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So I came across a question which asked to calculate the $\mathrm{pH}$ at which an indicator of given $\mathrm{K_{a}}$ would change colour. The solution was given in the following way

Since when the color change occurs $\ce{[HIn]=[In^{-}]}$,

As $\mathrm{K_a}= \left(\frac{[\ce{H^+} ][\ce{In^-}]}{[\ce{HIn}]}\right)$,

At $\mathrm{K_a}=[\ce{H+}]$ the colour change would occur.

And consequently from the $\ce{H+}$ concentration the $\mathrm{pH}$ was calculated, the solution imposes that during the colour change the equilibrium is attained, but how can that be since to observe colour change you add an acid or base which disturbs the initial equilibrium for which the $\mathrm{K_{a}}$ was defined and hence the now won't hold true.

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  • $\begingroup$ Hint: Indicators change their color not at Ka but rather pKa+1 and pKa-1. Sort this out by searching more on the web about the indicators. The dissociation constant is indeed a constant, it does not change with pH. It is only affected by temperature and pressure. $\endgroup$
    – ACR
    Commented Oct 6, 2020 at 15:22
  • $\begingroup$ I totally get that and that isn't my question whether or not the Ka changes but is rather why is the author taking the condition of equilibrium to plug the values of HIn and In-. $\endgroup$
    – Frustrated_lewispair
    Commented Oct 6, 2020 at 17:53
  • $\begingroup$ Ka value has to remain the same, regardless of the pH. The system is at equilibrium all the time no matter whether you add acid or base to an indicator solution. The rate of reaction is instantaneous. $\endgroup$
    – ACR
    Commented Oct 6, 2020 at 19:59
  • $\begingroup$ Note that indicator range is much dependent of the color transition and related eye sensitivity. Reported intervals are AFAIK always narrower than 2 pH units, like 1.3 ( methyl orange 3.1-4.4) - 1.8 ( phenolphthalein 8.2-10.0). The perceived center pH is biased toward the less intense colour. $\endgroup$
    – Poutnik
    Commented Oct 7, 2020 at 6:36

1 Answer 1

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First of all, the indicator does not change colour instantaneously at $\mathrm{pH = pK_a}$. The indicator is a different colour than it's conjugate base, and exists in an equilibrium with it in solution. For the colour change to be noticed, the concentration of the base has to be at least 10 times more than that of the conjugate acid, or vice versa. This is why there is a range of colour change(look at these pH charts), with the upper and lower bounds having a $\mathrm{pH}$ difference of $2$. As an example, here's a simplified reaction for Phenolphthalein ($\ce{Ph}$ is the phenolphthalein anion) $$\ce{\underset{\text{colourless}}{H2Ph} <=> 2H+ + \underset{\text{violet}}{Ph^2-}}$$

If you take a simpler, weak monobasic acid indicator $\ce{HIn <=> H+ + In-}$, the $\mathrm{K_a}$ is $$\mathrm{K_a} = \frac{\ce{[H+][In^-]}}{\ce{[HIn]}}$$

The relative concentration of the indicator and it's conjugate base is only determined by the concentration of the $\ce{H+}$ ion, as $\mathrm{K_a}$ is a constant.

Answering your second query, the author takes the condition of equilibrium, simply because the system is in equilibrium, that is $\mathrm{Q_a = K_a}$. It is assumed that the reaction is instantaneous or that titration is performed slowly enough so that the system is in equilibrium at every single instant.

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  • $\begingroup$ It still doesn't clear my confusion, I know that the system is assumed to go through the process slowly and that causes the system to be in equilibrium all the time but the Ka was defined for the concentration of the species that were a result of dissociation of the indicator only. When the pH changed due to adding of extra acid or base the concentration of the species also changed so how can you take Qa=Ka. $\endgroup$
    – Frustrated_lewispair
    Commented Oct 8, 2020 at 11:34
  • $\begingroup$ @buddy001 Because the extra acid reacted with the indicator and attained equilibrium, so $\mathrm{Q_a = K_a}$ $\endgroup$
    – Aniruddha Deb
    Commented Oct 8, 2020 at 12:30

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