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My teacher taught us that higher the ionic character in a compound, higher is its solubility in polar solvents like water. To illustrate this statement, he also gave us the following example-

LiOH< NaOH< KOH< RbOH< CsOH (solublity in water)

According to this example, if the anion is same in all compounds then ionic character increases as we go down the group and thus the compounds become more soluble.

But then our teacher also gave us an exception. He said that if we another example with alkaline earth metals, then ideally according to Fajan's rule BaSO4 should be more soluble than MgSO4. But this is not the case because as we go down the group the decrease in hydration energy is greater than the decrease in lattice energy and thus solublity decreases.

My question is that shouldn"t this explanation also apply for the alkali metals? Shouldn't the decrease in hydration energy be greater than the decrease in lattice energy here as well?

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  • $\begingroup$ Convenient reference for text/formula formatting: Notation basics / Formatting of math/chem expressions / upright vs italic // For more: Math SE MathJax tutorial. // Keep CH SE Q titles in plain text. $\endgroup$
    – Poutnik
    Oct 4, 2022 at 17:20
  • $\begingroup$ @Poutnik yes, I did mean that. Have edited that bit. $\endgroup$
    – Pumpkin_Star
    Oct 4, 2022 at 17:25
  • $\begingroup$ @Poutnik Also, 'Keep CH SE Q titles in plain text.' - the title had been typed in plain text at my end. Does it show something else at your end? $\endgroup$
    – Pumpkin_Star
    Oct 4, 2022 at 17:28
  • $\begingroup$ Note that barium hydroxide IS much more soluble than magnesium hydroxide, which is practically insoluble (and the exact opposite for sulfates). $\endgroup$
    – Poutnik
    Oct 4, 2022 at 17:48
  • $\begingroup$ @Poutnik may I ask why it is the opposite for sulphates. Also, I have edited my question. I checked my notebook and indeed, the exception was related to sulphates and not hydroxides. Sorry. $\endgroup$
    – Pumpkin_Star
    Oct 4, 2022 at 17:50

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