Silver ion, Ag+, can be recovered from photographic solutions by adsorbing onto

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.

Explanation / Answer

Ag+ +Ex Ag-Ex       logKL =12.30 ,   KL = 1012.30 = 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

KL =[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

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