I'm answering a problem where $\ce{KOH}$ and $\ce{KOAc}$ dissociate nearly as much as each other in separate aqueous solutions at equal concentrations. The problem essentially assumed both completely dissociate.

It's my understanding that weak bases are not meant to dissociate completely, yet $\ce{KOAc}$ is a weak base. However, $\ce{KOAc}$, and $\ce{KOH}$, are both salts.

I understand that a salt is the combination of a charged anion and cation.

The rules that determine solubility state that Group I elements and acetates are always soluble. Thus, it does make sense that $\ce{KOAc}$ would completely dissolve.

Is it always true that if a weak base is also a highly soluble salt, it will still dissociate completely into its component ions, like a strong base would? Would this trend be generalizable to a weak acid that is also a highly soluble salt?

edit: I was trying to simplify the problem, but I made an error in 'transcribing' it. The problem stated that separate solutions of $\ce{KOAc}$ and $\ce{KOH}$ essentially produced the same amount of ions in solution (ie., they conducted electricity to an equal extent). After reading on solubility and dissociation, both compounds dissolved equally as much, producing ionic compounds. But only the strong base $\ce{OH-}$ dissociates completely. Although relative ratios of ions are different in each solution, the amount of ions remains the same.

My apologies for a badly worded question.

  • 2
    $\begingroup$ What do you mean by "dissociation of the weak base"? There are two separate but coupled equilibria here, and I think you are confused because you're mixing the two processes together. $\endgroup$
    – Zhe
    Apr 1, 2019 at 14:50
  • $\begingroup$ I'm aware that weak bases do not fully dissociate. But, if the weak base is a highly soluble salt, then it does dissociate completely. I'm assuming you mean the 2 separate but coupled equilibria are acid-base and salt dissociation? What I'm wondering is why the fact that $\ce{KOAc}$ is a soluble salt takes precedence over the fact that it is a weak base in determining whether dissociates. $\endgroup$
    – chompion
    Apr 1, 2019 at 15:08
  • 1
    $\begingroup$ You seem to be confusing solubility with dissociation. In solution both the KOAc and K$^+$ and OAc$^-$ exist in equilibrium, but for KOH only K$^+$ and OH$^-$ exist. See chemistry.stackexchange.com/questions/60068/… and chemistry.stackexchange.com/questions/100346/… for more details. $\endgroup$
    – porphyrin
    Apr 1, 2019 at 15:15
  • $\begingroup$ "I'm aware that weak bases do not fully dissociate." This statement doesn't quite make sense, and I think is at the root of your confusion. $\endgroup$
    – Zhe
    Apr 1, 2019 at 16:58
  • 1
    $\begingroup$ I give you acetate. It's an anion. It doesn't dissociate into anything. $\endgroup$
    – Zhe
    Apr 1, 2019 at 17:25

1 Answer 1


Weak bases, that are not salts, do not dissociate, but partial react with water as $$ \ce{R-NH2 + H2O <<=> R-NH3^+ + OH-}$$

$\ce{KOAc}$ is not weak base, but a salt.
The weak base is $\ce{OAc-}$, that is created by dissociation of the salt $\ce{KOAc}$ in water, which undergoes protonization: $$\begin{align} \ce{KOAc &-> K+ + OAc-} \\ \ce{OAc- + H2O &<=> HOAc + OH-} \\ \end{align}$$

$\ce{KOH}$ is not a salt, but a base. If it were a salt, I would be curious what are the related base and an acid that create this salt.

  • $\begingroup$ $\ce{KOH}$ is a potassium salt. It dissociates into ions $\ce{K+}$ and $\ce{OH-}$ the same way salt $\ce{KOAc}$ does. $\endgroup$
    – chompion
    Apr 1, 2019 at 15:02
  • 1
    $\begingroup$ Salts dissociate, but not everything, what dissociates, is a salt. Would you consider sulfuric acid a salt ? Rather, $\ce{KOH}$ is a ionic compound. $\endgroup$
    – Poutnik
    Apr 1, 2019 at 15:12
  • 1
    $\begingroup$ @chompion If you consider $\ce{KOH}$ as a salt, what is the acid and the base that mutually react to create the "salt" $\ce{KOH}$ ? $\endgroup$
    – Poutnik
    Apr 1, 2019 at 15:18
  • $\begingroup$ The strong bases we usually work with are confusing in terms of the Brønsted-Lowry definition. It would be better to start with something like $\ce{CH3CH2O-}$ or butyl lithium as the first strong base (clearly proton acceptors) and then discuss NaOH (makes hydroxide, does not accept protons unless you count self-exchange) later. $\endgroup$
    – Karsten
    Apr 1, 2019 at 18:51

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