304 ELEMENTARY GENERAL SCIENCE CHAP. 



Hence we find in the above supposititious experiment that '86 

 gram of caustic potash can neutralise as much sulphuric acid as 

 616 gram of caustic soda, &c. Calculate from your own ex- 

 periments what quantity would neutralise as much of each acid 

 as 1 gram of caustic soda. The result should be found to be the 

 same for each acid, viz., that 1'4 grams of caustic potash neu- 

 tralises the same quantity of any acid, as does 1 gram of caustic 

 s< tda. 



EXPT. 307. Weigh out also some concentrated sulphuric 

 acid, which may be regarded as containing 98 per cent, of the 

 pure acid, and make up a solution so that 1 litre contains 

 about 50 grams of sulphuric acid. Again perform the neutra- 

 lisation experiment, and, knowing the weight of actual acid 

 present in the 25 c.c. and the weight of caustic soda present in 

 the solution of caustic soda used, calculate the weight of acid 

 which neutralises 1 gram of caustic soda. You should find that 

 1-225 grams of acid are necessary. 



As the strength of the laboratory solution of the acids varies, 

 you cannot weigh these, but if a solution of known strength is 

 given, you could, in a similar manner, prove that the 1 gram of 

 caustic soda or 1'4 grams of caustic potash can neutralise 

 1'575 grams of nitric acid and '91 gram of hydrochloric acid. 

 If we now calculate the results for the molecular weight of 

 caustic soda we find 40 grams of caustic soda are equivalent 

 to 56 grams of caustic potash and neutralise 49 grams of 

 sulphuric acid, 63 grams of nitric acid or 36 '4 grams of hydro- 

 chloric acid. 



We may represent the changes occurring in these reactions 

 by the following equations, and the reader should write in words 

 the exact meaning of these (and all other) equations, and should 

 see that the quantitative relations expressed by the equations 

 are those actually found. 



2KOH + H 2 S0 4 = K 2 S0 4 + 2H 2 O. 

 2NaOH + H 2 S0 4 = Na 9 SO 4 + 2H 2 O. 

 KOH + HN0 3 - KN0 3 + H 2 O. 

 NaOH + HNO 3 = NaNO 3 + H 2 O. 

 KOH + HC1 = KC1 + H 2 O. 

 NaOH + HC1 - NaCl + H 2 O. 



We thus see that hydroxides as well as oxides may react with 



