1. Calculate moles of acids mixed

• For $\mathrm{HCl}$: $n(\mathrm{HCl}) = 0.01 \times 0.200 = 0.002\ \text{mol}$

Since $\mathrm{HCl}$ is a strong monoprotic acid, it gives: $0.002\ \text{mol of } \mathrm{H}^+$

• For $\mathrm{H_2SO_4}$: $n(\mathrm{H_2SO_4}) = 0.01 \times 0.400 = 0.004\ \text{mol}$

In this level of problem, $\mathrm{H_2SO_4}$ is treated as a strong dibasic acid, so it gives: $2 \times 0.004 = 0.008\ \text{mol of } \mathrm{H}^+$

2. Total moles of $\mathrm{H}^+$

$n(\mathrm{H}^+)_{\text{total}} = 0.002 + 0.008 = 0.010\ \text{mol}$

3. Total volume after mixing

$V_{\text{total}} = 200\ \text{mL} + 400\ \text{mL} = 600\ \text{mL} = 0.600\ \text{L}$

4. Concentration of $\mathrm{H}^+$ in mixture

$[\mathrm{H}^+] = \frac{0.010}{0.600} = \frac{1}{60} = 1.667 \times 10^{-2}$

5. Calculate pH

$\mathrm{pH} = -\log[\mathrm{H}^+] = -\log\left(\frac{1}{60}\right) = \log 60$

Now, $\log 60 = \log(6 \times 10) = \log 6 + 1$ $\log 6 = \log 2 + \log 3 = 0.30 + 0.48 = 0.78$

Therefore, $\mathrm{pH} = 1 + 0.78 = 1.78$

6. Option check

• A: $1.14$ ❌
• B: $1.78$ ✅
• C: $2.34$ ❌
• D: $3.02$ ❌

Hence, the correct answer is B.