Water Determination in Paper with Potentiometric Method

Equipment

Potentiometer = 751 GPD Titrino / Titroprocessor
Electrodes = 6.0338.100 double Pt electrode; input Pol

Reagents

HYDRANAL® Composite 5; D0

Sample

20 -30 mL methanol, conditioned to complete dryness
1.000 mL sample solution
extract app. 1.0g of paper (cut in small pieces) in 50 mL methanol (extraction medium)

Remarks

Calculations:

Water=(EP1-C38)*C39*C01/C00/C02;2;%

smpl size 0.02003 g
EP1 0.2260 ml
Water 5.29 %
Titer 5.0058 mg/ml
Blank 0.0145 ml
Water = content of water in paper in %
C00 = sample size / aliquot (1.0015 g / 50 mL = 0.02003 g/mL)
C01 = factor for % (0.1)
C02 = divisor (1)
C38 = blank value in mL of “Blank_KF“ method (0.0145 mL)
C39 = titer of HYDRANAL® Composite 5

Answer

Water = (0.2260-0.0145) x 5.0058 x 0.1 / 0.02003 / 1 = 5.29%

Mean from 3 determinations.
This method can be generally used for KF titrations with blank values.

· Adjust the calculation values C01 and C02 according the desired result unit and your sample size:

Unit RS Smpl size

Sample preparation:

Stir sample solution for app. ½ h to extract the water and take aliquots (see calculation).

Literature

Water determination by Karl Fisher titration, G. Wieland, GIT Verlag, Darmstadt (Germany)
HYDRANAL®, practical course, Water reagents according to Eugen Scholz, Riedel de Haën, Seelze (Germany)
Metrohm Application Bulletin No. 77: KF water determination