Chromatography Forum

Hello all:


I'm new to lab & am trying to understand how the calculation is performed by the Titrator. I understand acid-base titration calculations but need some help with KF Titration because I'm not sure how to convert mg of KF reagent into mole of KF reagent.


These are the info according to the print out:


25 mg of H2O sample was titrated with 5.1248 mg/mL of KF reagent. 4.9185 mL of the KF reagent was required for neutralization. I need to calculate the concentration of the H2O sample.


4.9185 mL x 5.1248 mg/mL = 25.2063 mg of KF reagent


Next step: (1) How to convert 25.2063 mg of KF reagent into moles? (2) After the conversion, what is the ratio between the components for the calculation?


Thanks in advance.

My experience has been that KF reagent is expressed as mg H2O / mL KF reagent, though that does not appear to be the case here as the amount of water found would be greater than your sample amount. So it appears the KF reagent factor is expressed as mg KF reagent / mL KF reagent. Rather than tell you everything, a less than exhaustive internet search on "karl fischer" gave this web site:
http://www.emdchemicals.com/analytics/l ... Basics.pdf
This will tell you the "active" ingredient in KF reagent and allow you to calculate your result.

Hi,


I agree with you that the concentration of the KF reagent is expressed in mg of H2O per mL of KF reagent.


I will read the info carefully and try the calculation.


Thanks.

We calculate the titer of the KF titrant in mg H2O/ml KF, using weighed water as the standard. Then we assay the sample which has a known weight, and we know the ml KF titrant consumed, which gives us the mg H2O in the sample. From the mg H2O in the sample and the sample weight, % H2O is calculated. We have a Brinkmann titrator which spits out the % water automatically.

CPG: That's the way I've always run KF's but if you do that in Liza's example, you "find" more water (25.2 mg) than the sample weight (25 mg). I thought maybe in her case that the factor was expressed as mg I2 / mL reagent. That way you would get less than 100 % water. Can't say that I've heard of it being done that way, though.

Thanks for both of your replies. Now I understand the calculation! However, my result doesn't match that of the titrator...


Info:
1 ml KF --> 5.1248 mg H2O
4.9185 ml KF --> ? mg H2O


My calculation:
mg H2O = 5.1248 x 4.9185 = 25.2063288 mg
% H2O = 25.2063288 mg / 25 mg x 100 = 100.8253152%, whereas the print-out states 100.62%.


What am I doing wrong? Please help.

Are you sure the correct factor is stored in the instrument? If you haven't changed it, a factor from a previous reagent may be in the instrument causing your result to differ.

The difference could be the reagent factor, as already suggested, however it may also be that the instrument has a "drift" correction factor that is automatic, and also used in the instrument report calculation. If that factor is wrong, the result will be slighly incorrect.

Bruce Hamilton

Thanks for your help. I appreciate it.

Hello again:


What exactly is the instrument "drift"? I understand that it indicates the stability of the reading and needs to be corrected for final %H2O calculation. But what is it that fluctuates and what needs to be stabilized exactly?


Thanks in advance.

Drift is the background titration of water over time when no sample has been introduced. For example, the Mettler DL18 Karl Fisher titrator has a "drift" setting that compensates for the slow ingress of water into the vessel during use.

Basically, you set up the titrator and allow it to automatically titrate any water that remains in the system for the length of the determination, say 2 minutes - without adding anything to the vessel. The instrument might determine a drift value of 10 ug water/minute. Multiple measurements of drift are used to obtain an average value.

You then perform a titation with water, and it takes 2 minutes, and the result that will be used will have had 20 ug of water subtracted.

Say you used reagent with exactly 2 mg/ml factor, and you added 1 mg of water, and the titation took 2 minutes, it would have titrated 510ul of reagent, but will subtract the 10 ul of drift and report 500ul used. If the titration took four minutes, it would subtract 20 ul of drift and report 490 ul used.

Thus the result for your sample can change with the duration of the titration, especially if samples take much longer than the calibration standards.

Most instrument manuals that have automatic drift compensation should explain this better than I can.

Please keep having fun,

Bruce Hamilton

If you are using a Mettler KF titrator, you may have a difficult time manually calculating the same result the titrator produces. Specifically on the DL38, you'll run into a problem where the titrator prints out a time in minutes:seconds format, but the problem is internally, the titrator actually calculates the time in decimal format that goes beyond the significant figures shown on the printed results. For example, 2 minutes 30 seconds may actually be 2.54 minutes.

Also, depending on your make and model, the drift may or may not be included at all. I understand on Brinkmanns you set a drift limit, and the titrator will not give you the option of starting a titration until it is below the set value, then it will not include the drift in the calculation since it is insignificant to your calculation.

Thanks very much for your replies.