# Chemical Forums

## Chemistry Forums for Students => Physical Chemistry Forum => Topic started by: hahahanna on August 04, 2009, 09:45:27 AM

Title: Buffert pH
Post by: hahahanna on August 04, 2009, 09:45:27 AM
Hi!

im going to do a buffer with C6H5COOH and with NaOH. I solved 1,32 g C6H5COOH in water then i establish 25 ml, 1,00M NaOH, and then i dilute with water too 200 ml. Now im going to calculate pH.

With the things i know I can calculate nC6H5COOH = m/Mw= 0,0109 mol
and nNaOH=c*V=0,0024 mol

How do i calculate the pH if my solvent was dilute with water too 200ml?
but what formula will i used after that?
pH=pKa+log [A]/[HA]??

Title: Re: Buffert pH
Post by: plankk on August 05, 2009, 12:44:50 AM
but what formula will i used after that?
pH=pKa+log [A]/[HA]??

Yes. And in this equation you have symbols like that [X]. It means the molar concentration of X's molecule/ion.
Title: Re: Buffert pH
Post by: hahahanna on August 05, 2009, 10:46:29 AM
CH3COOH + OH- :rarrow:H2O+CH3COO-

mCH3COOH=1,32 g
MwCH3COOH=122 g/mol

nCH3COOH=0,0108 mol

cNaOH=1,00 M
VNaOH=24*10-3

nNaOH=0,0024 mol

Because i have a new volume, 200ml i do like this:

n=c*V  :rarrow: c=n/V
0,0108 mol/200*10-3=0,054 M

and i do the same with NaOH

n=c*V  :rarrow: c=n/V
0,0024 mol/200*10-3=0,12 M

After that i looked up pKa for CH3COOH=4,19

pH=4,19+log (0,12/0,054) = 4,54

Is this correct??

Title: Re: Buffert pH
Post by: plankk on August 05, 2009, 07:38:57 PM
cNaOH=1,00 M
VNaOH=24*10-3

nNaOH=0,0024 mol

Here is a mistake. Try to caluculate one more time nNaOH.
Title: Re: Buffert pH
Post by: hahahanna on August 05, 2009, 09:56:11 PM
CH3COOH + OH- :rarrow:H2O+CH3COO-

mCH3COOH=1,32 g
MwCH3COOH=122 g/mol

nCH3COOH=0,0108 mol

cNaOH=1,00 M
VNaOH=24*10-3

nNaOH=0,024 mol

Because i have a new volume, 200ml i do like this:

n=c*V   c=n/V
0,0108 mol/200*10-3=0,054 M

and i do the same with NaOH

n=c*V   c=n/V
0,024 mol/200*10-3=0,12 M

After that i looked up pKa for CH3COOH=4,19

pH=4,19+log (0,12/0,054) = 4,54

--- I saw that i caluculate with right NaOH (0,024mol)
But i missed it in the beginning, but it is right now?
:)
Title: Re: Buffert pH
Post by: hahahanna on August 06, 2009, 12:14:56 PM
So is this right that  have done?

Because i have another question. The solution i calculated the pH will be called  "Solution A".
But now im going to calcultare the pH again but now i will have 20*10-3 ml of "solution A", and then i will add 1*10-3 , 1,00 M NaOH to make a new solution (solution B).

So, how will I do here?
I have n CH3COOH and i have n NaOH.
but how will I add this together to get out 20 ml solutions A and then mix it with NaOH again????
Title: Re: Buffert pH
Post by: AWK on August 06, 2009, 10:53:42 PM
20*10-3 ml
???
Title: Re: Buffert pH
Post by: hahahanna on August 06, 2009, 11:18:57 PM
Sorry,

i mean 20 ml Solution A och 1 ml NaOH.