What will be the value of pH of 0.01 mol dm-3 CH3COOH (Ka=1.74×10-5)?

(1) 3.4

(2) 3.6

(3) 3.9

(4) 3.0

Hint: [H+]=Ka.C
 
Given that, (Ka=1.74×10-5)
Concentration of CH3COOH=0.01 mol dm-3
[H+]=Ka.C=1.74×10-5×0.01=4.17×10-4pH=-log [H+]=-log (4.17×10-4)=3.4