Question

Explain how the endpoint is detected in a Karl Fischer titration in Figure 17-35.

Short answer

The process of detection of endpoint in Karl Fischer titration.

Step-by-step solution

Define Karl Fischer titration

The Karl Fischer titration, which measures traces of water in transformer oil, solvents, foods, polymers, and other substances, might be performed half a million times each day. The titration is usually performed by delivering titrant from an automated burette or by coulometric generation of titrant. The volumetric procedure tends to be appropriate for larger amounts of water (but can go as low as $~1{\mathrm{mgH}}_2\mathrm{O}$), and the coulometric procedure tends to be appropriate for smaller amounts of water.

Determine the Process End point in Karl Fischer titration

The anode solution (base, alcohol, ${\mathrm{SO}}_2,{\mathrm{I}}^{-}$) 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 1: 1, then 2 moles of ${\mathrm{e}}^{-}$corresponds to one mole of water $\mathrm{ROH}+{\mathrm{SO}}_2+\mathrm{B}\to {\mathrm{BH}}^{+}+{\mathrm{ROSO}}_2^{-}$.

${\mathrm{H}}_2\mathrm{O}+{\mathrm{I}}_2+{\mathrm{ROSO}}_2^{-}+2\mathrm{B}\to 2{\mathrm{BH}}^{+}{\mathrm{I}}^{-}$

lodine molecule is generated on oxidizing in anode compartment. Then Iodine molecule oxidizes ${\mathrm{SO}}_2$to form ${{\mathrm{ROSO}}_3}^{-}$. One mole of iodine molecule is required to consume one mole of water.

Determine the End point in Karl Fischer titration

• A constant current of 5 or$10\mathrm{\mu A}$is maintained between detector electrode and measures the voltages.
  • Before reaching equivalence point the solution has an iodide ion with athe trace of iodine molecule.
  • The cathode potential should be negative to reduce species in the solvent system and to sustain the current of$10\mathrm{\mu A}$
  • When the equivalence point is reached, excessive iodine molecule forms, and current at lower voltage is maintained by the following reaction
• Cathode: ${\mathrm{I}}_3^{-}+2{\mathrm{e}}^{-}\to 3{\mathrm{I}}^{-}$
• Anode:$3{\mathrm{I}}^{-}\to {\mathrm{I}}_3^{-}+2{\mathrm{e}}^{-}$