Question

What is the pH at the equivalence point in the titration of a 23.2 mL sample of a 0.376 M aqueous acetic acid solution with a 0.432 M aqueous barium hydroxide solution? pH =

Answer 1

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)     [ 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