Reversed size dependence in ion solubility My book says that solubility increases with interionic distance, because the attractive forces between ions gets smaller. However, if one of the ions in a binary ionic system is very large compared to the other, the trend is reversed. For example, CsF has greater solubility than CsCl. Why is this?
This can be explained with some crude thermodynamic models. The lattice enthalpy $\Delta H_\mathrm L$ of an alkali metal halide $\ce{MX}$ is given by

$$\Delta H_\mathrm L = \frac{k_1}{r_\ce{M} + r_\ce{X}}$$

and the hydration enthalpies by

$$\Delta H_\mathrm S(\ce{M}) = -\frac{k_2}{r_\ce{M}}\qquad  \qquad \Delta H_\mathrm S(\ce{X}) = -\frac{k_2}{r_\ce{X}}$$

where $k_1$ and $k_2$ are some positive constants and $r_\ce{M}$ and $r_\ce{X}$ are the ionic radii of $\ce{M+}$ and $\ce{X-}$ respectively.

The solubility is therefore a balance between these two quantities: larger (magnitudes of) solvation enthalpies increase solubility, but larger lattice enthalpies reduce solubility. The solubility $S$ is thus related to the difference of these terms. We ignore the entropic contribution, or to be precise, we assume that $\Delta S$ does not vary too much between different salts $\ce{MX}$.

$$S \sim \frac{k_2}{r_\ce{M}} + \frac{k_2}{r_\ce{X}} - \frac{k_1}{r_\ce{M} + r_\ce{X}}$$

Let us now assume that

$$r_\ce{M} = a+b \qquad \qquad r_\ce{X} = a - b$$

where $a,b$ are more positive constants. Here we have implicitly assumed that $r_\ce{M} \geq r_\ce{X}$, but it works the other way round too, so there is no loss of generality. Thus

$$\begin{align}
S &amp;\sim \frac{k_2}{a+b} + \frac{k_2}{a-b} - \frac{k_1}{2a} \\
&amp;= \frac{2ak_2}{a^2-b^2} - \frac{k_1}{2a}
\end{align}$$

This is clearly a minimum when $b = 0$, i.e. $r_\ce{M} = r_\ce{X}$. So, salts with similar cationic and anionic radii are less soluble, and vice versa.