When I was learning about salts that show exothermic dissolution, I noticed that the same salts used in Daniel Cell( due to similarity of size of anion and cation) were undergoing exothermic dissolution. Is there any relation between the similarity in size of anion/cation and the type of dissolution it undergoes?
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2 Answers
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There is no easy general pattern.
The dissolution can be formally divided into 2 steps:
- breaking the ion lattice into free ions in vacuum
- hydration these "naked" ions by water.
Both lattice forming and hydration enthalpies are negative and their absolute value increases with increasing ion charge and decreasing ion radius.
The relation of their values and the overall enthalpy sign depends on the particular element chemistry, that modifies hydration and lattice enthalpy values.
There are some trends, but hard to generalize.
You can compare sodium and potassium salts of big anions with delocalized charge, like $\ce{NO3-, ClO3-, ClO4-}$. Dissolution of respective potassium salts is significantly more endothermic, compared to sodium ones. Sodium ion is smaller and is more intensely hydrated with lower enthalpy of hydration.
This difference is bigger that difference of their lattice enthalpies. It is probably due anion size, packing of bigger potassium cation makes less difference than for chlorides.
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$\begingroup$ So there is no way to predict type of dissolution other than looking at their enthalpies? $\endgroup$ Commented May 22, 2020 at 15:53
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2$\begingroup$ There is always way of learning empirical facts and patterns.. Chemistry students often search for deriving predictions to avoid it, but chemistry is in many aspects a stubborn old empirical man. It could be predicted by intensive application of quantum chemistry, sometimes relativistic one, but learning empirics is much easier. $\endgroup$– PoutnikCommented May 22, 2020 at 15:56
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For simple ionic salts, the enthalpy of dissolution is given by $$\Delta H_{\text{disso}} = \Delta H_{\text{lattice}} + \Delta H_{\text{solvation}}$$
where $\Delta H_{\text{lattice}}$ is the lattice dissociation enthalpy (positive) and $\Delta H_{\text{solvation}}$ is the solvation enthalpy (negative).
For exothermic or endothermic dissolution, we have to compare the lattice and solvation enthalpies. Generally, salts with a small cation and anion (think Lithium Fluoride) have very low solvation enthalpies but also have very high lattice enthalpies (given by the Born-Lande equation), hence their enthalpies of dissolution are relatively higher than salts with larger cations/anions, such as Caesium Fluoride.
This graph1 illustrates the comparative hydration enthalpies of fluorides of $\ce{Na+, K+, Rb+, Cs+}$ (graph a) and also the hydration enthalpies of sodium salts of $\ce{F-, Cl-, Br-, I-}$ (graph b). Notice the decreasing trend: dissolution of $\ce{CsF}$ is more exothermic than $\ce{NaF}$ and similarily for $\ce{NaI}$ and $\ce{NaF}$.
Hydration enthalpies for gaseous ions are calculated as described by Latimer et al.2 in his paper.
Further reading:
- Wikipedia article on Enthalpy Change of Solution
- Chemistry LibreTexts: Enthalpy Change of Solution
References:
- Yıldıran, Hüseyin, et al. “A Theoretical Study on Calculation of Absolute Hydration Enthalpies for Some Univalent Ions.” Ionics, vol. 14, no. 6, Nov. 2008, pp. 541–43. doi:10.1007/s11581-008-0214-3.
- Latimer, Wendell M., et al. “The Free Energy of Hydration of Gaseous Ions, and the Absolute Potential of the Normal Calomel Electrode.” The Journal of Chemical Physics, vol. 7, no. 2, Feb. 1939, pp. 108–11. doi:10.1063/1.1750387.
answered May 22, 2020 at 15:57