When the [H+] in a solution is 1.7 × 10^-9 M, what is the pOH ?
A) 8.80 B) 5.20 C) 14 D) 8.60
@james1107
@Lexi.babyy
Aug 4, 2011 - Best Answer: 14 = pH + pOH -log(1.7x10^-9) = pH =(8.77) Calculate the [OH ‾ ], [H+], and the pH of 0.17 M solutions of each of the following ...
according to this it would be a
so C according to that?
No 14 isnt the answer but thats okay :3 thank you
look at everything in the parentheses (8.77)
Do it as i said earlier ? @ilovebmth1234
ohhh lol and yeah but its kinda confusing like i said
did you see what i ws talking about
Aug 4, 2011 - Best Answer: 14 = pH + pOH -log(1.7x10^-9) = pH = 8.77 ... Calculate the [OH ‾ ], [H+], and the pH of 0.17 M solutions of each of the following ...
pH=8.77
when you add them together how would i do that exactly?
type in just the question so you can looh at the first site that comes up i think its like answers .com or something its the very first website
look at the
can you take a screen shot and post it? i cant get on yahoo its blocked :/
how do you do that on a laptop
press your start button click all programs and go to accessories and find snip tool and click on it
or try control+ alt + prnt scrn(aka- end)
Do u want me to go thru it again ? @ilovebmth1234
if you make it to where i can understand it
okai
When the concentration of H^+ is given we can't directly use an equation to find the pOH To find we need to know the pH of the solution.
\[pH = - \log[H ^{+}]\] \[pH= -\log[1.7 * 10^{-9}]\]
find that value first !
@ilovebmth1234 hey are u working it out ?
okay im trying to
I'll make it easy for u -log10^-9 = 9 -log1.7 u need to find the log value for 1.7 and put a minus charge in front !
what does -log mean?
log values !!!!!!!!!!!!!!!!!!!!!!!! -log is the minus log value
damn my bad sorry i dont quit understand this like you do v.v
lol so did u find the value now ?
\[-\log10^{-9} = 9\] \[-\log1.7 = - 0.23\] \[pH = 9 - 0.23 = 8.77\]
A
or D
pH + pOH = 14 Since pH = 8.77 pOH = 14 - pH = 14 - 8.77 pOH = 5.23
pH is not pOH
right right
so B would be the answer
yes yes
great thanks :3 could you give @james1107 a medal for his help too?
ah yes sure :)
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