To be honest, I just don't get it. I had this question while reading a Chemguide article, on this topic: enter image description here

I think my confusion is caused by the way I look at pKa of acid-base indicator. Should you think of it more as a reaction quotient. Because technically, at equilibrium, the two concentration aren't the same so cannot be cancelled out.

  • $\begingroup$ Would you add a link to the Chemguide page so that we can see if the way it is being explained there is part of what is tripping you up? $\endgroup$ – Cohen_the_Librarian May 2 '15 at 2:07
  • $\begingroup$ Just search up acid base indicator Chemguide $\endgroup$ – most venerable sir May 2 '15 at 9:37

Look at the following equation:

$$\mathrm{pH} = \mathrm pK_{\mathrm a} + \log\frac{[\ce{A^-}]}{[\ce{HA}]}$$

At the half equivalence point, say we have 10 moles of $\ce{WA},$ and so there will be 5 moles of $\ce{SB}$ as the name suggests ("half equivalence"). So we know that the 10 moles will be neutralized to 5 moles of $\ce{WA}$ by the $\ce{SB}.$ If we do this then the $\ce{CB}$ is also now equal to the $\ce{WA}$ since it will increase by 5 moles.

$$\ce{[WA]~ =~ [CB]}.$$


\begin{align}\mathrm{pH} &= \mathrm pK_{\mathrm a} + \log 1\\ &=\mathrm pK_{\mathrm a}\;.\end{align}

So this is why the the $\mathrm{pH} = \mathrm pK_{\mathrm a}$ at the Half-Equivalence Point.


  • SB : Strong Base. (eg. $\ce{NaOH}$)
  • CB : Conjugate Base. (eg. $\ce{Ac^{-}}$)
  • WA : Weak Acid. (eg. $\ce{HAc}$)

NOTE: We can never be 100% accurate here, this is just used to derive that $\mathrm{pH} = \mathrm pK_{\mathrm a}$

  • $\begingroup$ No I am talking about the endpoint-which in most case is the equivalence point. $\endgroup$ – most venerable sir May 1 '15 at 23:26
  • $\begingroup$ Yeah but the pKa equals the pH at the H-E point $\endgroup$ – Asker123 May 1 '15 at 23:29
  • $\begingroup$ But we are talking about an indicator which itself is a weak acid. I understand what you said is correct, for other situations, but isn't exactly what I am looking for. $\endgroup$ – most venerable sir May 1 '15 at 23:31
  • $\begingroup$ OK, I will delete my answer or turn it into a community wiki. $\endgroup$ – Asker123 May 1 '15 at 23:32
  • $\begingroup$ Wait, what I need to know is what causes the two concentration to be equal, when the equivalence point is reached, equal moles of acid and base are added and react. $\endgroup$ – most venerable sir May 1 '15 at 23:34

Why does pKa of a acid-base indicator equal to the pH when the equivalence point is reached?

Consider titrating an acid with NaOH. There two pKa values to consider:

  • The indicator itself has a pKa value
  • The acid being used has a pKa value

So titrating acetic acid (pKa = 4.76) with NaOH, then phenolphthalein (pka = 10) is a good indicator to use since it will be colored after the neutralization reaction is complete. However bromophenol blue (pKa = 4.75) wouldn't work because it would turn blue before the neutralization reaction is complete.

So the idea is that the indicator should indicate when the reaction is complete, not the equivalence point of the reaction.

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well, weak acids ionize partially n thus we should expect a half ionization . if the acid is xmole at the start, at the end point we should expect the conjugate base to be x/2mole leaving behind x/2mole of the acid. hasselbach equation applies to weak acid n bases

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  • 1
    $\begingroup$ This doesn't really answer the question. It is also very poorly written. Please use proper syntax and avoid non-standard abbreviations like 'n' for 'and'. $\endgroup$ – bon May 12 '16 at 11:57

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