Question

What is the OH^- if the pH is 3.2, H3O^+ is 0.0006 M and the pOH is 10.8?
Please explain how you found.

Answer 1

\[[H30+][OH-] = 1x10^-14 at 25C \]

\[pH = -Log[H+] = -\log[0.0006] = 3.22 \]

\[1x10^-14/[H3O+] = [OH-]

1x10^-14/(6x10^-4) = (1.7x10^-11) = pOH

(1.7x10^-11)(6x10^-4) = 10^-14

Answer 2

The pH+pOH = 14 we have 3.2+10.8 = 14
also the [H3O+][OH-] = 10^-14 at 25 degrees celsius
we know that the pH = -log[H] so the -log[0.0006] = 3.2
so we take 0.0006 = [H3O+] and we divide this by 1.0x10^-14 to get the concentration of OH- ions. oh in the last line it's 1.0x10^-14

Answer 3

pOH = -Log[OH-]

Answer 4

Ok so the H3O and OH together equal 1*10^-14 so, we do what we do with pH and pOH only instead of subtracting the one you have by what they equal together it's dividing?

Answer 5

Yeah so if you find one you know the other one b/c the product of both = 1.0x10^-14 at 25C, of course if you change the temperature this value changes.

Answer 6

FYI it's called Kw

Answer 7

Ok that totally makes sense (as if it wouldn't) Thank you for explaining that!! Feel kinda silly for even asking

Answer 8

nah its' totally fine.. feel free to ask me anything

Answer 9

Another thing is that your Kw will change depending on the temp, but they usually give you the one at 25*C.

Answer 10

Good to know will make a note of that! Thanks again

Answer 11

NP