Chemical Forums
Chemistry Forums for Students => Physical Chemistry Forum => Topic started by: hahahanna on August 04, 2009, 03:45:27 PM
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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]??
Thanks for answers.
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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.
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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
Answer: pH 4,45
Is this correct??
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cNaOH=1,00 M
VNaOH=24*10-3
nNaOH=0,0024 mol
Here is a mistake. Try to caluculate one more time nNaOH.
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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
Answer: pH 4,45
--- I saw that i caluculate with right NaOH (0,024mol)
But i missed it in the beginning, but it is right now?
:)
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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????
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20*10-3 ml
???
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Sorry,
i mean 20 ml Solution A och 1 ml NaOH.