1. Calculate moles of acetic acid added

Given mass of acetic acid $=3\text{ g}$, molar mass $=60$.

$n(\text{CH}_3\text{COOH})=\frac{3}{60}=0.05\text{ mol}$

This is made up to $500\text{ mL}$, so concentration is

$[\text{CH}_3\text{COOH}]=\frac{0.05}{0.5}=0.1\text{ M}$

2. Calculate concentration of HCl after dilution

Initial HCl: $250\text{ mL}$ of $0.1\text{ M}$.

Moles of HCl:

$n(\text{HCl})=0.25\times 0.1=0.025\text{ mol}$

After making total volume $500\text{ mL}$,

$[\text{HCl}]=\frac{0.025}{0.5}=0.05\text{ M}$

So the final mixed solution contains:

• Acetic acid: $0.1\text{ M}$
• HCl: $0.05\text{ M}$

3. Take 20 mL of this solution

For $20\text{ mL}=0.02\text{ L}$:

• Moles of acetic acid: $n(\text{CH}_3\text{COOH})=0.1\times 0.02=0.002\text{ mol}$

• Moles of HCl: $n(\text{HCl})=0.05\times 0.02=0.001\text{ mol}$

4. Add NaOH

Volume of NaOH $=\frac12\text{ mL}=0.0005\text{ L}$, concentration $=5\text{ M}$.

Moles of NaOH added:

$n(\text{NaOH})=5\times 0.0005=0.0025\text{ mol}$

5. Neutralization sequence

NaOH will first neutralize strong acid HCl completely:

$\text{HCl}+\text{NaOH}\to \text{NaCl}+\text{H}_2\text{O}$

HCl consumes $0.001\text{ mol}$ NaOH.

Remaining NaOH:

$0.0025-0.001=0.0015\text{ mol}$

This remaining NaOH neutralizes acetic acid:

$\text{CH}_3\text{COOH}+\text{NaOH}\to \text{CH}_3\text{COONa}+\text{H}_2\text{O}$

Initial acetic acid = $0.002\text{ mol}$.

After reaction with $0.0015\text{ mol}$ NaOH:

• Acetic acid left: $0.002-0.0015=0.0005\text{ mol}$

• Acetate formed: $0.0015\text{ mol}$

So we now have a buffer with:

• acid = $0.0005$
• salt = $0.0015$

6. Use Henderson–Hasselbalch equation

$\text{pH}=\text{p}K_a+\log\frac{[\text{salt}]}{[\text{acid}]}$

Since volume change is neglected, use mole ratio directly:

$\text{pH}=4.75+\log\frac{0.0015}{0.0005}$

$\text{pH}=4.75+\log 3$

Given $\log 3=0.4771$,

$\text{pH}=4.75+0.4771=5.2271$

$\boxed{\text{pH}\approx 5.23}$

7. Comparison with stored correct answer

Stored correct answer: $5.22$ to $5.24$

Our calculated answer is $5.23$, which lies in this range.