mol of acetic acid=0.0232L *0.376mol/L=0.00872mol
The neutralization rxn of acetic acid with barium hydroxide is
shown below:
2CH3COOH +Ba(OH)2 ---->Ba( CH3COO)2 + 2H2O
2 mol acetic acid reacts with 1 mol barium hydroxide to form 1
mol barium acetate.
mol of Ba(OH)2 added at eqvalence point=1/2*mol of acetic
acid=1/2*0.00872mol=0.00436 mol
mol of salt barium acetate. formed=0.00436 mol
mol of acetate formed=2*0.00436=0.00872 mol
At equivalence point all the acetic acid is neutralized,so only
salt is present that hydrolyses as follows:
Ba( CH3COO)2 ---->Ba2+ +2CH3COO-
CH3COO- +H2O <----> CH3COOH +OH-
kb=kw/ka=[CH3COOH] [OH-]/[CH3COOH]
ka (acetic acid)=acid dissociation constant=1.8*10^-5
kb(acetate)=kw/ka=10^-14 /(1.8*10^-5)=0.555*10^-9
0.555*10^-9=kb=[CH3COOH] [OH-]/[CH3COOH]
ICE table
|
[CH3COO-] |
[CH3COOH] |
[OH-] |
| initial |
0.00872mol |
0 |
0 |
| change |
-x |
+x |
+x |
| equilibrium |
0.00872-x |
x |
x |
0.555*10^-9=kb=x*x/(0.00872-x)
or, 5.55*10^-10=x^2/(0.00872)Â Â Â Â [
x<<0.0436]
x=[OH-]=0.0484*10^-5M
[H3O+]=10^-14/[OH-]=10^-14/0.22*10^-5M=4.545*10^-9
pH=-log [H3O+]=-log (4.545*10^-9M)=8.3
pH=8.3
