To find the pH of the solution containing sodium acetate and acetic acid, we will use the Henderson-Hasselbalch equation.$\\$The Henderson-Hasselbalch equation is given by:$\\ pH = pK_a + \log \left( \frac{[Salt]}{[Acid]} \right) \\$Given:$\\$• $$Volume of sodium acetate solution$= 50$ mL$\\$• $$Concentration of sodium acetate$= 0.10$ M$\\$• $$Volume of acetic acid solution$= 50$ mL$\\$• $$Concentration of acetic acid$= 0.01$ M$\\$• $$p$K_a$ of $CH_3COOH = 4.57 \\$Since the volumes of both solutions are equal, the concentrations can be directly used in the equation.$\\$Calculate the pH:$\\$pH$= 4.57 + \log \left( \frac{0.10}{0.01} \right) \\$Simplify the expression inside the logarithm:$\\ \frac{0.10}{0.01} = 10 \\$Substitute back into the equation:$\\$pH$= 4.57 + \log(10) \\$Since $\log(10) = 1,$ we have:$\\$pH$= 4.57 + 1 \\$Therefore, the pH of the solution is $5.57. \\$This corresponds to Option 1.