Ion exchange - Principles of Chemical Separations with Environmental Applications - page 230

4

3

y = − 0.76x + 15.3

2

1

0

15 17 19 21 23 25

− 1

− − −

V (L)

Figure 8.9 Ion exchange in waste treatment, Example 8.1.

Example 8.2 [15]

Problem:
A certain ion-exchange resin used for treating wastewater contains a finite quantity
of charged groups. Therefore, the equilibrium can be expressed in the same way that
an adsorption equilibrium is described: with an isotherm. Laboratory analysis of this
resin shows that it follows the Langmuir isotherm:

aX

Y =

X ,

(8.15)

b +

where Y = amount exchanged (mass contaminant / volume resin)

X = concentration in solution (mass contaminant / volume water)

cm 3

a = 70 mg /

b = 50 mg / L.

/

L contaminant,
how much fresh resin is necessary to adsorb 90% of the contaminant? Solve this
part algebraically.
(b) How many countercurrent stages would be required to adsorb 95% of the contam-
inant in the same feed as part (a) using 6.25 cm 3

(a) If you have 1.5 L of an aqueous waste stream containing 220 mg

of pure resin? Solve this part

graphically.
Solution:
(a) A schematic diagram is shown in Figure 8.10.
Contaminant balance: 1.5(220) = SY 1 + 1 . 5(22) .

Y 1 .

∴ S = 297 /

cm 3 Equilibrium: Y 1 = 70 X 1 / (50 + X 1 ) where Y 1 is in mg / and X 1 is in mg / L.

cm 3

At X 1 = 22 mg / L ⇒ Y 1 = 21.4 mg / resin.