Solving PH/PoH

[StephanieIb]

Regarding solving for PH, POH, etc do we just need to know how to do the 6B textbook problems?

[Valerie Tran 2B]

I think so. If you can do those problems confidently then you should be fine for anything on the final.

[Evonne Hsu 1J]

I believe the final will definitely test us on concepts and questions in regards to pH, pOH, [H+],[OH-], etc. Doing the textbook questions will definitely benefit you as well.

[505612629]

I believe the textbook problems are up for grabs on the final, so I think it's safe to say that they're good practice!

[Juliet Carr 1F]

I think the important takeaways from solving problems with PH and POH are:
[H3O+][OH-] = 1 x 10^-14
pH + pOH = 14
pH = -log[H3O+]
pOH = -log[OH-]

So being able to apply these concepts and doing the textbook problems are probably sufficient enough.

[Kamille Kibria 2A]

i think you should be fine if you just know all the calculations for pH and pOH and you could most likely plug and chug for this concept. I'm not sure if it's posted on chem community but in the TA review session on acids and bases w Katherine she posted a rly helpful square w all the calculations that makes it super easy to visualize
pH + pOH = 14
pH = -log[H+]
pOH = =log[OH-]
[OH-][H+] = 1e14

[Leyla Anwar 3B]

This is what the square looked like and helps if only given one value you're trying to find the other.

[Lucy Weaver 1K]

It helps me to remember these rules
pH: -log of (H+ concentration)
pH: 14-pOH
pOH:14-pH
pOH: -log of (OH- concentration)
H+: 10^-pH
OH-: 10^-pOH

[David Y]

If given the molarity of an H+ ion, then you take the -log of that molarity to gain the pH. Then, you do 14-pH in order to give you the pOH

[Hannah Chang 3K]

I pretty much just memorized the formulas and did lots of practice problems.

[Andrew Yoon 3L]

Yeah just know the relationship between pH, pOH, [H+], [OH-]. If you know the relationships among them well, you should be good!

[Halle Villalobos 3E]

Hi! It is important to remember these basic formulas:
pH + pOH = 14
pH = -log[H+]
[H+] = 10^(-pH)
pOH = -log[OH-]
[OH-] = 10^(-pOH)