ON SOME N.S.W. TA.N-SIJBSTANCE8. 263 



A quantity, about 100 c.c. of decoction (it is desirable to be 

 uniform, but absolute accuracy in the quantity is unnecessary), 

 is placed in a lOoz. beaker, about a tablespoonful of hide-powder 

 added (it is not necessary to previously moisten, as sometimes 

 stated), well stirred with a glass-rod, and allowed to stand about 

 20 hours.* At the end of this period the liquid is filtered, and 

 is always quite transparent and usually colourless. It filters 

 with the utmost facility. It is not even necessary to previously 

 moisten the filter paper with distilled water, especially as the 

 moisture thus introduced would dilute the filtrate and affect 

 the result. 



20 c.c. of indigo solution are placed in J litre of water in the 

 Berlin dish, as usual, and 10 c.c. of the above filtrate run in. 

 Permanganate is then run in with all the precautions, and when 

 the pure yellow colour arrives, the quantity of permanganate is 

 noted. Call it S. Repeat, and call it T. The mean (U) is 

 therefore the quantity of permanganate necessary to oxidise 

 10 c.c. of filtrate + 20 c.c. of indigo. Subtracting M (the 

 quantity necessary to oxidise 20 c.c. of indigo) from XT, we have 

 the amount of permanganate necessary to oxidise 10 c.c. of 

 filtrate alone. Call this Y. 



Therefore N — V = obviously the quantity of permanganate 

 necessary to oxidise the tannic acid which has been taken up by 

 the hide powder. Y represents the oxidisable impurities which 

 hide powder is incapable of absorbing. 



An actual example is given in the following form (which I 

 have adopted as being very convenient), in which the letters 

 M to Y referred to above are used : — 



Standards. 

 (1) 20 c.c. indigo. 



M 16.9 c.c. KM11O4 required. 



(2) Gallo-tannic Acid = 9-8438. 

 (See separate determination below.) 



20 c.c. indigo. 



10 c.c. decoction (5 grins, to 1 litre). 



P 24-2 KMn0 4 required 1st test. 



Q 24-1 do. 2nd do. 



R 24-15 do. for 10 c.c. of decoction - 16-9 = 



7-25 N. 



* I have proved by experiment that the hide-powder may remain in 

 the liquid for a much longer period, without absorbing any further tannin. 



