A student needs to prepare a buffer solution of propanoic acid and its sodium salt with pH 4. The ratio of $$\frac{[CH_3CH_2COO^-]}{[CH_3CH_2COOH]}$$ required to make buffer is Given: $$K_a(CH_3CH_2COOH) = 1.3 \times 10^{-5}$$

Required pH of the buffer is 4 and the $$K_a$$ of propanoic acid ($$CH_3CH_2COOH$$) is $$1.3 \times 10^{-5}$$. Using the Henderson-Hasselbalch equation:

$$\text{pH} = \text{p}K_a + \log\frac{[\text{CH}_3\text{CH}_2\text{COO}^-]}{[\text{CH}_3\text{CH}_2\text{COOH}]}$$

We calculate $$\text{p}K_a$$ as follows: $$\text{p}K_a = -\log(1.3 \times 10^{-5}) = -\log(1.3) - \log(10^{-5}) = -0.114 + 5 = 4.886$$

Substituting into the Henderson-Hasselbalch equation gives $$4 = 4.886 + \log\frac{[\text{CH}_3\text{CH}_2\text{COO}^-]}{[\text{CH}_3\text{CH}_2\text{COOH}]}$$, so $$\log\frac{[\text{CH}_3\text{CH}_2\text{COO}^-]}{[\text{CH}_3\text{CH}_2\text{COOH}]} = 4 - 4.886 = -0.886$$.

Thus, $$\frac{[\text{CH}_3\text{CH}_2\text{COO}^-]}{[\text{CH}_3\text{CH}_2\text{COOH}]} = 10^{-0.886} = \frac{1}{10^{0.886}}$$, and since $$10^{0.886} \approx 7.7$$, we obtain $$\frac{[\text{CH}_3\text{CH}_2\text{COO}^-]}{[\text{CH}_3\text{CH}_2\text{COOH}]} \approx \frac{1}{7.7} \approx 0.13$$.

The correct answer is Option B (0.13).