Question

pH and related question:
if H3O+ is 1*10^3 then what is OH-, pH respectively, & is it acid base or neutral?

Answer 1

clarification:
\(H_{3}O^{+}=1\times10^{3}\)\[OH^{-}=?\]\[pH=?\]Acid, base, neutral?

Answer 2

What I've done so far: \[pH=-\log{H_{3}O^{+}}=-\log{(1\times10^{-3}M)}=3\]
\[pOH=14-3=11\]\[ OH^{-}=\frac{K_{w}}{H_{3}O^{+}}=\frac{1\times10^{-14}}{1\times10^{-3}}=1\times10^{-11}\]
So, I ended up with these answers:\[\color{steelblue}{pH=3}\]\[\color{olive}{OH^{-}=1\times10^{-11}}\]

Answer 3

Ok so all your current answers look good. Now you are left with the question "is it acid base or neutral?". Any ideas on this one?

Answer 4

Oh, I used the right formulas? It was guess-risk there ahah ^-^;

Answer 5

And I believe that since the pH is 3 it's an acid?

Answer 6

yes it would be acidic.

Answer 7

to determine [OH-] you could have also done \[[OH^-]=10^\left( -pOH \right)\]

Answer 8

the way you did it works absolutely fine though

Answer 9

Alright
And there is another one (this is a chart but I won't ask for all the things) that says \(H_{3}O^{+}\) is 0.001? I'm not sure what that means

Oh, right. That confused me though so I used the \(K_{w}\) version lol

Answer 10

Im assuming when they say H3O+ is .001 they are referring to ".001" being the concentration of H3O+. not sure what you are asking tbh.

Answer 11

Lol, yea if they are all "M units" probably referring to concentration then.

Answer 12

ARGH LAG

*original post*
I asked the professor, but she was distracted at the time and gave a vague answer of "they are all M units" ... eh, so it's 0.001 M?

-
Alright x_x

Answer 13

Yea if she is saying "m units" im guessing she is talking molarity. so the concentration of the H3O+ ions is .001M.

Answer 14

So basically the same as the previous ... ?
I mean... \(\large{0.001=1\times10^{-3}}\)

Answer 15

yep

Answer 16

I feel like it's not that but the professor said as much -- unless she wasn't focusing on the issue at hand.

Answer 17

well in terms of the questions revolving around teh subject of acids, i doubt the .001 value is representing a pH value.

Answer 18

Oh, the 0.001 is under the \(H_{3}O^{+}\) column in the table. idk

And if I have \(OH^{-}=2\times10^{-5}\) I would just reverse to find \(pOH\), then calculate \(pH\) by subtracting \(pOH\) from 14, and then calculate the rest off that?

Answer 19

Bleh, sorry for the big load of questions. x_x

Answer 20

yea to find pOH you would do \[pOH=-\log(2\times 10^\left( -5 \right))\]then given the pOH you use.\[pH+pOH=14\] and just isolate the pH term. And from there you can calculate mostly anything in terms of these questions. so yes you are correct.

Answer 21

and its no problem. for some reason i cant sleep :/

Answer 22

might as well do some chem :P

Answer 23

I would be thankful but now I feel bad, regardless of the fact that you can't sleep hahah ; -;

Answer 24

haha, its no worries. I dont mind at all.

Answer 25

And I could sleep, but I am half unwilling and half forcing as I study less effectively in the daytime (there's a heatwave here)

Answer 26

Yea i feel you on that one. Florida summers. sooo hot

Answer 27

Random question though, how would I get an antilog when calculating for \(H^{+}\) or \(OH^{-}\)?
i.e. \([H^{+}]=10^{-pH}=\log^{-1}pH\)

Answer 28

Same in California ; -;

Answer 29

*as in how do I input antilog on the calculator*

Answer 30

Ok well say you are taking \[\log_{10}(1.25) \] the only way i know to take the antilog on the calc would be to use "10^1.25".

Answer 31

normal log on a calculator = \(\LARGE{log_{10}}\)?

Answer 32

depending on your calculator it may be possible that it has a log^-1 key or something of the sorts. however, my ti89 doesnt

Answer 33

yes sometimes the calculators default base for logarithms is "e" so just take note of that.

Answer 34

but it's most likely 10 like you said

Answer 35

So, antilog, or \(\log^{-1}\), is basically flipping the whole shebang around and getting a 10 exponential. :p

Answer 36

Okay ☺

Answer 37

Yea, you got it.

Answer 38

Woo! *victory dance* :D

One down, eight million more to go --
jk, still a lot though x_x

Answer 39

ok typo

Answer 40

I meant to say for \[\log_{4}x \] the antilog is \[4^x\]

Answer 41

Ah

I see o: