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Derive (15-44) and use it to solve the following problem. Sulfate ion is to be removed from 60 L of water by exchanging it with
- Derive (15-44) and use it to solve the following problem. Sulfate ion is to be removed from 60 L of water by exchanging it with chloride ion on 1 L of a strong-base resin with relative molar selectivities listed in Table 15.6 and an ion-exchange capacity of 1.2 eq∕L of resin. The water to be treated has a sulfate-ion concentration Trim Size: 8.5in x 11in Seader c15.tex V2 - 10/16/2015 11:06 A.M. Page 515 Exercises 515 of 0.018 eq∕L and a chloride-ion concentration of 0.002 eq∕L. Following the attainment of equilibrium ion exchange, the treated water will be removed and the resin will be regenerated with 30 L of 10 wt% aqueous NaCl. (a) Write the ion-exchange reaction. (b) Determine the value of KSO2− 4 , Cl− . (c) Calculate equilibrium concentrations cSO2− 4 , cCl− , qSO2− 4 , and qCl− in eq/L for the initial ion exchange step. (d) Calculate the concentration of Cl− in eq/L for the regenerating solution. (e) Calculate cSO2− 4 , cCl− , qSO2− 4 , and qCl− upon reaching equilibrium in the regeneration step. (f) Are the separations sufficiently selective?
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