Verified Solution • 2025–26 Session moderate

QJEEM20Ch282e: JEE Main 2020 Chemistry - Class 11 JEEM20Ch282e Solution: 3 g of acetic acid is added to 250 mL of 0.1 M HCL and the…

3 g of acetic acid is added to 250 mL of 0.1 M HCL and the solution made up to 500 mL. To 20 mL of this solutions 12 mL of 5 M NaOH is added. The pH of the solution is __________.

[Given : pKa of acetic acid = 4.75, molar mass of acetic of acid = 60 g/mol, log 3 = 0.4771] Neglect any changes in volume.

Step-by-Step Verified Solution

milimole of acetic acid in 20 ml
= 360×1000×20500 = 2

milimole of HCl in 20 ml = 25 × 20500 = 1

milimole of NaOH 20 ml = 12×5 = 2.5

NaOH + CH3COOH → CH3COONa + H2O
1.5 2 0 0
0 0.5 1.5


pH = pKa + log1.50.5

= 4.74 + log 3 = 5.22

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