CHEMICAL ANALYSIS 13 



full, and that the lowest part of the meniscus is opposite 

 the zero mark. 



Now add the acid I c.c. at a time, stirring after each 

 addition, with a glass rod. Continue the process until the 

 colour of the solution changes to red. The result will be 

 accurate to I c.c. With practice the end reaction can be 

 judged more accurately ; but, as already mentioned, a 

 certain quantity of acid is required for change of colour, 

 and this is greater with litmus than with some other 

 indicators. 



Say that 13 c.c. of N/i H 2 S0 4 were required. How 

 much NaOH was present in the 1 c.c. of sample ? 



1 c.c. N/i H 2 S0 4 =1 c.c. N/i NaOH 



But N/i NaOH = 40 grm. per litre 



Then 1 c.c. N/i NaOH = 0-040 grm. 



Hence 1 c.c. N/i H 2 S0 4 = 0-040 grm. NaOH 

 Then 13 c.c. ,, ,, = 0-52 grm. NaOH. 



But 13 c.c. were required to neutralize 1 c.c. of sample. 



Therefore 1 c.c. of sample contains o # 52 grm. NaOH, 

 or 100 c.c. ,, will contain 52 grm. NaOH. 



The discrepancy between 50 grm. in 100 c.c. and 52 grm. 

 may be due to : 



(1) Inaccuracy in making the soda solution originally ; 



(2) Inaccuracy in measuring the 1 c.c. of soda solution ; 



(3) Inaccuracy in the strength of the normal acid 



solution ; 



(4) Inaccuracy in titration. 



2. Repeat the experiment, using 1 drop methyl-orange 

 as indicator. 



3. Repeat the experiment, using 1 drop phenolphthalein 

 as indicator. 



4. Titrate 5 c.c. 25 per cent H 2 SO 4 diluted with a little 

 water, with N/i NaOH, using a few drops of litmus as 

 indicator. 



5. Repeat the experiment, using 1 drop methyl-orange 

 as indicator. 



6. Repeat the experiment, using 1 drop phenolphthalein 

 as indicator. 



