What is the pH of a 0.1 M solution of acetic acid?

Acetic acid ionizes partially, so you need the or k value
of acetic acid (in data book), this should be given in the question. SO YOU NEED TO
CALCULATE THE CONCENTRATION OF H+ (OR
H3O+)

Calculation

CH3COOH(aq)
+ H2O(l) -----> CH3COO-(aq) + H3O+(aq)

K
IS DISSOCIATION CONSTANT WHICH IS THE PRODUCT OF CONCENTRATION OF PRODUCTS/PRODUCT OF
CONCENTRATION OF REACTANTS

K= [CH3COO-] * [H3O+] , CH3COOH
AND H20 ARE not included because its change in concentration is negligible. After
partial dissociation, [CH3COOH] IS STILL APPROXIMATELY
0.1M.

[CH3COO-] = [H3O+]

K of
CH3COOH=1.74 * 10^-5 (from data book)

=>1.74 * 10^-5
= [H30+]^2 / 0.1 [H30]^2 = 1.74*10^-5 * 0.1

= 1.74*10^-6
[H30+]

=
1.32*10^-3

pH=-log[H30+]

pH=
-log(1.32*10^-3)

pH=
2.88

=>pH of 0.1M acetic acid is
2.88

hOPE IT HELPS..!