Question

not getting these questions right. Last attempt. Help!
Find the pH of 0.113 M phenol, C6H5OH. The H bonded to the O is acidic, that is,

C6H5OH(aq) + H2O(l) ↔ H3O+(aq) + C6H5O-(aq). Ka = 1.0 x 10-10 M.

i took the sqrt(1.0*10^-10 * .113)
then took the -log of that.

Calculate the approximate [OH-] and [NH4+] in a 0.66 M ammonia solution, NH3(aq).

NH3(aq) + H2O(l) ↔ OH-(aq) + NH4+(aq). Kb = 1.75 x 10-5M.
i did the same thing i did in the problem above, but after i took the -log I subtracted that answer from 14, but It's wrong. WHY?

Answer 1

I don't think you subtract from 14. My understanding (and I may be wrong, I am not a science major) is that the negative log is the pH. Also, hydroxide (OH-) is a base, so should have a pH of above 7.

Answer 2

yeah but they are asking for the ph which is the -log of the H+ concentration. They aren't asking for the POH. So the equation for find ph is PH+POH=14..I get the POH from taking the -log of my sqrted answer. It would have a ph of 7 only if it was a neutral reaction I believe, but thanks anyways.

Answer 3

My understanding to get the pH, given molarity is -log[H+]. Both OH- and NH4+ (Ammonium) are basic.

Answer 4

As I understand, pH is just the concentration of the Hydrogen Ion, and you figured out the molarity already. Like I said, science isn't in my major, so I'm just going from that which I learned in general chem, and it may not be accurate in this case. Sorry! Just trying to help!
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Answer 5

that's ok at least you tried! I really appreciate it :) I got the right answer. You're right I didn't have to subtract from 14 since they're asking for the concentration not the ph or poh. so I multiplied .66 and 1.85*10^-5 and took the square root of that answer and got 3.39*10^-3. Thank you!