Separation Process Principles- 2n - Seader & Henley - Solutions Manual

0610001000 2625 x 100 813 wt interpolate the given

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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. (15-40), 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. (15-39), 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. (15-40), 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 ion-exchange 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: Ion-exchange resin of Dowex 50 cross-linked 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. (15-38) applies at equilibrium, with n = 1. Let x = eq of Ag in resin at eq...
View Full Document

Ask a homework question - tutors are online