Ionic Equilibrium Part 1

3gm Acetic Acid + 250 ml 0.1 M HCl +Water → made to 500 ml solution.
⇒ 500 ml solution has 25 meq of HCl
∴ 20 ml solution has 50 meq of CH3COOH, 1 meq of HCl, 2 meq of CH3COOH
We have added 2.5 meq. of NaOH (5M,

1

2

ml)
Finally, NaOH & HCl are completely consumed and we are left with 0.5 meq of CH3COOH and 1.5 meq of CH3COONa
pH=pKa+log‌

1.5

0.5

= 4.75 + log 3 = 4.75 + 0.4771 = 5.2271