millimole of acetic acid in 20mL=2
millimole of HCl in 20mL=1
millimole of NaOH=2.5

pH=PKa+log23/2
=4.74+log3
=4.74+0.48=5.22
JEE Main 2020 — Chemistry Physical Chemistry
3g of acetic acid is added to 250mL of 0.1MHCl and the solution made up to 500mL. To 20mL of this solution 21mL of 5MNaOH is added. The pH of the solution is ________
[Given: pKa of acetic acid =4.75, molar mass of acetic acid 60g/mol,log3=0.4771 , Neglect any changes in volume]
Held on 7 Jan 2020 · Verified 6 Jul 2026.
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