911

γ = 10^-(0.51*(1^2)*sqrt(0.5) / ( 1 + (300* sqrt(0.5)/305))) =0.6127

8.52*10^-17= Y_Ag+ *[ Ag+ ] * Y_I- * [-]

8.52*10^-17=0.5911  *[ Ag+ ] * 0.6127* [I-]

assume Ag+ = I- = X

8.52*10^-17=0.5911  *X * 0.6127* X

X^2 = (8.52*10^-17)/(0.5911 *0.6127)

X^2 = 2.3525*10^-16

x = sqrt(2.3525*10^-16) = 1.533*10^-8 M

[Ag+] = 1.533*10^-8 M