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Please explain to me why $\ce{KSO4^-}$ doesn't exist as a molecular ion in (aqueous) solution. I couldn't find a reason why.

I assumed if you took one $\ce{SO4^{2-}}$ and one $\ce{K^+}$ ion, then only that one would form the molecular ion $\ce{KSO4^-}$.

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You can make an aqueous solution of $\ce{KSO_4^{-}}$ (and no other cations), in the sense that you can make a regular $\ce{K2SO4}$ solution which lacks some potassium ions. By doing so, you are producing a solution with a net electric charge. The higher the net electric charge, the less stable the solution is, as electrons will start redistributing into the surroundings in order to correct the charge imbalance to a high precision. While a net charge is possible, it seems unlikely to be able to produce a solution which contains even close to a 1 µM excess of an ion, and I do not know if electrically imbalanced solutions can be prepared quantitatively. Take at look at the answers and comments to this previous question for more information.

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In solution, $\ce{K2SO4}$ is going to fully dissociate into $\ce{2K^+}$ and $\ce{SO_4^2-}$, see the Wikipedia article on dissociation.
In solution, the ions will orient towards each other so as to functionally neutralize the charges, but you are still going to have two $\ce{K+}$ ions orienting towards each $\ce{SO4^2-}$ ion, unless there is something else in solution that is "distracting" one of the $\ce{K+}$ ions.

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  • $\begingroup$ Ok but what if I remove the other K^+ ion? There will be only one ion left, oriented towards the SO4^2-. $\endgroup$ – guest Apr 24 '14 at 4:40
  • $\begingroup$ @guest How are you planning on removing it? Cuz if you were to add something like NaCl (a simple ionic compound) and one of the K+ ions went with the Cl, there would be Na+ which would orient towards the anion. The key principle is that the charges have to balance, even in solution, unless you are applying a charge to the solution. $\endgroup$ – Cohen_the_Librarian Apr 24 '14 at 5:02
  • $\begingroup$ You can mask $\ce{K+}$ with crown ether removing it from the equilibrium. However, you are right. But consider $$\ce{K2SO4 <=>[H_2O] K+ + KSO4-},\\\ce{KSO4- <=>[H_2O] K+ + SO^{2-}4},$$ if it is concentrated enough, you will always have all three sulphate molecules present. (I am completely unsure about the degree of dissociation.) $\endgroup$ – Martin - マーチン Apr 24 '14 at 7:52
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You could have something like $\ce{KSO4^-}$ in the form of ion pairs, in a suitably concentrated potassium sulfate solution. But, of course, such ion pairs are easily dissociated. Add $\ce{HCl}$, for instance, and the ion pair releases its potassium ion as the weakly basic sulfate ion is protonated to bisulfate.

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