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The principle of the Karl Fischer moisture analyzer

2021-12-01View Original

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The Karl Fischer method was developed by Karl Fischer in 1935; it involves using reagents prepared from I2, SO2, pyridine, and anhydrous CH3OH (with a water content of less than 0.05%). The water equivalent of these reagents is determined, and after the reagents react with the water present in the sample, the water content in the sample is calculated by determining the amount of reagent consumed. The International Organization for Standardization has designated this method as the standard for measuring trace amounts of water, and we **also consider this method to be** the standard for measuring trace water content. Principle: In the presence of water, the water in the sample undergoes redox reactions with SO2 and I2 present in the Karl Fischer reagent. I2 + SO2 + 2H2O → 2HI + H2SO4, but this is a reversible reaction; the reverse reaction occurs when the sulfuric acid concentration exceeds 0.05%. If we want the reaction to proceed in a positive direction, it is necessary to add an appropriate alkaline substance to neutralize the acid produced during the reaction. Experiments have shown that the addition of pyridine to the system allows the reaction to proceed in the right direction. 3 C5H5N + H2O + I2 + SO2 → 2C5H5NHI (pyridinium diiodide) + C5H5NSO3 (pyridine sulfonic anhydride). Pyridine sulfonic anhydride is unstable and can react with water, consuming some of the water and thus interfering with the measurement. To stabilize it, we can add anhydrous methanol. C5H5NSO3 (pyridine sulfate anhydride) + CH3OH (anhydrous) → C5H5N·HSO4CH3 (methyl pyridinium sulfate). We can write these three reactions together as a single overall reaction: I2 + SO2 + H2O + 3 pyridine + CH3OH → 2 C5H5NHI (pyridinium diiodide) + C5H5N·HSO4CH3 (methyl pyridinium sulfate). From this reaction equation, it can be seen that 1 mol of water requires 1 mol of iodine, 1 mol of sulfur dioxide, 3 mol of pyridine, and 1 mol of methanol in order to produce 2 mol of pyridinium diiodide and 1 mol of methyl pyridinium sulfate. These are theoretical values; in practice, however, excessive amounts of SO2, pyridine, and CH3OH are used. After the reaction is complete, the excess free iodine appears red-brown, indicating that the endpoint has been reached. I2︰SO2︰C5H5N = 1︰3︰10

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