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In the case of compounds of polar molecules dissolved in water, I know it's to do with the formation of permanent dipole force of interactions between the water molecules and the polar molecules. But how exactly does dissolution happen? I've thought about it, and the only reasoning that comes to mind is that the London forces between the polar molecules break, as it is weaker than the force between the water and polar molecule, although I'm not sure if this is the reason why..

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    $\begingroup$ Not only London forces (induced-induced), but Debye ( permanent-induced) and Keesom ( permanent-permanent) forces as well - being together van der Waals forces. $\endgroup$
    – Poutnik
    Jan 19, 2022 at 8:48
  • $\begingroup$ It might read like a nit pick, however (in standard lab chemistry), we dissolve compounds (which might consist of polar/apolar molecules) in solvents. But dissolving e.g., glucose in water, does not dissolve the molecules into their parts. In case of succrose, though, the split into glucose and fructose is a reaction (hydrolysis), a different process (break of a chemical bond). $\endgroup$
    – Buttonwood
    Jan 19, 2022 at 9:14
  • $\begingroup$ Opposite charges attract each other (situation in the solid, undissolved state). Water (high dielectric constant) is able to attenuate this attractive forces drastically (the state of solution). Then it is thermodynamics' book keeping of enthalpy and entropy to favour one, or the other state, or an equilibrium. $\endgroup$
    – Buttonwood
    Jan 19, 2022 at 9:19

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Dissolving a substance in water means breaking hydrogen bonds. It requires energy. If enough new hydrogen bonds are made between solute molecules and water, this dissolution is energetically favorable. This happens when the solute is polar. If not enough new hydrogen bonds are made, the entropy effect is not sufficient to counterbalance the enthalpy effect : the substance will not be soluble in water.

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