preface to fifteenth edition

11.40 SECTION 11
distribution coefficient value of 12 or lower is required to elute an ion completely from a column
containing about 10 g of dry resin using 250 to 300 mL of eluant. A larger volume of eluant is
required only when exceptionally strong tailing occurs. Ions may be eluted completely by 300 to
400 mL of eluant from a column of 10 g of dry resin at D g values of around 20. The first traces of
an element will appear in the eluate at around 300 mL when its D g value is about 50 to 60.
Example Shaking 50 mL of 0.001 M cesium salt solution with 1.0 g of a strong cation exchanger
in the H-form (with a capacity of 3.0 mequiv · g 1 ) removes the following amount of cesium. The
selectivity coefficient, k Cs/H , is 2.56, thus:
[Cs ] r[H ]
2.56
[Cs ][H ] r
The maximum amount of cesium which can enter the resin is 50 mL 0.001 M 0.050 equiv.
The minimum value of [H ] r 3.00 0.05 2.95 mequiv, and the maximum value, assuming
complete exchange of cesium ion for hydrogen ion, is 0.001 M. The minimum value of the distribution
ratio is:
[Cs ] r (2.56)(2.95)
(D c) Cs 7550
[Cs ] 0.001
Amount of Cs, resin phase (7550)(1.0 g)
151
Amount of Cs, solution phase 50 mL
Thus, at equilibrium the 1.0 g of resin removed is:
100% x 151
x
with all but 0.66% of cesium ions from solution. If the amount of resin were increased to 2.0 g, the
amount of cesium remaining in solution would decrease to 0.33%, half the former value. However,
if the depleted solution were decanted and placed in contact with 1goffresh resin, the amount of
cesium remaining in solution would decrease to 0.004%. Two batch equilibrations would effectively
remove the cesium from the solution.