Can someone help me with this pH calculation?
- Thread starter ddimebag
- Start date
hotrodharley
Well-Known Member
You will not be able to reach that pH with simple OTC bicarb. Try some lye. Just a bit. Cautious when using as it can eat a hole right through you.
ddimebag
Active Member
Thanks for your reply. If you re-read the question, you will see that bicarbonate was never mentioned (i know it won't go high enough with that)...I wanted to know how much NaOH (sodium hydroxide, aka lye) i need to add to 50 ml H2O to get pH 11.7. This is pure NaOH I am talking about...not an aqueous solution. Also, I need to calculate it...eyeballing it isn't accurate enough since this is a log scale...it has to be between 11.5 and 11.9, with 11.7 being preferable.
Figong
Well-Known Member
Hint #1: Balance the acid/base in an equation
Hint #2: -log is required
Hint #3: Mole value is used directly with the -log
Hint #4: Calculate for pOH
Hint #5: Subtract pOH from 14 to get your pH
Yes, I could have just as easily given the answer...but learning something new while solidifying chemistry background never hurts. If questions, ask.
Hint #2: -log is required
Hint #3: Mole value is used directly with the -log
Hint #4: Calculate for pOH
Hint #5: Subtract pOH from 14 to get your pH
Yes, I could have just as easily given the answer...but learning something new while solidifying chemistry background never hurts. If questions, ask.
AimAim
Well-Known Member
Hint #1: Balance the acid/base in an equation
Hint #2: -log is required
Hint #3: Mole value is used directly with the -log
Hint #4: Calculate for pOH
Hint #5: Subtract pOH from 14 to get your pH
Yes, I could have just as easily given the answer...but learning something new while solidifying chemistry background never hurts. If questions, ask.
Yeah but if one does not have a chemistry background to "solidify" your helpful information, isn't helpful.
ddimebag
Active Member
I have university education, and I know more than enough about chemistry to work with the stuff...the problem is that Im terrible at math... also, it now turns out that my calculation was correct in the first place, I just thought the number was too small...
ddimebag
Active Member
Hint #1: Balance the acid/base in an equation
Hint #2: -log is required
Hint #3: Mole value is used directly with the -log
Hint #4: Calculate for pOH
Hint #5: Subtract pOH from 14 to get your pH
Yes, I could have just as easily given the answer...but learning something new while solidifying chemistry background never hurts. If questions, ask.
Thats pretty much what I did, came out with 10mg for 50 ml, which I thought was too small. Looked at your reply and rechecked it, seems to be right. Thanks
Figong
Well-Known Member
Thats pretty much what I did, came out with 10mg for 50 ml, which I thought was too small. Looked at your reply and rechecked it, seems to be right. Thanks
Not a problem, figured you had a decent background or you wouldn't be asking about cutting NaOH to a specific pH
Glad to help, have fun man!
