Question

3. Household vinegar is an aqueous acetic acid solution. A particular sample of vinegar has a pH of 2.90. Assuming that the vinegar contains only acetic acid (CH3COOH), with a ki of 1.8 x 10-5 at this temperature, calculate the initial concentration of acetic acid in the vinegar in units of molarity.

Answer 1

Consider a weak acid CH3COOH. It dissociates into water as shown below.

CH3COOH (aq) + H2O (l) $1584050021272_blob.png$ H3O + (aq) + CH3COO - (aq)

The dissociation constant for above reaction is Ka = [H3O +] [ CH3COO -] / [ CH3COOH ] = 1.8 $\times$ 10 -05

$\therefore$ [ CH3COOH ] = [H3O +] [ CH3COO -] / Ka

According to reaction, we have [H3O +]= [ CH3COO -]

We know that, pH = - log [H3O +]                          

Therefore, [H3O +] = 10 - pH = 10 -2.90 = 1.26 $\times$ 10 -03 M = [ CH3COO -]

Let's use ICE table.

M	CH3COOH $1584050021272_blob.png$ H3O + + CH3COO -
I	X
C	-1.26 $\times$ 10 -03	+ 1.26 $\times$ 10 -03	+ 1.26 $\times$ 10 -03
E	X - 1.26 $\times$ 10 -03	1.26 $\times$ 10 -03	1.26 $\times$ 10 -03

Substituting above values in  [ CH3COOH ] = [H3O +] [ CH3COO -] / Ka , we get

[ CH3COOH ] = ( 1.26 $\times$ 10 -03) ( 1.26 $\times$ 10 -03 ) / 1.8 $\times$ 10 -05

[ CH3COOH ] = 0.0882 M = X - 1.26 $\times$ 10 -03

Initial concentration of acetic acid ( X ) = 0.0882 M + 1.26 $\times$ 10 -03 M

Initial concentration of acetic acid = 0.08946 M

ANSWER : [ CH3COOH ] = 0.08946 M