Chapter 17, Problem 17.43P

Interpretation Introduction

Interpretation:

The process of detection of end point in Karl Fischer titration has to be explained.

Concept introduction:

Karl Fischer Titration of water:

Figure 1

The anode solution (base, alcohol, ${\text{SO}}_{\text{2}}\text{,}{\text{I}}^{\text{-}}$) is poured in main section in above figure and cathodic solution which has reagents to be reduced at cathode is poured in coulometric generator.  Current is passed till the end point.  An unknown solution is passed via septum and the moisture consumption is monitored by coulometer.  When the ratio of water and iodine is $\text{1:1}$, then $\text{2}\text{moles}\text{of}{\text{e}}^{\text{-}}$ corresponds to one mole of water.

$\begin{array}{l}\text{ROH}\text{+}{\text{SO}}_{\text{2}}\text{+}\text{B}\underset{}{\to }{\text{BH}}^{\text{+}}\text{+}{\text{ROSO}}_{\text{2}}{}^{\text{-}}\\ \\ {\text{H}}_{\text{2}}\text{O}\text{+}{\text{I}}_{\text{2}}\text{+}{\text{ROSO}}_{\text{2}}{}^{\text{-}}\text{+}\text{2B}\stackrel{}{\to }{\text{2BH}}^{\text{+}}{\text{I}}^{\text{-}}\end{array}$

Iodine molecule is generated on oxidizing ${\text{I}}^{\text{-}}$ in anode compartment.  Then Iodine molecule oxidizes ${\text{SO}}_{\text{2}}$ to form ${\text{ROSO}}_{\text{3}}{}^{\text{-}}$.  One mole of iodine molecule is required to consume one mole of water.