Question

A sample of 25.00 ml of 0.1007 M acetic acid (HC2H2o2 was titrated with 0.1004 M NAOH what is the pH of the sample at the equicalence point of the titration? The Ka for accetic acid is 1.8 x 10^-5

Answer 1

The equivalence point will occur at 25.0 mL added NaOH(aq) solution. Picture this as a two step process. The first step involves stoichiometric reaction between the NaOH and the acetic acid; the second step involves hydrolysis of the acetate formed by the first step.

The NaOH(aq) completely reacts with the HC₂H₃O₂(aq). Mixing the two solutions causes the following reaction to occur:

HC2H3O2(aq)

+

NaOH(aq)

=

C2H3O2-(aq)

+ H2O(l)

moles before reaction	0.00250		0.00250		0
change in moles	-0.00250		-0.00250		+0.00250
moles after reaction	0		0		0.00250
molarity after reaction	0		0		0.0500

1. The C₂H₃O₂^-(aq) hydrolyzes.

   	C2H3O2-(aq)	+ H2O(l) =	HC2H3O2(aq)	+	OH-(aq)
   molarity before equilibrium	0.050		0		~0
   change in molarity	- x		+ x		+x
   molarity at equilibrium	0.050 - x		x		x

The equilibrium constant expression is

K_b = [OH^-(aq)][HC₂H₃O₂(aq)]/[C₂H₃O₂^-(aq)] = x²/(0.050 - x)

and solving this for x with

K_b = K_w/K_a

= 1.0x10^-14/1.8x10^-5

= 5.6x10^-10 gives x

= [OH^-(aq)] = 5.27x10^-6.

Then [H₃O⁺(aq)] = K_w/[OH^-(aq)]

= 1.0x10^-14/5.27x10^-6

= 1.9x10^-9, and pH ~ -log [H₃O⁺(aq)]

= 8.72