To determine the pH of the solution containing sodium acetate and acetic acid, we need to consider the acid-base equilibrium of the acetic acid and its conjugate base, acetate ion.

The dissociation reaction of acetic acid can be represented as:
CH3COOH ⇌ CH3COO- + H+

Given that the pKa of acetic acid is 4.57, we can determine the equilibrium constant (Ka) using the formula:
Ka = 10^(-pKa)

Substituting the given pKa value, we get:
Ka = 10^(-4.57)

Now, we need to determine the concentrations of acetic acid and acetate ion in the solution.

For acetic acid:
Concentration (C1) = 0.01 M

For acetate ion:
Concentration (C2) = 0.10 M

Since acetic acid and sodium acetate are present in equal volumes (50 mL each), the initial concentrations can be considered the same as the final concentrations.

Next, we need to calculate the concentrations of H+ and CH3COO- ions in the solution.

Let's assume that x moles of acetic acid dissociate to form x moles of H+ ions and x moles of CH3COO- ions.

Therefore, the equilibrium concentrations can be written as:
[H+] = x
[CH3COO-] = x

The equilibrium concentration of acetic acid can be written as:
[CH3COOH] = C1 - x

Using the equilibrium constant expression, we can write:
Ka = [CH3COO-][H+]/[CH3COOH]

Substituting the values, we get:
10^(-4.57) = x*x/(C1 - x)

Since the initial concentration of acetic acid is much greater than x, we can assume that C1 - x ≈ C1.

Simplifying the equation, we get:
10^(-4.57) = x^2/C1

Now, solve for x:
x^2 = 10^(-4.57) * C1
x = √(10^(-4.57) * C1)

Substituting the values of C1 and solving, we find:
x ≈ 0.000316 M

Since the concentration of H+ ions is equal to x, the pH of the solution can be calculated as:
pH = -log[H+]
pH = -log(0.000316)
pH ≈ 5.50

Therefore, the pH of the solution containing 50 mL each of 0.10 M sodium acetate and 0.01 M acetic acid is approximately 5.50, which is closest to option 'D' (5.57).