The value of pH of 0.01 mol dm-3 CH3COOH (Ka=1.74×10-5) is-

1. 3.4

2. 3.6

3. 3.9

4. 3.0

Hint: [H+]=Ka.C
 
Step 1:
 
Calculate the concentration of H+ ion as follows:
 
Given that, (Ka=1.74×10-5)
 
Concentration of CH3COOH=0.01 mol dm-3
 
[H+]=Ka.C=1.74×10-5×0.01=4.17×10-4Step 2:pH=-log [H+]=-log (4.17×10-4)=3.4