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28.4-5. Equilibrium in an Ion Exchange of NH4+for H+. For the case where the cation NH4+(A) replaces H+(B) in a polystyrene resin with 8% DVB,

image text in transcribed 28.4-5. Equilibrium in an Ion Exchange of NH4+for H+. For the case where the cation NH4+(A) replaces H+(B) in a polystyrene resin with 8% DVB, calculate the equilibrium constant KA,B. The total resin capacity Q=2.0 equiv/L wet bed volume. The total concentration C=0.20N in the solution. Calculate at equilibrium the equivalents of NH4+in the resin when the concentration of NH4+in solution is 0.04N. Ans. qNH4R=0.6684 equiv/I

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