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I am reading Ionic Equilibrium by James N. Butler. Using silver(I) chloride in excess aqueous chloride as an example it speaks of several forms: $\ce{AgCl (s)}$ i.e. the ionic crystalline salt, $\ce{Ag+(aq)}$ and $\ce{Cl^{-}(aq)}$ the aqueous ions from dissociation, and finally $\ce{AgCl(aq)}$, which Butler purports to be "dissolved covalent silver chloride molecules as distinct from solid salt or dissolved ions."

Does this "molecule" of $\ce{AgCl}$ really exist in solution? If so, is there a way to quantify the concentrations of this covalent species? Equilibrium solubility product constants seem to give the concentration of the aqueous ions only.

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  • $\begingroup$ For your second question: If you can determine the quantity of ions present in solution, you should easily determine the other quantity, i.e. the "covalent dissolved" AgCl. $\endgroup$ – CHM Sep 8 '12 at 1:57
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Some amount of $[Ag Cl (H_2O)]$ particles should exist in solution from pure thermodynamic point of view. See, $Ag^+$ ion can coordinate two ligands, and in solution both $Cl$ and $H_2O$ are present. However, it should be really hard to detect trace amounts of this particle, as $AgCl$ is almost insoluble. In case one can get concentrated salt solutions, similar particles may exist and hints for their existence may be got from conductivity measurements and osmotic pressure. Sometimes a direct study using NMR is possible, which can give some structural information if used properly (see 2D-NMR methods). As I remember, $Cr(CH_3COO)_2$ hydrate is slightly soluble in water but undergoes little to no dissociation.

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  • $\begingroup$ So...It sounds like the AgCl (aq) exists but a near or non-detectible concentration. So there is no need to account for it in equilibrium calcs? $\endgroup$ – Jason Waldrop Sep 9 '12 at 0:08
  • $\begingroup$ It depends, there is no way to figure it without problem and numeric data for it. Usually, when particle concentration is very small, you can drop it's concentration from equation where it is part of sum with other much bigger operand, but not from other equations. $\endgroup$ – permeakra Sep 9 '12 at 5:58

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