# Is pK_{In} = pH at equivalence point still true for weak basic organic indicators?

I am reading Ostwald Theory of Titration which says that indicators are organic weak acids or bases. To prove the relation that $$\mathrm{p}K_\mathrm{In} = \mathrm{pH}$$, the textbook uses the following equilibrium as an example:

$$\ce{HIn(aq) <=> H^{+}(aq) + In^{-}(aq)}$$

which of course produces the the required result, but a thing I noted that it was for acid type indicator, so I decided to fill in the gap and do it for base type indicator myself, but it did not proceed as imagined:

Consider a weak organic base indicator $$\ce{InOH}$$, for it, the equilibrium will be:

$$\ce{InOH (aq) <=> In^{+} (aq) + OH^{-} (aq)}$$

and it's equilibrium constant then would be:

$$\displaystyle{K_\mathrm{In} = \frac{[\ce{In^+}][\ce{OH^-}]}{[\ce{InOH}]}}$$

then following the reasoning as done in previous case, for a sudden color change, $$[\ce{In^+}] = [\ce{InOH}]$$, so that,

$$\displaystyle{K_\mathrm{In} = [\ce{OH^-}]}$$

so that,

$$\displaystyle{\mathrm{p}K_\mathrm{In} = \mathrm{pOH}}$$

which inturn implies that,

$$\displaystyle{\mathrm{pH} = 14 - \mathrm{p}K_\mathrm{In}}$$

which doesn't seem to match with the previous result! Am I missing something here?

Note: I doubt that the condition $$[\ce{In^+}] = [\ce{InOH}]$$ might be the trouble, but I think it shouldn't be as in theory, the different color arises due to the respective indicator and their respective conjugate acid or bases.

## 3 Answers

It is true that acid-base indicators ($$\mathrm{pH}$$-indicators) are either weak acid or weak base. What I understood reading your question is that you have wrong impression about bases. To clear your view, not all bases contain $$\ce{OH-}$$ ions. Specially, most organic bases are weak and contain electronegative ion (e.g., $$\ce{N}$$ or $$\ce{O}$$) with at least one lone pair. This fact is well cleared by Poutnik's answer, thus, I'm not going to elaborate it more. Instead, I'd like to explain the action of a $$\mathrm{pH}$$-indicator (e.g., $$\ce{HIn}$$ in acidic form) to some depth.

Since Indicators ($$\ce{HIn}$$) are weak acids (or weak bases), in aqueous medium, it's in an equilibrium:

$$\ce{HIn + H2O <=> H3O+ + In-} \tag1$$ $$\therefore \ K_\mathrm{aIn} = \frac{[\ce{H3O+}][\ce{In-}]}{[\ce{HIn}]} \tag2$$ Taking log on both side followed by simplification gives: $$-\log [\ce{H3O+}] = -\log K_\mathrm{aIn} + \log \left(\frac{[\ce{In-}]}{[\ce{HIn}]}\right)$$ $$\text{Hence}, \ \mathrm{pH} = \mathrm{p}K_\mathrm{aIn} + \log \left(\frac{[\ce{In-}]}{[\ce{HIn}]}\right) \tag3$$

This is a resemblance of Henderson-Hasselbalch equation. The equation $$(3)$$ shows that when $$[\ce{HIn}] = [\ce{In-}]$$, $$\mathrm{pH} = \mathrm{p}K_\mathrm{aIn}$$.

Let's consider the indicator, Litmus, which is we commonly use in labs. Litmus is a weak organic acid (see the structure in bottom box of the diagram; $$\mathrm{p}K_\mathrm{aIn} = 6.5$$). The $$\color{green}{\text{green circle}}$$ shows the $$\ce{H}$$ of $$\ce{HIn}$$ molecule. The un-ionized litmus ($$\ce{HIn}$$) is $$\color{red}{\text{red}}$$ (below $$\mathrm{pH}$$ 4.5), whereas the ion part $$(\ce{In-})$$ is $$\color{blue}{\text{blue}}$$ (above $$\mathrm{pH}$$ 8.2):

Usually color change happens within a range of $$\mathrm{pH}$$. It is a rule of thumb that this range falls between the $$\mathrm{p}K_\mathrm{aIn}\pm 1$$. That means, acid color persists if $$[\ce{HIn}] = 10 \times [\ce{In-}]$$ and base color dominates when $$10 \times [\ce{HIn}] = [\ce{In-}]$$. For example, apply these values in the equation $$(3)$$:

$$\mathrm{pH} = \mathrm{p}K_\mathrm{aIn} + \log \left(\frac{[\ce{In-}]}{10 \times [\ce{In-}]}\right) = \mathrm{p}K_\mathrm{aIn} + \log \frac{1}{10} = \mathrm{p}K_\mathrm{aIn} -1$$

Similarly,

$$\mathrm{pH} = \mathrm{p}K_\mathrm{aIn} + \log \left(\frac{10 \times [\ce{HIn}]}{[\ce{HIn}]}\right) = \mathrm{p}K_\mathrm{aIn} + \log 10 = \mathrm{p}K_\mathrm{aIn} +1$$

Hence, where $$\mathrm{pH} = \mathrm{p}K_\mathrm{aIn}$$ is the most intense point during color change. For litmus, color change happens between the range, $$\mathrm{pH} = 5.5-7.5$$ since $$\mathrm{p}K_\mathrm{aIn} = 6.5$$ for litmus.

Let's see litmus in basic solution where $$\ce{In-}$$ ions are predominant ($$\color{blue}{\text{blue}}$$ solution). Now you start adding acid to the solution and you can see what would happen to the solution (equation $$(1)$$) by using Le Chatelier's Principle. Since excess hydronium ions disturb the equilibrium, it'd shift to left hand side of equilibrium to reduce excess hydronium ions by reacting $$\ce{In-}$$ ions. Eventually, solution changes the color to $$\color{red}{\text{red}}$$. Opposite would happens if you added hydroxide ions to the $$\color{red}{\text{red solution}}$$ in equilibrium.

The scaning UV spectra in the upper box of the diagram shows how color change happens in solution with $$\mathrm{pH}$$. The indicator here is bromothymol blue, which is yellow in the acidic medium absorbing light at about $$\pu{430 nm}$$. When $$\mathrm{pH}$$ has changed systematically with time (by adding known aliquot of $$\ce{OH-}$$ solution with in fixed time), the solution changed the color accordingly, and bromothymol blue has changed to its basic color blue, absorbing light at about $$\pu{620 nm}$$. Note that the peak at $$\pu{430 nm}$$ decreases while the peak at $$\pu{620 nm}$$ increases while $$\mathrm{pH}$$ increases.

A basic indicator is usually a Broensted-Lawry base ( accepting protons ) rather than Arrhenius base ( releasing hydroxide ions ):

$$\ce{B + H2O <=> BH+ + OH-}$$
or $$\ce{BH+ + H2O <=> B + H3O+}$$

depending on if the indicator is used in its base form or conjugate acid form (the latter is usually more stable and more soluble).

Regardless of the applied indicator form, $$\mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log{ \frac{\ce{[B]}}{\ce{[BH+]}}}$$

where $$K_\mathrm{a}$$ is the acidity constant of the conjugate acid.

The indicators are never in the form "InOH" with a covalence between "In" and OH, producing $$\ce{OH^-}$$ ions in water. If a molecule contains one OH group attached to a Carbon atom, it would be an acid, an alcohol, en enol or a phenol. These sorts of molecules are never releasing $$\ce{OH^-}$$ ions in water. On the contrary, they are weak acids.