This is an acidic buffer solution made of a weak acid (acetic acid) and its salt with a strong base (sodium acetate). The pH of an acidic buffer is calculated using the Henderson-Hasselbalch equation: $\text{pH} = \text{p}K_a + \log\frac{[\text{Salt}]}{[\text{Acid}]}$

Given: Volume of sodium acetate ($V_1$) = $50\text{ mL}$ Molarity of sodium acetate ($M_1$) = $0.10\text{ M}$ Number of millimoles of salt = $50 \times 0.10 = 5\text{ mmol}$

Volume of acetic acid ($V_2$) = $50\text{ mL}$ Molarity of acetic acid ($M_2$) = $0.01\text{ M}$ Number of millimoles of acid = $50 \times 0.01 = 0.5\text{ mmol}$

Total volume of the mixture = $50\text{ mL} + 50\text{ mL} = 100\text{ mL}$

Concentration of salt in the mixture = $\frac{5\text{ mmol}}{100\text{ mL}} = 0.05\text{ M}$ Concentration of acid in the mixture = $\frac{0.5\text{ mmol}}{100\text{ mL}} = 0.005\text{ M}$ (Alternatively, since equal volumes are mixed, the ratio of their concentrations is the same as the ratio of their initial molarities: $\frac{0.10}{0.01} = 10$)

Now, substitute the values into the Henderson-Hasselbalch equation: $\text{pH} = 4.57 + \log\left(\frac{0.05}{0.005}\right)$ $\text{pH} = 4.57 + \log(10)$ $\text{pH} = 4.57 + 1 = 5.57$

Thus, the pH of the solution is $5.57$.