91Ó°ÊÓ

The intensity of light absorbed by a substance dissolved in a totally transmitting solvent is directly proportional to the substance's concentration and the route length travelled by the light while travelling through the solution, according to Beer-law. Lambert's

It is mathematically expressed as.

$$\begin{gathered} \mathit{A}\mathbf{}\mathbf{=}\mathit{l}\mathit{o}\mathit{g}\mathbf{}\left(\frac{I_0}{I}\right) \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathit{Î\mu }\mathit{I}\mathit{c} \end{gathered}$$

Here,

$$\begin{gathered} \mathbf{A}\mathbf{=}\mathbf{Absorbance} \\ {\mathbf{l}}_{\mathbf{0}}\mathbf{=}\mathbf{intensithy}\mathbf{}\mathbf{of}\mathbf{}\mathbf{incident}\mathbf{}\mathbf{radiation} \\ \mathbf{l}\mathbf{=}\mathbf{intensity}\mathbf{}\mathbf{of}\mathbf{}\mathbf{trasmitted}\mathbf{}\mathbf{radiation}\mathbf{} \\ \mathbf{Î\mu }\mathbf{=}\mathbf{molar}\mathbf{}\mathbf{extinction}\mathbf{}\mathbf{co}\mathbf{-}\mathbf{efficient} \\ \mathbf{I}\mathbf{=}\mathbf{pathlength}\mathbf{}\mathbf{of}\mathbf{}\mathbf{the}\mathbf{}\mathbf{light} \\ \mathbf{c}\mathbf{=}\mathbf{concentration}\mathbf{}\mathbf{of}\mathbf{}\mathbf{the}\mathbf{}\mathbf{solution} \end{gathered}$$

(a)

Consider the indicator Hln, it dissociates as,

$\mathbf{HIn}\mathbf{}\stackrel{{\mathbf{K}}_{\mathbf{a}}}{\mathbf{⇌}}{\mathbf{H}}^{\mathbf{+}}\mathbf{}\mathbf{+}\mathbf{}{\mathbf{In}}^{\mathbf{-}}$

Here,

$$\begin{gathered} \mathit{Î\mu }\mathbf{=}\mathbf{2080}{\mathit{M}}^{\mathbf{-}\mathbf{1}}\mathit{c}{\mathit{m}}^{\mathbf{-}\mathbf{1}}\mathbf{}\mathit{f}\mathit{o}\mathit{r}\mathbf{}{\mathbf{HIn}}^{\mathbf{-}\mathbf{1}} \\ \mathit{Î\mu }\mathbf{=}\mathbf{14200}{\mathit{M}}^{\mathbf{-}\mathbf{1}}\mathbf{}\mathbf{for}\mathbf{}{\mathbf{In}}^{\mathbf{-}\mathbf{1}} \\ \mathit{l}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{}\mathit{c}\mathit{m} \end{gathered}$$

$$\begin{gathered} \mathit{p}\mathit{H}\mathbf{=}\mathbf{}{\mathbf{pK}}_{\mathbf{a}}\mathbf{+}\mathbf{log}\frac{\left|{\mathrm{In}}^{-}\right|}{\left[\mathrm{HIn}\right]} \\ \mathbf{6}\mathbf{.}\mathbf{23}\mathbf{=}\mathbf{}{\mathbf{pK}}_{\mathbf{a}}\mathbf{+}\mathbf{log}\frac{\left(1.84×{10}^{-4}\right)\mathbf{-}\left(1.44×{10}^{-4}\right)}{\left(1.44×{10}^{-4}\right)} \\ {\mathbf{pK}}_{\mathbf{a}}\mathbf{=}\mathbf{6}\mathbf{.}\mathbf{23}\mathbf{-}\mathbf{log}\frac{\left(1.84×{10}^{-4}\right)\mathbf{-}\left(1.44×{10}^{-4}\right)}{\left(1.44×{10}^{-4}\right)} \end{gathered}$$

Now the concentration of ${}^{\mathbf{HIn}}$and ${}^{{\mathbf{In}}^{\mathbf{-}}}$are ${}^{\left[\mathrm{HIn}\right]}$and ${}^{\left[{\mathrm{In}}^{-}\right]}$respectively.

