Method for the estimation of Tannin content

Method for the estimation of  Tannin content

Reagent Required:

Indigo carmine solution, Potassium permanganate

Glass wares and Apparatus Required:

1000mL conical flask, Burette and measuring cylinder.

Procedure:

Weigh accurately about 0.10gm of sample and dissolve it in 50mL of distilled water, add 750mL of distilled water and shake well. Add 25mL indigo carmine solution. Shake well and titrate against N/10 KMnO4 up to golden yellow colour end point. Repeat the experiment with the same quantity of reagents and in the same manner but omitting the substance. The difference between the two titration represent the indigo carmine solution required to neutralize the tannin. Each mL of 0.1N KMnO₄ is equivalent to 0.004157 gm of Tannin (as gallotannic acid).

Calculation:

Calculate the percentage of Tannin (as gallotannic acid) (w/w) as :

             (A – B)  X  0.004157 X 100 x N

          -------------------------------------------

                              W  X 0.1

A  = Volume of 0.1N KMnO₄ consumed in titration (Test)

B  = Volume of 0.1N KMnO₄ consumed in titration (Blank)

W = Weight of material taken in gm

N  = Normality of Potassium permanganate

Reagent Preparation:

Indigo Carmine Solution:

Dissolve 1.5 gm Indigo carmine in 1000mL of distilled water containing 50mL Sulphuric acid.

N/10 KMnO4 Solution :

Dissolve  about 3.3 gm KMnO4 in 1000mL of water, heat on a water bath for one hour and allow to stand for two days and filter through whatman No.1 filter paper. Standardize the solution as follows.

Standardize the solution as follows:

Weigh accurately about 0.200 gm of sodium oxalate, previously dried at 110oC, and dissolve it in 250mL of distilled water. Add 7mL of sulphuric acid, heat to about 70oC and then slowly add the permanganate solution from a burette, with constant shaking, until a pale pink colour which persists for 15 seconds, is produced. The temperature should not be less than 60oC during titration. Calculate the normality of 0.1N KMnO4.  Each 6.70 mg of sodium oxalate is equivalent to 1mL of 0.1 N KMnO4.

For Raw material : Take 10mL of the filtrate from the estimation of water soluble extractive. Rest the method is same as above.