9IO 
Hydrocyanic  Acid. 
(  Am.  Jour.  Pharm. 
(     December,  1920. 
error  on  a  2  per  cent,  acid  would  only  affect  the  percentage  found, 
by  a  unit  in  the  second  decimal,  if  about  20  Cc.  of  N/io  AgNOa  were 
used.  Evidently  one  is  quite  safe  if  the  excess  of  soda  is  kept  at 
about  the  equivalent  of  i  or  2  Cc.  of  normal  NaOH. 
A  study  of  the  reactions  involved  explains  the  difference  between 
Liebig's  and  the  other  processes  in  above  list. 
(a)  When  a  solution  of  HCN  or  of  an  alkaline  cyanide  is  added 
slowly  to  an  excess  of  silver  solution,  precipitation  of  AgCN  is  im- 
mediate. The  continued  addition  of  HCN  only  increases  the  precip- 
itate till  all  the  silver  is  converted  into  AgCN.  But  continued 
addition  of  alkaline  cyanide  causes  the  precipitate  first  formed  to 
redissolve  as  the  double  cyanide. 
(b)  Reversing  the  order  of  mixing:  AgNOs  added  slowly  to  an 
alkaline  cyanide  gives  no  permanent  precipitate  till  all  the  cyanide 
is  changed  to  the  double  salt,  then  the  next  drop  of  AgNOs  solution 
causes  the  beginning  of  precipitation  of  AgCN  (the  end-point  in 
Liebig's  process).  The  final  complete  precipitation  of  cyanide  as 
AgCN  is  accomplished  by  adding  exactly  as  much  silver  again  as 
that  used  to  cause  the  first  sign  of  precipitation. 
The  equations  representing  a  and  b  are : 
(a)  HCN  +  AgNOs  =  AgCN  -f  HNO3 1  complete  precipitation  of 
KCN  +  AgNOa  =  KNO3  +  AgCN  /       both  CN  and  Ag 
(b)  2KCN  -f  AgNOs  =  KCN,  AgCN  +  KNO3 
No  precipitate  till  this  stage  (6)  is  complete.  End-point  of 
Liebig's  process. 
The  addition  of  more  AgNOs  precipitates  AgCN. 
Another  molecule  of  AgNOs  gives  the  same  final  result  as  (a) : 
2KCN  +  2AgN03  =  2Ag  CN-f  2KNO3  \  complete  precipita- 
?.^.,KCN  -f  AgNOs  =  AgCN  +  KNO3     /  tionof  both  AgandC. 
A  point  obvious  from  these  equations  which  seems  to  have 
escaped  notice  is  that  since,  in  Liebig's  process. 
2HCN  or  2KCN  is  equivalent  to  AgN03, 
similarly  2KOH  or  2NaOH  is  equivalent  to  AgNOs. 
And  therefore,  the  amount  of  alkali  required  for  exact  conversion 
of  the  HCN  into  cyanide  is  twice  as  many  Cc.  of  N/io  NaOH  (or 
KOH)  as  the  number  of  Cc.  of  silver  solution  used. 
i.e.,  No.  of  Cc.  of  N/io  silver  required  X  2  =  Cc.  of  N/io 
soda  that  should  be  used  to  convert  the  HCN  into  cyanide. 
No.  of  Cc.  of  N/i  silver  X  2  =  Cc.  ol  N / 1  soda  required. 
