[shipitpls]

This problem says: Find the pH of the solution prepared by dilution of a solution of known acidity.

Original solution                 Volume taken (ml)                  Diluted to a volume of (ml)             New pH
a) ph=2.204                                   50                                          150
b) [H3O+]=1.41x10^-4  M                 25                                           100
c) [OH-]=2.90x10^-3  M                   100                                           200

I did them, but they turned out wrong... can anyone give me some insight on what to do?

[Borek]

Show your work, it is hard to tell what you did wrong.

Besides, if you know only pH of the starting solution and don't know anything about the substance present - question doesn't make sense. Different solutions may behave differently.

[shipitpls]

Well I used McVc=MdVd (c is the concentrated d is the diluted). Md= pH [H3O+] as well.

So I used:

(for a)

2.204(pH)x0.05L (volume stated)=0.15L*Md
Md=0.735=pH

Did the same for the others.

However the teacher said "You cannot use pH in the dilution equation, convert to H3O+"
It seems weird because of "Md= pH [H3O+]" which was written on the board along with "McVc=MdVd".

As to solving them, you know the volume before and after, which is enough.

[Borek]

the teacher said "You cannot use pH in the dilution equation, convert to H3O+"

And teacher was right. McVc=MdVd holds if you use concentrations. pH is not concentration - it is minus log of the concentration. So what you have to do is calculate concentration from pH, calculate new concentration after dilution, take log and change sign to get new pH.

[shipitpls]

Yeah that's what I thought... How do I go from pH to concentration though?  I mean how do I calculate  it from 2.204=-log(x)... mathy is a bit fuzzy right now.

EDIT: Nvm. lol, I just forgot log properties, figured it out, it's 10^2.204 to get x.

[shipitpls]

So, I got the following:

a) pH=2.68
b) pH=4.45
c) Do you just work with the M of OH here? If so I did this:
2.90*10^-3M*0.1L=Md*0.2L
Md=0.00145
pH=2.84

I guess they should be right, but can anyone confirm this?

[AWK]

2.84 is called pOH
pH = 14 - pOH

[shipitpls]

Oops, yeah that's what I meant 11.16.

Also, so I avoid starting a new thread is this  right?

a) Find the pH of pure water at 50.0 degrees (Celsius) if Kw=5.47*10-14

I just took the square root of that and then -log of it to get the pH?

[AWK]

OK

pH of pure (neutral) water above 25 C is below 7

[shipitpls]

Yeah, that's what I got, 6.63.

Ty.

[Borek]

See the table here:

http://www.chembuddy.com/?left=pH-calculation&right=water-ion-product