Question

The pH of a solution containing 0.1M of acetic acid and 0.05 M of sodium acetate is ________ (rounded off to 2 decimal places).
The $pK_a$ value of ionization of acetic acid is 4.76.

Answer 1

Correct Answer: 4.46

Step 1: Apply the Henderson-Hasselbalch equation. \[ \text{pH} = \text{pKa} + \log \left(\frac{\text{[base]}}{\text{[acid]}}\right) \] Step 2: Substitute the given values. \[ \text{pH} = 4.76 + \log \left(\frac{0.05}{0.1}\right) \] Step 3: Calculate the pH. \[ \log \left(\frac{0.05}{0.1}\right) = -0.3010 \] \[ \text{pH} = 4.76 - 0.3010 = 4.459 \] Rounded to two decimal places, the pH is 4.46