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Phenolphthalein $(\ce{HIn})$ is slightly acidic. Why doesn't it completely break down into $\ce{H+}$ and $\ce{In-}$ and also reacting with all of the $\ce{OH-}$ ion present, unlike $\ce{NaOH}$ and $\ce{HCOOH}$ reacting completely to form $\ce{HCOONa}$ and $\ce{H2O}$ only?

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    $\begingroup$ Who told you that it is not neutralized in basic medium? It is deprotonated, indeed. The pink color is that of the anion. Look at the Wikipedia chart. It explains it beautifully. en.wikipedia.org/wiki/Phenolphthalein $\endgroup$
    – AChem
    Oct 3, 2022 at 18:51
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    $\begingroup$ @AChem unless you go too far, and the phenolphthalein overreaction and loses its color. This is also explained in Wikipedia. We can't trust sodium hydroxide, bring on milk of magnesia! $\endgroup$ Oct 3, 2022 at 19:27
  • $\begingroup$ what would happen with the other half? @Poutnik $\endgroup$ Oct 4, 2022 at 5:38
  • $\begingroup$ HPh is weaker. the basic medium is stronger. now why wont all of them react altogether unlike 30 ml 1M HCOOH with 30 ml 1M NaOH completely neutralize system. we dont use pKa in the last case maybe. @Poutnik $\endgroup$ Oct 4, 2022 at 6:23
  • $\begingroup$ so it is theoretically possible for more amount of strong base to react with less amount of weak acid to take all of its H+ away and leave only the anions in the system but the exact same scenario wont be for phenolphthalein practically for some other reason? $\endgroup$ Oct 4, 2022 at 6:54

1 Answer 1

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Depending of the weak acid $\mathrm{p}K_\mathrm{a}$, the degree of the acid neutralization $\alpha$ depends on $\mathrm{pH}$, following famous equation:

$$\mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log {\left(\frac{[\ce{A-}]}{[\ce{HA}]}\right)}$$

related to the chemical equilibrium reaction:

$$\ce{HA(aq) + OH-(aq) <=> A-(aq) + H2O}$$

$\mathrm{pH}$ $\alpha$
$\mathrm{p}K_\mathrm{a} - 3$ $\approx 0.001$
$\mathrm{p}K_\mathrm{a} - 2$ $\approx 0.01$
$\mathrm{p}K_\mathrm{a} - 1$ $\frac {1}{11}$
$\mathrm{p}K_\mathrm{a}$ $0.5$
$\mathrm{p}K_\mathrm{a} + 1$ $\frac {10}{11}$
$\mathrm{p}K_\mathrm{a} + 2 $ $\approx 0.99$
$\mathrm{p}K_\mathrm{a} + 3 $ $\approx 0.999$

For HCOOH, $\mathrm{p}K_\mathrm{a}=3.75$
For phenolphthaleinum, $\mathrm{p}K_\mathrm{a} \approx 9$

Note that the colour of ionized phenolphthaleinum disappears at very high $\mathrm{pH}$ by slow transition due nucleophilic $\ce{OH-}$ addition.

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  • $\begingroup$ Not sure that last change is irreversible. My reading is it's just slow. $\endgroup$ Oct 4, 2022 at 10:14
  • $\begingroup$ @OscarLanzi I have always thought until recently it is the second transition, but I have recently (not handy) read it is some structural change. So, perhaps it is both, I have not done any testing. $\endgroup$
    – Poutnik
    Oct 4, 2022 at 10:22
  • $\begingroup$ Wikipedia says the structural change is that at high pH, a hydroxide ion adds as a nucleophile instead of extracting another proton; such a nucleophilic addition is in principle reversible but expected to be slower than proton transfer. $\endgroup$ Oct 4, 2022 at 10:54
  • $\begingroup$ @OscarLanzi You may be right. $\endgroup$
    – Poutnik
    Oct 4, 2022 at 11:00

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