Question

Why does the color change of an indicator take place over a range of about 2 pH units?

Short answer

As,$\text{log(10)}\text{=}\text{1 or log(.1)}\text{=}\text{- 1}$ can be used as the $\text{log}\frac{\text{HIn}}{{\text{In}}^{\text{-}}}$. As a result, $\mathrm{pH}={\mathrm{pK}}_{\mathrm{a}}\pm 1$.

Step-by-step solution

Define indicators

A material that can be added to the reaction mixture to show the titration's equivalence point. pH is frequently measured with coloured indicators. In a basic environment, these indications are a different colour than in an acidic environment.

Explanation

In order to see a distinct colour in a mixture of two colours, one of the colours must be 10 times as intense as the other.

The concentration ratio is calculated as follows:

$\frac{\text{[HIn]}}{\text{[In - ]}}$

The concentration ratio$\frac{\text{[HIn]}}{\text{[In - ]}}$ for one of these to be the case must be either larger than ten or less than one tenth.

$\text{i.e.}\frac{\text{[HIn]}}{\text{[In - ]}}\text{> 10}$or$\frac{\left[\mathrm{HIn}\right]}{[\mathrm{In}-]}<1/10$

$\text{pH}\text{=}{\text{pK}}_{\text{a}}\text{+ log}\left(\frac{\text{[HIn]}}{\text{[In - ]}}\right)$

Then, when the $\text{[HIn]/[In - ] = 10}$,

$\text{pH}\text{=}{\text{pK}}_{\text{a}}\text{+ 1}$

And when the $\text{[HIn]/[In - ] = 10}$,

$\text{pH}\text{=}{\text{pK}}_{\text{a}}\text{- 1}$

Therefore,role="math" localid="1663363455551" $\mathrm{pH}={\mathrm{pK}}_{\mathrm{a}}\mathrm{+\; /\; -\; 1}$ gives a pH range of around 2 units.