To determine the volume of $$0.2 \mathrm{~M}$$ acetic acid needed to create a buffer solution with a pH of 4.94, we'll use the Henderson-Hasselbalch equation, which is:

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

Here:

$$\mathrm{pH} = 4.94$$
$$\mathrm{p}K_{\mathrm{a}}$$ of acetic acid $$= 4.76$$
$$\mathrm{[Salt]}$$ is the concentration of sodium acetate
$$\mathrm{[Acid]}$$ is the concentration of acetic acid

Given the volume and concentration of sodium acetate:

Volume of sodium acetate, $$V_{\text{salt}} = 100 \mathrm{~ml}$$
Concentration of sodium acetate, $$C_{\text{salt}} = 0.4 \mathrm{~M}$$

First, let's find the ratio of the concentrations of salt to acid using the Henderson-Hasselbalch equation:

$$4.94 = 4.76 + \log \left(\frac{[\mathrm{salt}]}{[\mathrm{acid}]}\right)$$

$$4.94 - 4.76 = \log \left(\frac{[\mathrm{salt}]}{[\mathrm{acid}]}\right)$$

$$0.18 = \log \left(\frac{[\mathrm{salt}]}{[\mathrm{acid}]}\right)$$

To remove the logarithm, we raise 10 to the power of both sides:

$$10^{0.18} = \frac{[\mathrm{salt}]}{[\mathrm{acid}]}$$

$$1.51 = \frac{[\mathrm{salt}]}{[\mathrm{acid}]}$$

Let's denote the volume of acetic acid to be added as $$V_{\text{acid}}$$ in milliliters. The concentration of acetic acid is given as $$0.2 \mathrm{~M}$$. The total volume of the buffer solution will be $$100 \mathrm{~ml} + V_{\text{acid}}$$. Therefore, the concentrations of salt and acid in the buffer are:

$$[\mathrm{Salt}] = \frac{0.4 \mathrm{~M} \times 100 \mathrm{~ml}}{100 \mathrm{~ml} + V_{\text{acid}}}$$
$$[\mathrm{Acid}] = \frac{0.2 \mathrm{~M} \times V_{\text{acid}}}{100 \mathrm{~ml} + V_{\text{acid}}}$$

Now, substituting these into the ratio we found:

$$1.51 = \frac{\frac{0.4 \mathrm{~M} \times 100 \mathrm{~ml}}{100 \mathrm{~ml} + V_{\text{acid}}}}{\frac{0.2 \mathrm{~M} \times V_{\text{acid}}}{100 \mathrm{~ml} + V_{\text{acid}}}}$$

The $$100 \mathrm{~ml} + V_{\text{acid}}$$ terms cancel out, so we have:

$$1.51 = \frac{0.4 \times 100}{0.2 \times V_{\text{acid}}}$$

$$1.51 = \frac{40}{0.2 \times V_{\text{acid}}}$$

$$1.51 \times 0.2 \times V_{\text{acid}} = 40$$

$$0.302 \times V_{\text{acid}} = 40$$

Finally, solving for $$V_{\text{acid}}$$:

$$V_{\text{acid}} = \frac{40}{0.302} \approx 132.45 \mathrm{~ml}$$

However, this value does not match any of the given options exactly. Typically, slight approximations may occur due to rounding off values during intermediate steps. Considering the options, the closest value is:

Option A: 132.1 ml

Ionic Equilibrum

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