2
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ 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? $\endgroup$ May 2, 2015 at 2:07
  • $\begingroup$ Just search up acid base indicator Chemguide $\endgroup$ May 2, 2015 at 9:37
  • $\begingroup$ This is the best website I've found re explaining this concept re why indicator pH needs to be within +/- one pH unit of equivalence point: groups.chem.ubc.ca/courseware/pH/section15/index.html check it out $\endgroup$ Aug 1, 2021 at 8:35

4 Answers 4

3
$\begingroup$

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.

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

Thus,

\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.


Vocabulary:

  • 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 $\mathrm{pH} = \mathrm pK_{\mathrm a}$

$\endgroup$
9
  • $\begingroup$ No I am talking about the endpoint-which in most case is the equivalence point. $\endgroup$ May 1, 2015 at 23:26
  • $\begingroup$ Yeah but the pKa equals the pH at the H-E point $\endgroup$
    – Asker123
    May 1, 2015 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. $\endgroup$ May 1, 2015 at 23:31
  • $\begingroup$ OK, I will delete my answer or turn it into a community wiki. $\endgroup$
    – Asker123
    May 1, 2015 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. $\endgroup$ May 1, 2015 at 23:34
2
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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

It does not. The equivalence point depends on the reaction you are monitoring with the indicator. The $\mathrm{p}K_\mathrm{a}$ of a given indicator is given.

So you would first figure out what the pH at the equivalence point is. Then you would choose an indicator that changes color near that pH. Now, you can ask about the relationship of the pH of color change and the $\mathrm{p}K_\mathrm{a}$ of the indicator.

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.

The pH indicator reaction reaches equilibrium after less than a couple of seconds. So if you stop titrating and swirl the reaction mixture a bit, you are at equilibrium and the indicator color reflects the pH. This happens before, at and after the equivalence point (of the titration reaction).

When you are just in the middle of the transition between one and another color of the indicator, you have the same concentration of protonated and deprotonated indicator. Under those circumstances, the two concentrations do cancel out and the pH is equal to the $\mathrm{p}K_\mathrm{a}$.

To summarize, you choose an indicator such that at the equivalence point of your titration reaction, the indicator reaction is roughly halfway between colors. You also make sure the concentration of the indicator is much much lower than that of the analyte so that analyte and titrant determine the pH and the indicator reports on it without skewing the result.

$\endgroup$
-2
$\begingroup$

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

$\endgroup$
1
  • 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, 2016 at 11:57

Your Answer

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

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