Medal will be awarded :) workings please -Calculate the pH of the following aqueous solutions at 25°C, with the [OH–] equal to: 9.3 x 10^-10

Question

aaronq · Answer

use:  pOH=-log$[OH^-]$; pH+pOH=14

anonymous · Answer

what does pOH stand for again @aaronq

aaronq · Answer

it's the negative logarithm of the hydroxide ion concentration (essentially a measure of the hydroxide ion concentration).

anonymous · Answer

umm hmm in simpler terms? or could you possibly show me the workings ?

aaronq · Answer

just plug it into the formula, like you would for "pH=-log$[H_3O^+]$"

anonymous · Answer

i know that formula and i did that but my answer is app not correct :/

aaronq · Answer

pOH=-log$(9.3*10^{-10})$

then, since pH+pOH=14, pH=14-pOH

anonymous · Answer

thats what i did but the answer says 4.79

aaronq · Answer

i got 4.968482948553936

anonymous · Answer

how what did you plug in where?

aaronq · Answer

read what i wrote just before

anonymous · Answer

ok  thanks

aaronq · Answer

no problem. let me know if you still don't get it