Question

In a coulometric Karl Fischer water analysis, 25.00 mL of pure "dry" methanol required 4.23 C to generate enough I₂ to react with residual H₂O in the methanol. A suspension of 0.8476 g of finely ground polymeric material in 25.00 mL of the same "dry" methanol required 63.16 C. Find the wt %H₂O in the polymer.

Short answer

The percentage of water in polymer in Karl Fischer water analysis is 0.6491 wt%.

Step-by-step solution

Karl Fischer titration

The major application of Karl Fischer titration, is in the measurement of traces of water in transformer oil, polymers, solvents, foods, and other substances etc. This procedure is expected to be performed half a million times each day.

The titration is done by transferring the titrant from an automated burette or by coulometric generation of titrant. For large amount of water (but can go as low as $~1{\mathrm{mgH}}_2\mathrm{O}$ ), the volumetric procedure is considered to be suitablewhereas for very little amount of water the coulometric procedure is suitable.

Determine percentage of water in polymer

Determinig the percentage of water in polymer using Karl Fischer water analysis.

The required current for water in 0.8476g polymer$\left(63.16-4.23\right)=58.93\mathrm{C}$

No of moles reacted $=\frac{58.93\mathrm{C}}{96485\mathrm{C}/\mathrm{mol}}=0.6108\mathrm{m}{\mathrm{mole}}^{-}$which correspomds to $0.5\times 0.6108=0.3054\mathrm{m}{\mathrm{moll}}_2$

$\%\mathrm{of}\mathrm{water}=100\times \frac{5.502{\mathrm{mgH}}_2\mathrm{O}}{847.6\mathrm{mg}\mathrm{polymer}}=0.6419\mathrm{wt}\%$