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Unformatted text preview: =
=
= 0.00271 mol/L or 0.00542 eq/L
4
zSO=
2
4 cCl = CxClzCl = 0.02(1 − 0.271)
= 0.01458 mol/L or 0.01458 eq/L
1 From Eq. (1540),
QySO= 1.2(0.629)
4
qSO= =
=
= 0.377 mol/L or 0.754 eq/L
4
zSO=
2
4 qCl = QyClzCl = 1.2(1 − 0.629)
= 0.445 mol/L or 0.445 eq/L
1 (8) Exercise 15.13 (continued)
Analysis: (continued)
(d) The regenerating solution is 30 L of 10 wt% aq. NaCl. From Perry's Handbook,
assuming a temperature of 25oC, the density is 1.0689 g/cm3 or 1068.9 g/L. The MW of NaCl =
58.45. Therefore, the concentration of chloride ion = cCl = 0.10(1068.9)/58.45 = 1.829 eq/L.
(e) From Part (c), 1 L of resin includes: 0.445 eq of Cl and 0.754 eq
From Part (d), 30 L of regenerating solution includes: 30(1.829) = 54.870 eq. ClThus, we have a total of 30 L of solution and 1 L of resin containing a total of:
0.754 eq. of SO4= and 54.870 + 0.445 = 55.315 eq. of ClLet: a = eq of SO4= on the resin at equilibrium. The equivalent fractions of the ions for
1.0 L of resin, 30 L of aq. solution, and n = 2 are:
0.574 − a
xSO= =
= 0.01046 − 0.01822a
(9)
4
54.870
a /1 a
ySO= =
=
= 0.833a
(10)
4
1.2 1.2
Combining Eqs (1), (5), (9), and (10), with C = 1.829 eq/L and
Q = 1.2 eq/L, with n = 2,
KA,B C
=
Q n −1 ySO= xCl  n 4 xSO= yCl  n 1829
.
= 015 =
.
12
. 4 2 −1 0.833a 1 − 0.01046 + 0.01822a
0.01046 − 0.01822a 1 − 0.833a Solving Eq. (11) with a nonlinear solver, a = 0.00123 eq.
From Eqs. (9) and (10),
xSO= = 0.01046 − 0.01822(0.00123) = 0.01044
4 ySO= = 0.833(0.00123) = 0.001025
4 From Eq. (1539),
CxSO= 1.829(0.01044)
4
cSO= =
=
= 0.00955 mol/L or 0.0191 eq/L
4
zSO=
2
4 cCl = CxClz Cl  = 1.829(1 − 0.01044)
= 1.810 mol/L or 1.810 eq/L
1 2
2 (11) Exercise 15.13 (continued)
Analysis: (e) (continued)
From Eq. (1540), qSO= =
4 QySO=
4 zSO=
4 qCl = QyClzCl = = 1.2(0.001025)
= 0.000615 mol/L or 0.00123 eq/L
2 1.2(1 − 0.001025)
= 1.199 mol/L or 1.199 eq/L
1 (f) For the ionexchange step, (0.755/1.080) x 100% = 70% of the sulfate ion is transferred from
the solution to the resin. This is reasonably high considering the low and unfavorable chemical
equilibrium constant of 0.15.
The regeneration step is very effective, with (0.754  0.00123)/0.754 x 100% = 99.84% of the
sulfate ion removed from the resin. Exercise 15.14
Subject:
ion. Equivalent fractions at equilibrium for the ion exchange of silver ion with sodium Given: Ionexchange resin of Dowex 50 crosslinked with 8% divinylbenzene. Exchange of
silver ion in methanol with sodium ion in the resin. Wet capacity of the resin = 2.5 eq/L. Resin
is initially saturated with sodium ion. Experimental data for the molar selectivity coefficient as a
function of equivalent fraction of sodium ion in the resin:
xNa+
0.1
0.3
0.5
0.7
0.9
Kag+, Na+
11.2 11.9 12.3 14.1 17.0
Find: Equivalent fractions if 50 L of 0.05 M Ag+ in methanol is treated with 1 L of wet resin.
1 L of resin contains 2.5 eq of Na+.
50 L of solution contains 0.05(5) = 2.5 eq Ag+
Because both ions have a charge of one, Eq. (1538) applies at equilibrium, with n = 1.
Let x = eq of Ag in resin at eq...
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 Spring '11
 Levicky
 The Land

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