Chapter 19, Problem 19.9P

Interpretation Introduction

Interpretation:

The concentrations of $\text{HIn}$ and ${\text{In}}^{\text{-}}$ and ${\text{pK}}_{\text{a}}$ for $\text{HIn}$ has to be calculated.

Concept Introduction:

Henderson-Hasselbalch equation:

Henderson- Hasselbalch equation is rearranged form of acid dissociation expression.

For base:

Consider a basic reaction:

${\text{BH}}^{\text{+}}\stackrel{{\text{K}}_{\text{a}}{\text{=K}}_{\text{w}}{\text{/K}}_{\text{b}}}{\rightleftharpoons }{\text{B+H}}^{\text{+}}$

The Henderson- Hasselbalch equation for a basic reaction can be given as,

${\text{pH=pK}}_{\text{a}}\text{+log}\frac{\left[\text{B}\right]}{\left[{\text{BH}}^{\text{+}}\right]}$

Where , ${\text{pK}}_{\text{a}}$= acid dissociation constant for weak acid ${\text{BH}}^{\text{+}}$

To calculate the concentrations of $\text{HIn}$ and ${\text{In}}^{\text{-}}$ and ${\text{pK}}_{\text{a}}$ for $\text{HIn}$

Answer to Problem 19.9P

The concentrations of $\text{HIn}$ and ${\text{In}}^{\text{-}}$ is $\text{4}{\text{.36×10}}^{\text{-7}}$ and $\text{7}{\text{.94×10}}^{\text{-7}}$ respectively.

The ${\text{pK}}_{\text{a}}$ for $\text{HIn}$ is $\text{4}\text{.00}$.

Explanation of Solution

The quantity of $\text{HIn}$ is small when compared to Aniline and Sulphanilic acid.

The ICE table can be written as,

$\begin{array}{l}\begin{array}{cccc}\text{B}\text{+}& \text{H}\stackrel{}{\rightleftharpoons }& {\text{BH}}^{\text{+}}\text{+}& {\text{A}}^{\text{-}}\end{array}\\ \text{Initial}\text{moles}\text{2}\text{.00}\text{1}\text{.500}\text{-}\text{-}\\ \\ \text{Final}\text{moles}\text{2}\text{.00-x}\text{1}\text{.500-x}\text{x}\text{x}\\ \end{array}$

$\begin{array}{l}\text{K=}\frac{{\text{K}}_{\text{a}}{\text{K}}_{\text{b}}}{{\text{K}}_{\text{w}}}\\ \\ \text{K=}\frac{\left({\text{10}}^{\text{-3}\text{.232}}\right)\left({\text{K}}_{\text{w}}{\text{/10}}^{\text{-4}\text{.601}}\right)}{{\text{K}}_{\text{w}}}\text{=23}\text{.39}\\ \\ \frac{{\text{x}}^{\text{2}}}{\left(\text{2}\text{.00-x}\right)\left(\text{1}\text{.500-x}\right)}\text{=23}\text{.39}\\ \\ \text{x=1}\text{.372}\text{mmol}\end{array}$

$\begin{array}{l}\text{pH=4}\text{.601+log}\frac{\text{2}\text{.00-1}\text{.372}}{\text{1}\text{.372}}\\ \\ \text{pH=4}\text{.26}\end{array}$

For $\text{HIn}$,

Absorbance = $\text{0}\text{.110=}\left(\text{2}{\text{.26×10}}^{\text{4}}\right)\left(\text{5}\text{.00}\right)\left[\text{HIn}\right]\text{+}\left(\text{1}{\text{.53×10}}^{\text{4}}\right)\left(\text{5}\text{.00}\left[{\text{In}}^{\text{-}}\right]\right)$

Substitute,

$\left[\text{HIn}\right]\text{=1}{\text{.23×10}}^{\text{-6}}\text{-}\left[{\text{In}}^{\text{-}}\right]$ gives

$\left[{\text{In}}^{\text{-}}\right]\text{=7}{\text{.94×10}}^{\text{-7}}$

$\left[\text{HIn}\right]\text{=4}{\text{.36×10}}^{\text{-7}}$

The concentrations of $\text{HIn}$ = $\text{4}{\text{.36×10}}^{\text{-7}}$

The concentrations of ${\text{In}}^{\text{-}}$= $\text{7}{\text{.94×10}}^{\text{-7}}$

Henderson – Hasselbalch equation for $\text{HIn}$ is given as,

$\begin{array}{l}{\text{pH=pK}}_{\text{HIn}}\text{+log}\frac{\left[{\text{In}}^{\text{-}}\right]}{\left[\text{HIn}\right]}\\ \\ \text{4}{\text{.26=pK}}_{\text{HIn}}\text{+log}\frac{\text{7}{\text{.94×10}}^{\text{-7}}}{\text{4}{\text{.36×10}}^{\text{-7}}}\\ \\ {\text{pK}}_{\text{HIn}}\text{=4}\text{.00}\end{array}$

The ${\text{pK}}_{\text{a}}$ for $\text{HIn}$= $\text{4}\text{.00}$

Conclusion

The concentrations of $\text{HIn}$ and ${\text{In}}^{\text{-}}$ and ${\text{pK}}_{\text{a}}$ for $\text{HIn}$ were calculated and found to be ,

The concentrations of $\text{HIn}$ = $\text{4}{\text{.36×10}}^{\text{-7}}$

The concentrations of ${\text{In}}^{\text{-}}$= $\text{7}{\text{.94×10}}^{\text{-7}}$

The ${\text{pK}}_{\text{a}}$ for $\text{HIn}$= $\text{4}\text{.00}$