Dilution of 0.01M HCN Calculating Volume of Water to Add

Click For Summary
SUMMARY

The discussion focuses on diluting a 0.01M solution of HCN to achieve a pH of 6. The user calculates that the required concentration of HCN in the new volume is 2.08*10^-3M, leading to a total volume of 192cm3. Consequently, the volume of water to be added is determined to be 152cm3. The calculation is confirmed using the BATE tool.

PREREQUISITES
  • Understanding of molarity and dilution principles
  • Knowledge of acid dissociation constants (Ka)
  • Familiarity with pH calculations
  • Experience using the BATE tool for chemical calculations
NEXT STEPS
  • Study the principles of acid-base chemistry and pH calculations
  • Learn about the use of the BATE tool for chemical analysis
  • Explore the implications of dilution on chemical equilibrium
  • Research the properties and applications of HCN in laboratory settings
USEFUL FOR

Chemistry students, laboratory technicians, and anyone involved in chemical solution preparation and pH adjustment.

josephcollins
Messages
59
Reaction score
0
I have a short question concerning the dilution of a 0.01M solution of HCN. I need to dilute 40cm3 of the above to give a solution with a pH of 6. Given the Ka of HCN is 4.8*10^-10 I obtain the concentration of HCN necessary in the new volume:2.08*10^-3M. Therefore the volume has to be multiplied by 0.01/2.08*10^-3 to give a new volume of 192cm3. Is it therefore that the volume of water necessary to add is 192-40=152cm3? Thx, Joe
 
Chemistry news on Phys.org
Yes, you are correct. I have checked it using BATE :wink:
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
1
Views
3K
Replies
1
Views
3K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 2 ·
Replies
2
Views
4K
Replies
14
Views
7K
  • · Replies 3 ·
Replies
3
Views
14K