Question

calculate the ph of 0.1 solution of acetic acid. Ka for acetic acid is 1.8 x 10^-5 at 25°C.

Answer 1

heya....

your answer is in attachment ...✌ ✌

Answer 2

HELLO FRIEND HERE IS UR ANSWER ♥️;-

Acetic acid is weak acid
CH3COOH(aq) -----> CH3COO-(aq) + H+(aq) 

Ka ( Dissociation constant for acid) represented as
Ka= [CH3COO-] x [H+] / [CH3COOH] 
  CH3COOHCH3COO-H+Initial conc0.1 00Change in conc-x+x+xequilibrium conc0.1-x+x+x
Ka=  [x] [x] / [0.1-x] 
1.8 x 10-5 =  [x] [x] / [0.1-x] 
Now acetic acid is a very weak acid so its extent of dissociation is vey low and x should be neglected from denominator.

Considering  the above aspect the obtained equation is given as  :
1.8 x 10-5 =  [x] [x] / 0.1 
x2 = 1.8 x 10-5 x 0.1 
  = 1.8 x 10-6 
x = 1.34 x 10-3 = [H+]

pH= -log[H+] 
pH= -log(1.34 x 10-3) 
pH=  2.87
pH of 0.1M acetic acid = 2.87
$hope \: helps$