Q19.40P - Chemistry: Molecular Nature Of Matter And Change

Question

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

Step-by-Step Solution

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 1: 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.

Step 2: 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,  $\mathrm{pH} \mathrm{ = } {\mathrm{pK}}_{\mathrm{a}}\mathrm{ +\; /\; -\; 1}$ gives a pH range of around 2 units.