200 mL of 0.01 M HCl is mixed with 400 mL of 0.01 M $$H_2SO_4$$. The pH of the mixture is

We have 200 mL of 0.01 M HCl mixed with 400 mL of 0.01 M $$H_2SO_4$$. We need to find the pH of the mixture.

HCl is a strong monoprotic acid, so the moles of $$H^+$$ from HCl = $$0.01 \times 0.2 = 0.002$$ mol = 2 mmol.

$$H_2SO_4$$ is a strong diprotic acid, contributing 2 moles of $$H^+$$ per mole. So the moles of $$H^+$$ from $$H_2SO_4$$ = $$2 \times 0.01 \times 0.4 = 0.008$$ mol = 8 mmol.

Total moles of $$H^+$$ = $$0.002 + 0.008 = 0.010$$ mol. The total volume is $$200 + 400 = 600$$ mL = 0.6 L.

The concentration of $$H^+$$ in the mixture is: $$[H^+] = \frac{0.010}{0.6} = \frac{1}{60}$$ M.

The pH is: $$\text{pH} = -\log\left(\frac{1}{60}\right) = \log 60 = \log(6 \times 10) = \log 6 + 1 = 0.778 + 1 = 1.778 \approx 1.78$$

Hence, the correct answer is Option B.