Am,  Jour,  Pharm. ) 
Oct.  1S75.  J 
Chemical  Analysis  of  Potash. 
465. 
(e.)  Estimation  of  the  potassa  and  soda  :  From  the  lastly-named  50 
c.c.m.  evaporate  10  c.c.m.  to  dryness,  weigh  the  saHne  residue,  and 
subtract  the  sulphate  of  potassium  contained  therein  ;  as  it  is  already 
known  (by  experiment  b)  how  much  SO3  is  contained  in  10  c.c.m.  of 
the  solution,  it  is  only  necessary  to  obtain  the  weight  of  the  sulphate  of 
potassium  by  calculating  from  that  SO3.  The  alkaline  chlorides  are 
contained  in  the  remaining  salt.  Dissolve  the  salt  again  in  water,  pre- 
cipitate with  nitrate  of  silver,  and  calculate  the  chlorine  form  the 
chloride  of  silver  (previously  dried  at  ioo°C.) by  subtracting  the  chlorine 
from  the  two  chlorides  the  weight  of  both  metals  is  found. 
The  fourth  treatment  is  executed  in  the  so-called  "  Indirect  Analy- 
sis," *  as  follows  :  To  find  the  weight  of  the  potassium,  multiply  the 
weight  of  the  two  metals  with  2*5416  (the  quotient  of  the  division  from 
the  equivalent  of  sodium  into  the  equivalent  of  chloride  of  sodium), 
subtract  the  product  from  the  weight  of  the  two  chlorides,  and  divide 
the  remainder  by  0*6355  (the  difference  between  2*5416  and  1*9061 
[the  quotient  of  the  division  from  the  equivalent  of  potassium  into  the 
equivalent  of  chloride  of  potassium.]) 
To  find  the  weight  of  the  sodium,  multiply  the  weight  of  the  two 
metals  by  1*9061,  subtract  the  weight  of  the  two  chlorides  from  the 
product  and  divide  the  remainder  by  0*6355. 
EXAMPLE. 
Grams. 
The  weight  of  the  two  chlorides  is      =  0*902 
"       "  "  metals  is      =  0*442 
0*902  —  (0*442  X  2-5416) 
As  K  -  _  0.533^  =  0*348  grams. 
^   _  (0*442  X   1*9061)  —  0*902 
iNa—  .  0^55   =0*094  grams. 
Sodium  and  potassium  are  next  to  be  converted  into  potassa  and 
soda  after  the  proportions  : 
for  KO  490  :  590  —  0*348  :   X  X  ==  0*419 
"  NaO  288  :  388  =  0*094  :   X  X  0*127 
B.    The  part  insoluble  in  water. 
{a.)  Estimation  of  the  insoluble  silica.  Cover  that  part  of  potash, 
which  was  not  dissolved  by  the  water,  in  a  retort  with  HCl  (sp.  gr. 
1*12),  any  effervescence  thereby  caused  indicates  carbonate  of  calcium, 
*  See  theory  of  this  calculation  in  "  Frickhingefs  Katechismiis  der  Stoechiomctrie.'' 
30 
