QNEET-2022-Ch-408: NEET 2022 Chemistry - Class 11 NEET-2022-Ch-408 Solution: The pH of the solution containing 50 mL each of 0.10 M sodium…

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 : pH = p K a + log _ ( [ Conjugate Base ] [ Acid ] ) $\text{pH} = \text{p}K_a + \log\left(\frac{[\text{Conjugate Base}]}{[\text{Acid}]}\right)$ pH = p K a _ + lo g ( [ Acid ] [ Conjugate Base ] _ ) Here, sodium acetate acts as the conjugate base (acetate ion, C H 3 C O O _ $CH_3COO^-$ C H 3 _ CO O _ ) and acetic acid (CH_COOH) is the acid. Given the p K a $pK_a$ p K 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 : pH = 4.57 + log _ ( 0.10 0.01 ) $\text{pH} = 4.57 + \log\left(\frac{0.10}{0.01}\right)$ pH = 4.57 + lo g ( 0.01 0.10 _ ) 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