Soil-solution cations in competition for exchange sites

Question

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 $\ce{Na+}$?

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⁺