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Question: What will be the pH at the equivalence point during the titration of a 100 mL 0.2 M solution of CH3C...

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.

A

3 – log2\sqrt{2}

B

3 + log2\sqrt{2}

C

3 – log 2

D

3 + log 2

Answer

3 – log2\sqrt{2}

Explanation

Solution

CH3COONa + HCI ⟶\longrightarrow NaCI + CH3COOH

t=0 20 m eq. 20 meq. teq.

– – 20 meq.

[CH3COOH] = 20200\frac{20}{200} = 0.1 M

pH = 12\frac{1}{2} [pKa – log C] = 12\frac{1}{2} [5 – log2 + 1]

= 12\frac{1}{2} [6 – log2] = 3 – log 2\sqrt{2}