To find the pH of the solution containing a mixture of sodium acetate and acetic acid, we can use the Henderson-Hasselbalch equation. This equation is given by :

$ \text{pH} = \text{p}K_a + \log\left(\frac{[\text{Conjugate Base}]}{[\text{Acid}]}\right) $

Here, sodium acetate acts as the conjugate base (acetate ion, $ CH_3COO^- $) and acetic acid (CH₃COOH) is the acid. Given the $ pK_a $ of acetic acid is 4.57, we can plug in the values.

The molarity of sodium acetate and acetic acid are given as 0.10 M and 0.01 M, respectively. Since the volumes of the solutions are equal, the molarities can be directly used in the equation :

$ \text{pH} = 4.57 + \log\left(\frac{0.10}{0.01}\right) $

Let's calculate the pH :

The pH of the solution containing 50 mL each of 0.10 M sodium acetate and 0.01 M acetic acid is approximately 5.57. Therefore, the correct option is :

Option A : 5.57