What is the pH of buffer solution prepared by mixing 0.01 M weak acid and 0.02 M salt of weak acid with strong base ? $\left(\mathrm{pK}_{\mathrm{a}}=4 \cdot 680\right)$

The Henderson-Hasselbalch equation for an acidic buffer is given by:

$\mathrm{pH} = \mathrm{pK}_{\mathrm{a}} + \log \left( \frac{[\text{Salt}]}{[\text{Acid}]} \right)$

Given $[\text{Acid}] = 0.01\ \text{M}$, $[\text{Salt}] = 0.02\ \text{M}$, and $\mathrm{pK}_{\mathrm{a}} = 4.680$:

$\mathrm{pH} = 4.680 + \log \left( \frac{0.02}{0.01} \right) = 4.680 + \log 2$

Since $\log 2 \approx 0.301$:

$\mathrm{pH} = 4.680 + 0.301 = 4.981$

The options are A) 4.379, B) 2.379, C) 4.981, D) 2.981, so the correct choice is $\boxed{\text{C}}$.