1. Identify the titration type

We have:

• $100\,\text{mL}$ of $0.1\,\text{M}$ HCl in a beaker
• $0.1\,\text{M}$ NaOH added gradually

This is a strong acid–strong base titration.

2. Initial moles of HCl

$n_{\text{HCl}} = 0.1 \times 0.100 = 0.010\,\text{mol}$

So initially there are $0.010$ mol of $\mathrm{H^+}$.

Initial concentration of HCl is $0.1\,\text{M}$, hence

$\text{pH} = -\log(0.1) = 1$

So the graph must start at about pH = 1.

3. Equivalence point

NaOH concentration is also $0.1\,\text{M}$.

Volume of NaOH needed to neutralize HCl:

$V = \frac{0.010}{0.1} = 0.100\,\text{L} = 100\,\text{mL}$

Thus, equivalence occurs when 100 mL NaOH has been added.

At equivalence for strong acid–strong base titration:

$\text{pH} = 7$

So the graph must show a sharp rise around $100\,\text{mL}$ added and pass through pH 7 at equivalence.

4. Nature of the curve before and after equivalence

Before equivalence

Acid is in excess, so pH increases slowly from 1.

For example, after adding $98\,\text{mL}$ NaOH:

• moles NaOH added: $0.1 \times 0.098 = 0.0098\,\text{mol}$
• excess HCl: $0.010 - 0.0098 = 0.0002\,\text{mol}$
• total volume: $100 + 98 = 198\,\text{mL} = 0.198\,\text{L}$
• $[\mathrm{H^+}]$: $\frac{0.0002}{0.198} \approx 1.01 \times 10^{-3}$
• pH: $\approx 3$

So just before equivalence, pH is still acidic.

At equivalence

$\text{pH} = 7$

Just after equivalence

After adding $102\,\text{mL}$ NaOH:

• moles NaOH added: $0.1 \times 0.102 = 0.0102\,\text{mol}$
• excess NaOH: $0.0102 - 0.010 = 0.0002\,\text{mol}$
• total volume: $0.202\,\text{L}$
• $[\mathrm{OH^-}]$: $\frac{0.0002}{0.202} \approx 9.9 \times 10^{-4}$
• pOH: $\approx 3$
• pH: $\approx 11$

So near equivalence, the pH changes very sharply from about $3$ to about $11$ in only a few mL.

5. Shape of the correct graph

Therefore, the correct graph must:

1. Start at pH $\approx 1$
2. Increase gradually at first
3. Show a very steep vertical rise near 100 mL of NaOH added
4. Pass through pH = 7 at the equivalence point
5. Level off in the basic region after equivalence

This is the standard strong acid–strong base titration curve.

Hence, Option C is correct.

6. Comparison with stored answer

Stored correct answer: C

Our derived answer: C

So they agree.