How could one determine the likelihood of one cation displacing another on a negatively charged exchange site of a clay particle?

For example, in a soil solution containing equal parts (by number of atoms) $\ce{Ca^2+}$, $\ce{Mg^2+}$, $\ce{K+}$, $\ce{Na+}$, $\ce{Al^3+}$, which cations will find the most exchange sites and which will ultimately be washed deeper into the soil? (Assuming there are not enough sites for all of them.)

Could two $\ce{Na+}$ cations displace an adsorbed $\ce{Ca^2+}$ cation simply because they "bump" into it? Or is there some mechanism where cations are dislodged? Is it easier for two $\ce{Na+}$ to replace $\ce{Ca^2+}$ or for $\ce{Ca^2+}$ to replace two $\ce{Na+}$?


1 Answer 1


Only half as many Ca2+ are needed as Na+ to neutralize the charge of the ion exchanger (here: clay).

The affinity of clay for bivalent ions is larger than that for monovalent ions:
Ca2+ > Na+

For most clays, the affinity for equally charged ions increases with the nonhydrated radius of the ions:
Cs+ > K+ > Na+ > Li+
Ba2+ > Sr2+ > Ca2+ > Mg2+


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