As$A=\mathrm{Î\mu }Ic$

Hence, the expression for the absorbance of a solution containing this indicator is given by ${}^{\mathit{A}\mathbf{=}\mathbf{}\mathbf{2080}\mathbf{\left[}\mathbf{HIn}\mathbf{\right]}\mathbf{+}\mathbf{14200}\mathbf{\left[}{\mathbf{In}}^{\mathbf{-}}\mathbf{\right]}}$.

(b)

The formal concentration of indicator ${}^{\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{84}\mathbf{×}{\mathbf{10}}^{\mathbf{4}\mathbf{-}}\mathbf{M}}$

Also,

$$\begin{gathered} \mathit{p}\mathit{H}\mathbf{=}\mathbf{6}\mathbf{.}\mathbf{23} \\ \mathit{A}\mathbf{}\mathbf{}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{868} \end{gathered}$$

Assume,

$$\begin{gathered} \left[HIn\right]\mathbf{}\mathbf{=}\mathit{x}\mathbf{}\mathit{a}\mathit{n}\mathit{d} \\ \left[I{n}^{-}\right]\mathbf{}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{84}\mathbf{×}{\mathbf{10}}^{\mathbf{-}\mathbf{4}}\mathbf{-}\mathit{x} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathit{A}\mathbf{=}\mathbf{2080}\left[HIn\right]\mathbf{+}\mathbf{14200}\left[I{n}^{-}\right] \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{868} \end{gathered}$$

Substitute the given values in above equation and get,

$\mathbf{0}\mathbf{.}\mathbf{868}\mathbf{=}\mathbf{2080}\mathit{x}\mathbf{+}\mathbf{14200}\left(1.84×{10}^{-4}-x\right)$

Next, solve for ‘x’,

$\mathit{x}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{44}\mathbf{×}{\mathbf{10}}^{\mathbf{-}\mathbf{4}}\mathit{M}$

Now, findpKa,

$$\begin{gathered} \mathit{p}\mathit{H}\mathbf{=}\mathbf{}{\mathbf{pK}}_{\mathbf{a}}\mathbf{+}\mathbf{log}\frac{\left|{\mathrm{In}}^{-}\right|}{\left[\mathrm{HIn}\right]} \\ \mathbf{6}\mathbf{.}\mathbf{23}\mathbf{=}\mathbf{}{\mathbf{pK}}_{\mathbf{a}}\mathbf{+}\mathbf{log}\frac{\mathbf{(}\mathbf{1}\mathbf{.}\mathbf{84}\mathbf{×}{\mathbf{10}}^{\mathbf{-}\mathbf{4}}\mathbf{)}\mathbf{-}\mathbf{(}\mathbf{1}\mathbf{.}\mathbf{44}\mathbf{×}{\mathbf{10}}^{\mathbf{-}\mathbf{4}}\mathbf{)}}{(1.44×{10}^{-4})} \\ {\mathbf{pK}}_{\mathbf{a}}\mathbf{=}\mathbf{6}\mathbf{.}\mathbf{23}\mathbf{-}\mathbf{log}\frac{\mathbf{(}\mathbf{1}\mathbf{.}\mathbf{84}\mathbf{×}{\mathbf{10}}^{\mathbf{-}\mathbf{4}}\mathbf{)}\mathbf{-}\mathbf{(}\mathbf{1}\mathbf{.}\mathbf{44}\mathbf{×}{\mathbf{10}}^{\mathbf{-}\mathbf{4}}\mathbf{)}}{(1.44×{10}^{-4})} \end{gathered}$$

Again, solve the above equation,

$\mathit{p}{\mathit{K}}_{\mathbf{a}}\mathbf{=}\mathbf{6}\mathbf{.}\mathbf{79}$

Hence, the value of${}^{\mathit{p}{\mathit{K}}_{\mathbf{a}}}$is${}^{\mathbf{6}\mathbf{.}\mathbf{79}}$ .