Ion Exchange, Equilibrium and Types of Contact

An acid solution containing 2 per cent by mass of NaNO3 and an unknown concentration of HNO3

is used to regenerate a strong acid resin. After sufficient acid had been passed over the resin for

equilibrium to be attained, analysis showed that 10 per cent of resin sites were occupied by sodium

ions. What was the concentration of HNO3 in the solution, if its density were 1030 kg/m3.

Hard water containing 500 ppm (by weight) MgCO3 and 50 ppm

NaCl is to be softened at 25C in an existing fixed, gel resin bed

with a cation capacity of 2.3 eq/L of bed volume. The bed is 8.5 ft

in diameter and packed to a height of 10 ft, with a wetted-resin void

fraction of 0.38. During loading, the recommended throughput is

15 gal/minute-ft2. During displacement, regeneration, and washing,

flow rate is reduced to 1.5 gal/minute-ft2. The displacement and

regeneration solutions are water-saturated with NaCl (26 wt%).

Determine: (a) feed flow rate, L/minute; (b) loading time to breakthrough,

h; (c) loading wave-front velocity, cm/minute; (d) regeneration

solution flow rate, L/minute; and (e) displacement time, h.

Equipment & Types of contact

Concentration of MgCO3 in feed = (500/106 )*1000/83.43 = 0.006 M = 0.012 eq/L

Similarly, Conc. Of Na+ = 0.000855 eq/L

Bed area = (3.14/4)*8.52 = 56.7 ft2

a. Flow rate = 15*56.7=851 gpm = 3219 L/min

bed volume = (56.7)(10) = 567 ft3 or 16,060 L

total bed capacity = 2.3(16,060) = 36,940 eq

Mg2+ absorbed by resin = 0.9961(36,940) = 36,796 eq;

Mg2+ entering bed in feed solution = 0.012(3,219) = 38.63 eq/minute

tL = 953 minutes or 15:9 h

uL =L/tL = 10/953 = 0.0105 ft/minute or 0.32 cm/minute.

b. XMg2+ = 0.012/(0.012+0.000855)= 0.9355

From table K = 1.65

C=0.012, Q=2.3

Substitute in equilibrium relation to get Y = 0.9961

Equilibrium

Process Definition & Applications

Reversible Chemical Reaction

An ion-exchange resin (sulphonated Styrene-DVB) with a maximum ion-exchange capacity of 4.90 meq/g of dry resin, is used to remove cupric ion from a waste stream containing 0.00975-M CuSO4 (19.5 meq Cu2+/L solution). The spherical resin particles range in diameter from 0.2 to just over 1.2 mm.The resin Capacity is 2.3 eq/L The equilibrium ion-exchange reaction is of the divalent–monovalent type:

K

K average = 2

X Cu2+ Y Cu2+

0.00113 0.135

0.0403 0.665

0.230 0.929

0.528 0.949

1.35

1.15

4.04

1.30

NaNO3 (M.wt 85 kg/kmol)

Concentration = 2 per cent by mass

= (0.02/85)(1030) = 0.242 kmole/m3

HNO3 (M.wt = 63 kg/kmol)

Concentration = p per cent

Concentration = ((p/100)/63)(1030)

= 0.163p kmole/m3

K=3.47/1.27 = 2.7

In the solution: XNa+ = 0.242/(0.242 + 0.163p)

a. From the data, compute the molar selectivity coefficient, K, at each value of c for Cu2+ and compare it to the value estimated from using the table

In the resin; Y=0.1

Please refer to last tutorial

from table, K =1.5

b. Predict the milliequivalents of Cu2+ exchanged at equilibrium from 10 L of 20 meq Cu2+/L, using 50 g of dry resin with 4.9 meq of H+/g.

Substitute to get P = 20%

b. C= 0.02 eq/L

Q= 2.3 eq/L

n=2

K=2

Solution: 0.1887 equivalents

Ion Exchange