166 Messrs. Gladstone and Hibbert 



With reference to these results we have made some obser- 

 vations. 



Buff-coloured body from Litharge. — The formula (S0 3 ) 3 

 (PbO)g can be written 2PbO + 3PbS0 4 . In other words, the 

 substance may either be a true sub-sulphate, or a mixture of 

 the ordinary sulphate and oxide of lead. When attempting 

 to prepare some of this compound, preliminary results sug- 

 gested the following experiment. 



Known quantities of litharge and sulphuric acid were 

 digested together in a flask. From time to time the flasks 

 were shaken vigorously and a small quantity of the clear 

 liquid subtracted. The strength of the acid was carefully 

 determined; and from this, together with the known volume of 

 acid, it was easy to find how much acid had combined with 

 the lead oxide. The curves given below, which are drawn 

 through the actual points of observation, show the course of 

 the action for 28 days, the ordinates giving the acid absorbed, 

 and the abscissae the time. 



In Exp. I., 33*41 grams of litharge and 150 cubic centim. 

 of 1 to 5 acid were used. This liquid contained 43*38 grams 

 of sulphuric acid. 



In Exp. II., 39*73 grams of litharge and 150 cubic centim. 

 of 1 to 6 acid =36*37 grams H 2 S0 4 were used. 



In Exp. III., 39*92 grams of litharge and 150 cubic centim. 

 of 1 to 10 acid =25 grams H 2 S0 4 were used. 



The acid required for total conversion of the litharge into 

 sulphate would be, 



In Exp. I. = 14*7 grams. 

 In Exp. II. = 17*45 „ 

 In Exp. III. = 17*54 „ 



Of course only three fifths of these quantities would have been 

 required to form the compound (S0 3 ) 3 (PbO) 5 . 



The salt gradually became white in colour. It is evident 

 that in each case the action was slow and progressive, without 

 any sudden change of rate to mnrk the formation of a sub-salt. 



At a later date we stopped the experiments and analysed 

 the substances produced in Exps. I. and II. 



In I. we found 31*26 per cent, of S0 4 , which indicates the 

 presence of 98*6 per cent. PbS0 4 and 1*4 per cent, of PbO. 



In II. we found 31*67 per cent, of S0 4 , which indicates 99*5 

 per cent. PbS0 4 and 0*5 per cent, of PbO. 



The general outcome is clearly in favour of the idea that the 



