Chromatography Forum

Could anyone please help me on answering the following question regarding calculating the characteristic properties of ion exchange resins:

Q: Find a) the exchange capacity of the resin, b) the selectivity coefficient of the resin and c) the distribution coefficient of sodium using the following information:

A column contains 0.005g of a stong acid ion-exchange resin in the H+ form. A 1.00M NaCL solution is passed through the column until the eluent shows a neutral reaction. The total eluent is titrated with 17.5mL of 0.05M NaOH. A 100mL portion of a 1.00 x 10-3 M NaOH solution is equilibrated with -0.150g of the same resin, and the sodium concentration in the supernatant solution is found to be 2.20 x 10-4M by flame spectroscopy.

(the concentration is expressed in meq/g for the resin and meq/mL for the solution)

Would really appreciate if anyone could help me out on this?!

Thank you

Some hints:

1. The ion exchange capacity of the resin has units of meq/g. You know how many grams of resin you have. All you have to do is to figure out how many meq of Na+ were taken up by the resin. You know how much Na+ went into the column, and you should be able to figure out how much was left at the end.

2. Along the same lines: what is the definition of distribution coefficient? And how would you go about figuring out the quantities involved?

Thanks Tom,

I'll try the calculations again taking your hints into account! I noticed an error in the question where it reads NaOH solution is equilibrated with .15g of the resin it should be NaCl, not NaOH. Apologies

Actually, for a cation exchange resin, the counterion (Cl- or OH-) is irrelevant to these calculations. What counts is the amount of Na+