Question

Given the ph of 7 explain how many hydroxide -OH and hydronium H30+ are present in the solution

Answer 1

It depends on the temperature, because Kw can vary. At SATP their concentrations are both equal to 1.0 x 10^{-7}.

Answer 2

It didn't say anything else but that...

Answer 3

Then go with 1.0 x 10^{-7} for each.

Answer 4

but I don't know how to do this stuff...at all :(

Answer 5

So,
\[pH = -\log_{10}[H+]\]
\[[H3O+] = 10^{-pH}\]
This gives you the concentration of Hydroxide and Hydronium ions present in a solution at 7pH

Answer 6

Well from both of these equations you can find Hydronium (H3O+) and hydroxide (OH) concentrations

Answer 7

Oh and the concentration is in Molarity (mol/L) hence the square brakets

Answer 8

you can convert mols to particles by using avagadros number which will give you particles per liter and an answer to your question (if you were given a volume you can figure out by multiplication how many particles of each you have in it)

Answer 9

oh okay!!!(:

Answer 10

OOPs

Answer 11

Use only the bottom equation I gave you

Answer 12

the H3O one?

Answer 13

You still need to find -OH in solution which is easy by the following relations

pH + pOH = 14
so you can find pOH then use pOH to find molarity of -OH with the following formula

\[pOH = -\log_{10}[-OH]\]

Answer 14

yeah the H3O one

Answer 15

After you find -OH concentration just do the same thing I described above to find particles of -OH

Answer 16

ahh! okay!! thank you!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!(:

Answer 17

No problem if you get lost let me know, getting off my bus now so I wont be around for a few hours, but I will check back later so post any problems if you run into them. You need to use algebra to solve for the concentrations but I have faith you can here have the logarithmic rules for help,

http://www.mathwords.com/l/logarithm_rules.htm

Answer 18

@Australopithecus Im so confused :(
I don't have it

Answer 19

[H3O+]=10^(−pH)
becomes
[H3O+] = 10^(-7)M =10^(-7)mol/L

pH + pOH = 14
7 + pOH = 14
pOH = 7

pOH=−log10[−OH]
[−OH] = 10^(-pOH)
[-OH] = 10^(-7)M = 10^(-7)mol/L

Avogadro number is 6.022141*10^(23)particles/mol
you can find particles/L by multiplication (units cancel)

if you have a volume for the solution you can figure out how many OH and H3O particles are present if you multiply particles/L by the volume of the solution

Number of -OH equals number of H3O also you should notice so yeah