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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.

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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):

pH Indicators

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.

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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.

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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.

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