Ionic EquilibriumHard

Question

What will be the pH at the equivalence point during the titration of a 100 mL 0.2 M solution of CH3COONa with 0.2 M solution of HCl ? Ka = 2 × 10-5.

Options

A.3 - log√2
B.3 + log√2
C.3 - log 2
D.3 + log 2

Solution

CH3COONa + HCI  → NaCI + CH3COOH
t=0    20 m eq.         20 meq.           
teq.     -            -                   20 meq.
[CH3COOH] = = 0.1 M
pH = [pKa - log C] = [5 - log2 + 1] = [6 - log2] = 3 - log√2

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