Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: snowie1010 on October 24, 2008, 05:43:46 AM
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our lecturer hasn't given any formulas and the text book doesn't seem to have a formula for any problem like this?? just need to get started..
1) The potassium ions in 250 mL of solution containing 14.6 g/L KCl are to be removed by passing the solution through a cation-exchange column in the hydrogen form. If the ion-exchange capacity of the resin is 8.04 meq/g (1 equivalent = 1 mol per unit charge, 1 meq = 10-3 mol per unit charge) for dry resin, what is the minimum weight of dry resin required?
.25*14.6=3.65g KCl
MW KLC = 74.6 g/mol 3.65g / 74.6 g/mol = 0.0489 mol KCl
MW K = 39.1 g/mol 0.0489 mol * 39.1 = 1.91g K
then im not exactly sure what to do??