Is this question asking for the equivalence point of the titration curve?

At what point does the concentration of acetic acid equal the concentration of the acetate ion? What is the pH of the equimolar, ideal buffer solution at this point.

At first, I thought this question was asking about the point right before the titration curve shoots up. However, I don't recall ever learning about such a point.

  • $\begingroup$ Welcome to Chemistry.SE! Take the tour to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax. This appears to be a homework question, please share your thoughts and attempts towards the solution - although I am not sure if it is just a very basic question. Have a look at the wikipedia page of buffer solutions, I am certain it will address this issue. $\endgroup$ Commented Feb 10, 2016 at 6:16
  • $\begingroup$ The equivalence point is would not be what you are looking for. You are trying to find the half-equivalence point, where instead of all the acid being neutralized and converted to acetate, only half is. If you have a titration curve, this occurs at half the equivalence volume. Based on MaxW's formula below, what would the pH be when the concentrations of acetic acid and acetate are equal? $\endgroup$
    – Tyberius
    Commented Mar 15, 2017 at 15:19

1 Answer 1


If you were titrating acetic acid with a base, then the equivalence point would be where as much base has been added (i.e. moles of base vs moles acid) to the solution as there was acetic acid. This is essentially the point where all the acetic acid has been neutralized.

In general the "pKa" is the term used to define the point at which the protonated and unprotonated forms are equal. The pKa is from the negative log of the acid dissociation constant as given by the reaction: $$\ce{HA <=> H+ + A-}$$ and the equation: $$ K_\mathrm{a} = \dfrac{\ce{[H+] [A- ]}}{\ce{[HA]}}$$

where $\ce{[H+] = [A- ]}$ then: $$K_\mathrm{a} = \dfrac{\ce{[H+]^2}}{\ce{[HA]}}$$

and $$ \ce{pKa} = -\log \ce{Ka} = -\log \dfrac{\ce{[H+]^2}}{\ce{[HA]}}$$


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