Question

Silver ion, Ag+, can be recovered from photographic solutions by adsorbing onto an ion exchange resin (Langmuir adsorption)

Ag+ +Ex↔ Ag-Ex logKL =12.30

(where Ex represents an adsorbing site on the resin). According to the manufacturer, the resin has 1.20 x10-5 moles of adsorbing sites per gram of resin.

a) If you have a liter of solution with [Ag+] = 0.030 mM, how much resin will you need to adsorb all the silver assuming 100% binding?

b) Calculate the predicted equilibrium dissolved [Ag+] if you add 10.00 g of resin to 1.00 liter of solution described in part a.

Answer 1

Ag+ +Ex↔ Ag-Ex       logK_L =12.30 ,   K_L = 10^12.30 = 2.00 x 10¹²

a.

[Ag+] x Vol.sol. = [Ag-Ex]x (resin mass)

0.030 x10^-3 mol/L x 1 L = 1.20 x10^-5 moles of adsorbing sites/g x m

m = 0.025x10² g = 2.5 g

b. [Ex] = 1.20 x10^-5 mol/g x 10 g / 1L = 1.20 x10^-4 M

Ag⁺                +       Ex                ↔ Ag-Ex      

0.30 x10^-4M        1.20 x10^-4 M          0                   initial

0                          0.90 x10^-4            0.30 x10^-4       adsorption

x                         0.90 x10^-4 + x        0.30 x10^-4 – x    equilibrium

K_L =[Ag-Ex] / ([Ag+][Ex])

2.00 · 10¹² = (0.30 x10^-4 – x)/ [ x( 0.90 x10^-4 -x)]

2.00 · 10¹² · x = (0.30 x10^-4 – x)/ ( 0.90 x10^-4 -x)

Assume that x << 1x 10^-4

2.00 · 10¹² · x = (0.30 x10^-4 / ( 0.90 x10^-4)

2.00 · 10¹² · x = 1/3

x = 1.7 ·10^-13

[Ag+] = 1.7 ·10^-13 M