- Joined
- Jun 12, 2008
- Messages
- 16
- Reaction score
- 0
- Points
- 0
- Pre-Dental
Gchem pH values
- Thread starter jedaerrow
- Start date
- Joined
- Sep 15, 2007
- Messages
- 228
- Reaction score
- 2
- Points
- 4,571
- Resident [Any Field]
Ok... so i'm not sure if I will explain this correctly, but here it goes:
Ca(OH)2 will give off 2 OH ions. So you need to multiply the concentration by 2. So the [OH] is now 2x10^-4.
pOH = -log [OH]
pOH = -log[2x10^-4]
since -log of 1x10^-4 = 4, then -log 2x10^-4 will be a little less than 4, ~ 3.8 or so.
pH = 14 - pOH, so pH = 10.2
They're not likely to give you another answer that is within decimal points of the correct one, so as long as you round reasonably then you should be able to get the right answer.
Ca(OH)2 will give off 2 OH ions. So you need to multiply the concentration by 2. So the [OH] is now 2x10^-4.
pOH = -log [OH]
pOH = -log[2x10^-4]
since -log of 1x10^-4 = 4, then -log 2x10^-4 will be a little less than 4, ~ 3.8 or so.
pH = 14 - pOH, so pH = 10.2
They're not likely to give you another answer that is within decimal points of the correct one, so as long as you round reasonably then you should be able to get the right answer.
Similar threads
