J. G. Dinwiddle — Hydrofluoric and Fluosilicic Acids. . 421 



Art. XLIII. — A Study of the Separation of Hydrofluoric 

 Acid and Fluosilicic Acid • by J. Gr. Dinwiddie. 



[Contributions from the Kent Chemical Laboratory of Yale Univ. — cclxxxiii.] 



The work to be described in this article was done with a 

 view towards finding some satisfactory method for the analysis 

 of a solution containing both hydrofluoric and fluosilicic acid. 



The importance of an accurate method for this determination 

 lies in the fact that commercial hydrofluoric acid very often 

 contains a6 an impurity fluosilicic acid, which is of no use in 

 etching, and which is a nuisance in analytical work. The value 

 of a solution of hydrofluoric acid is obtained often by direct 

 titration with alkali using phenolphthalein as indicator. Under 

 these conditions each molecule of fluosilicic acid, H 2 SiF 6 , 

 requires six molecules of alkali for neutralization and, there- 

 fore, will be counted as six molecules of hydrofluoric acid. 



Almost the only attempt to discriminate between these two 

 acids when present together has been made by Katz,* who 

 titrates first in water solution and then in alcoholic solution 

 together with potassium chloride in order to get a differential 

 equation. The reaction in water solution is 



(a) H 2 SiF 6 + 6NaOH = 6NaF + Si(OH) 4 + 2H,0 

 and in 50 per cent alcoholic-potassium-chloride solution, 



{b) H 3 SiF 6 + 2KC1 + 2NaOH = K 2 SiF 6 + 2NaCl + 2H a O. 



The hydrofluoric acid requires the same quantity of alkali for 

 neutralization whether titrated in water or in alcohol solution, 

 while the fluosilicic acid requires one-third of the alkali in 

 alcoholic-potassium-chloride solution that it does in water solu- 

 tion. 



Were there no complications in the above reactions, the cal- 

 culation of the results would be as follows : — 



Let x be the volume of alkali required for neutralizing the 

 hydrofluoric acid, and let y be that required by the fluosilicic 

 acid in the alcoholic solution. Then 2>y will be required by 

 the fluosilicic acid in water solution. 



Then, x + 3y = cc. req. in water titration = a 

 x + y = cc. req. in alcohol titration = b 



.-. 2y = (a -b) 



Thus the difference between the amount of alkali required for 

 the water and for the alcohol titration is equivalent to two- 

 thirds the amount which would be required for the fluosilicic 

 acid alone in water solution. 



*Katz, Chemiker Zeitung, xxviii, 356 and 387, 1904. 



Am. Jour. Sci. — Fourth Series, Vol. XLII, No. 251. — November, 1916. 

 29 



