100 mL of 0.15 M weak acid HA is mixed with 100 mL of 0.1 M NaOH . What is the pH of the resulting solution? [Given: pK _ a of HA = 4.70 , 2 = 0.30 , 3 = 0.48 ]

Step-by-step solution

Initial millimoles of HA = 100 0.15 = 15 mmol Initial millimoles of NaOH = 100 0.1 = 10 mmol The strong base NaOH completely reacts with the weak acid HA to form the salt NaA : HA + NaOH NaA + H ₂ O Millimoles of NaA formed = 10 mmol Millimoles of HA remaining = 15 - 10 = 5 mmol The resulting solution is an acidic buffer. According to the Henderson-Hasselbalch equation: pH = pK _ a + ( [ Salt ] [ Acid ] ) Since the volume is the same for both salt and acid in the mixture, we can use the ratio of their millimoles: p