BROMINE IODINE- FL UORINE. 239 



19. BROMINE IODINE FLUORINE. 



Bromine, Bro'mum, Br = 79.36. This element is found in sea- 

 water and many mineral waters, chiefly as magnesium, calcium, and 

 sodium bromides, which compounds, however, represent in all these 

 waters a comparatively small percentage of the total quantity of the 

 different salts present. Most of these salts are separated from the 

 water by evaporation and crystallization, and the remaining mother- 

 liquor, containing the bromides, is treated with chlorine, which liber- 

 ates bromine, the vapors of which are condensed in cooled receivers : 



MgBr 2 -f 2C1 = MgCl 2 + 2Br. 



Bromine is at common temperature a heavy, dark reddish-brown 

 liquid, giving off yellowish-red fumes of an exceedingly suffocating 

 and irritating odor; it is very volatile, freezes at about 24 C. 

 ( 11 F.), and has a specific gravity of 2.99; it is soluble in 33 

 parts of water, more freely in alcohol, abundantly in ether and bisul- 

 phide of carbon ; it is a strong disinfectant, and its aqueous solution 

 is also a bleaching agent ; it acts as a corrosive poison. 



Hydrobromic acid, Acldum hydrobromicum, HBr = 8O.36. 

 This acid cannot well be obtained by the action of concentrated sul- 

 phuric acid upon bromides, since the hydrobromic acid first formed 

 becomes readily decomposed with formation of sulphur dioxide and 

 free bromine. Thus : 



2NaBr + H 2 SO 4 = 2HBr + Na 2 SO 4 ; 

 2HBr + H 2 SO 4 = 2Br + SO 2 + 2H 2 O. 



If, however, dilute sulphuric acid is added to a warm solution of 

 potassium bromide, potassium sulphate is formed, a portion of which 

 crystallizes on cooling. From the remaining portion of the salt, the 

 hydrobromic acid may be separated by distillation. 



Hydrobromic acid may also be obtained by the formation of bromide of 

 phosphorus, PBr 5 (the two elements combine directly), and its decomposition 



by water : 



PBr 5 + 4H 2 = 5HBr + H 3 PO 4 . 



In the form of solution this acid may be prepared also by treating bromine 

 under water with hydrogen sulphide until the brown color of bromine has 

 entirely disappeared. The reaction is as follows : 



lOBr + 2H 2 S + 4H 2 = lOHBr + H 2 SO, + S. 



The liquid is filtered from the sulphur and separated from the sulphuric acid 

 by distillation, 



