2
$\begingroup$

I am trying to calculate the equilibrium constant of the following equilibrium:

$$\ce{CH3COOH (aq) + OH- (aq) <=> CH3COO- (aq) + H2O (l)}$$

I am aware that this reaction essentially goes to completion (the position of the equilibrium lies far to the right). However, I still would like to calculate the equilibrium constant.

I know that $K_\text{eq} = K_\mathrm{a}(\ce{CH3COOH}) \times K_\mathrm{b}(\ce{OH-)} / K_\mathrm{w}$ (I have derived this from the definitions $K_\mathrm{a}$ and $K_\mathrm{b}$). I know that $K_\mathrm{a}(\ce{CH3COOH}) = 1.8\times10^{-5}$ and that $K_\mathrm{w} = 10^{-14}$. However, I am struggling to determine what $K_\mathrm{b}(\ce{OH-})$ is.

I learnt that $K_\mathrm{b}$ is called the "base ionization constant." However, $\ce{OH-}$ is a base, but it is already an ion! So, how can something that is already an ion, ionize? According to this webpage, $\ce{OH-}$ has a $K_\mathrm{b}$ of 1.0. However, what is the equilibrium that led to this number? Is it $\ce{OH- + H2O <=> H2O + OH-}$?

Moreover, if the $K_\mathrm{b}$ of $\ce{OH-}$ is 1.0, is $\ce{OH-}$ considered to be a weak base? Generally, strong bases have a $K_\mathrm{b}$ that is much greater than 1.0. On the other hand, if bases that contain hydroxide ions, such as sodium hydroxide and potassium hydroxide, are strong bases, shouldn't the hydroxide ion itself also be a strong base?

Improve this question
$\endgroup$
2
  • $\begingroup$ Interesting question! Though in the third step you've written $K_\text{eq} = K_\mathrm{a}(\ce{CH3COOH}) \times K_\mathrm{b}(\ce{OH-)} / K_\mathrm{w}$ and said that "I have derived this from the definitions $K_\mathrm{a}$ and $K_\mathrm{b}$" That means you know the definition of $K_\mathrm{b}$ (base dissociation constnat) for $\ce{OH-}$ ion, don't you? If not, then how did you insert it into that formula in the first place? $\endgroup$
    – Gaurang Tandon
    Jun 7, 2018 at 3:59
  • $\begingroup$ Good point. I shouldn't have made that assumption. I have used the equation Ka(HA) * Kb(B) = Keq * Kw in the past for acid base reactions of the form HA + B <=> BH+ + A-, and I assumed that it would work for the acetic acid - hydroxide ion equilibrium. $\endgroup$
    – Michael Chu
    Jun 8, 2018 at 3:16

1 Answer 1

Reset to default
2
$\begingroup$

Firstly, to derive this equilibrium constant ($K_\text{eq}$), you don't need to bother about $K_\mathrm{b}$ of $\ce{OH-}$. There is a a far simpler way of doing it.

So from the equation, $\ce{CH3COOH(aq) + OH-(aq) -> CH3COO-(aq) + H2O(l)}$, you can easily write, $$K_{\text{eq}} = \frac{\ce{[CH_3COO^-]}}{\ce{[CH_3COOH][OH-]}} = \frac{\ce{[CH_3COO^-]}[\ce{H^+}]}{\ce{[CH_3COOH]}} \times \frac{1}{\ce{[H^+][OH^-]}}= \frac{K_{\mathrm{a}}(\mathrm{CH_3COOH})}{K_\mathrm{w}}$$ So, here no $K_\mathrm{b}$ of $\ce{OH^-}$ arises and therefore we don't have to bother about it.

Secondly, if we want to find out $K_\mathrm{b}$ of $\ce{OH^-}$, At $\pu{25^o C}$, $\mathrm{p}K_\mathrm{w} = 14$, and from the equation of dissociation of water, $\ce{H2O(l) <=> H+(aq) + OH-(aq)}$, we can consider water as a weak acid, and $K_\mathrm{a} = \ce{[H^+][OH^-]} = K_\mathrm{w}$, so, $\mathrm{p}K_\mathrm{a}= \mathrm{p}K_\mathrm{w}$, and from the relation $\mathrm{p}K_\mathrm{a} + \mathrm{p}K_\mathrm{b} = \mathrm{p}K_\mathrm{w}$ (note that in the relation the definitions are that $\mathrm{p}K_\mathrm{a}$ is the acid dissociation constant of a weak acid and $\mathrm{p}K_\mathrm{b}$ is the base dissociation constant of its conjugate base). The conjugate base of water is $\ce{OH-}$, so $\mathrm{p}K_\mathrm{b}=\mathrm{p}K_\mathrm{w} - \mathrm{p}K_\mathrm{a} = 0$ as, both are same for water. So, $K_\mathrm{w}= 10^{-\mathrm{p}K_\mathrm{a}} = 10^0 =1.$ Now, you can see why $K_\mathrm{b}$ of $\ce{OH-}$ is 1 at $\pu{25^o C}$

Thirdly, yes, $\ce{OH^-}$ can't be said as a strong base. It's a moderately strong base. And, the concept that $\ce{NaOH}$ or $\ce{KOH}$ being a strong base should imply $\ce{OH^-}$ should be strong is a totally wrong concept. There is no logic behind it. Those are ionic compounds and completely dissociates and, those are Arrhenius bases. But to become a base, $\ce{OH^-}$ should accept a proton, and hence its Brønsted base, and there can't be a comparison as such. If you have a weak Brønsted acid, its conjugate base must be strong and vice versa. But here the cases are completely different, and there can't be any direct implication.

Improve this answer
$\endgroup$
4
  • $\begingroup$ Are NaOH (aq) and KOH (aq) also Bronsted-Lowry bases, since technically OH- (aq) is a "part" of NaOH (aq)? $\endgroup$
    – Michael Chu
    Jun 8, 2018 at 2:32
  • $\begingroup$ Not at all. If they have to behave as Bronsted-Lowry Base, the whole molecule should undergo reaction like $$\ce{NaOH + H^+ <=> NaOH2^+}$$ Does such equilibrium ever exist ??? $\endgroup$
    – Soumik Das
    Jun 8, 2018 at 8:49
  • 1
    $\begingroup$ True, that equilibrium does not exist. But NaOH (aq) is not a molecule, right? Isn't it just a bunch of Na+ ions and OH- ions floating around in water? $\endgroup$
    – Michael Chu
    Jun 8, 2018 at 21:07
  • $\begingroup$ That's the point. Now, you see the definition of Bronsted Bases... Is it allowed to first dissociate into smaller molecules and then get protonated ?? $\endgroup$
    – Soumik Das
    Jun 9, 2018 at 5:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.