[hahahanna]

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.

[plankk]

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.

[hahahanna]

CH₃COOH + OH^- :rarrow:H₂O+CH₃COO^-

m_CH3COOH=1,32 g
Mw_CH3COOH=122 g/mol

n_CH3COOH=0,0108 mol

c_NaOH=1,00 M
V_NaOH=24*10^-3

n_NaOH=0,0024 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
0,0024 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??

[plankk]

c_NaOH=1,00 M
V_NaOH=24*10^-3

n_NaOH=0,0024 mol

Here is a mistake. Try to caluculate one more time n_NaOH.

[hahahanna]

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?
 

[hahahanna]

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?

[AWK]

20*10-3 ml

[hahahanna]

Sorry,

i mean 20 ml Solution A och 1 ml NaOH.