Acetic acid has a $K_\mathrm{a}$ of $\pu{1.8e-5}$. What is the equilibrium constant for the neutralization of this acid with $\ce{NaOH}$?

Given acetic acid

$$\ce{HC2H3O2 + H2O <=> C2H3O2- + H3O+} \qquad K_\mathrm{a} = \pu{1.8e-5}$$

$$\ce{HC2H3O2 + OH- <=> C2H3O2- + H2O}$$

So, if we do $K_\mathrm{w} = K_\mathrm{a}K_\mathrm{b}$, then we get $K_\mathrm{b} = \pu{5.55e-10}$. How do I use this to find $K_\mathrm{eq}$?

I know how to find $K_\mathrm{eq}$ using concentration, but I am unsure how to approach this further. The hint says consider the ion product of water, but what does this mean?


Sodium hydroxide is a strong base and is supposed to be fully dissociated. You then should've started by writing down the neutralization reaction itself for which you have to determine the equilibrium constant $K$ and unravel a tangle from there:

$$\ce{HOAc + OH- <=> OAc- + H2O}$$

$$K' = \frac{[\ce{OAc-}][\ce{H2O}]}{[\ce{HOAc}][\ce{OH-}]}$$

Since $[\ce{H2O}] = \text{const}$ (reaction medium), $K'[\ce{H2O}] = K = \text{const}$:

$$K = \frac{[\ce{OAc-}]}{[\ce{HOAc}][\ce{OH-}]}$$

By multiplying both numerator and denominator by $[\ce{H+}]$, you can find out that the constant for the neutralization of a weak acid solely depends on the relation between its dissociation constant $K_\mathrm{a}$ and ionic product of water $K_\mathrm{w}$:

$$K = \frac{\color{red}{[\ce{OAc-}][\ce{H+}]}}{\color{red}{[\ce{HOAc}]}[\ce{OH-}][\ce{H+}]} = \frac{\color{red}{K_\mathrm{a}}}{K_\mathrm{w}}$$

For acetic acid:

$$K = \frac{\pu{1.8e-5}}{\pu{1e-14}} = \pu{1.8e9}$$

  • $\begingroup$ wait so is keq always equal to ka/kw and would the same apply for kb/kw? From my understanding we setup the reaction kb and saw that it is similar to keq but is only lacking H+. So by multiplying in our H+ we see that it is also has our ka equation so we then can say ka/kw? Thank you for your help, truly I am so confused in this chapter and its hard stringing all these k's together. $\endgroup$ – Avarosa Feb 23 at 7:54
  • $\begingroup$ No, Keq should be determined from the equilibrium for each given situation. Remembering some formulas won't help, or at least you should understand on what premise they've been derived. That's why I edited the title of your question to cover the cases where this formula works: weak monobasic acids neutralized by strong alkali. $\endgroup$ – andselisk Feb 23 at 7:58
  • $\begingroup$ Oh I think I understand now, cuz keq=oh*h3o and that is equal to ka and then when we solve for the equation to isolate the k constant then we see its ka/keq. So I see how you thought about manipulating the variables and now it makes sense a lot more to me now. Thank you, you are the GOAT! $\endgroup$ – Avarosa Feb 23 at 18:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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