A buffer containing 0.2000 M of acid, HA, and 0.1500 M of its conjugate base, A-; has a pH of 3.35. What is the pH after 0.0015 mol NaOH

conjugate-base

#1

A buffer containing 0.2000 M of acid, HA, and 0.1500 M of its conjugate base, A-; has a pH of 3.35. What is the pH after 0.0015 mol NaOH is added to 0.500 L of this solution?

Answer:

pH = pKa + log 75/100 pKa = 3.35 - log 75/100= 3.35 + 0.125 =
pKa = 3.475
The 0.0015 moles = 1.5 mmoles of base will react with 1.5 mmoles of the acid form making an additional 1.5 mmoles of the buffer base form and reducing the acid form by 1.5 mmol
so new pH = pKa + log (75 + 1.5 mmoles) /(100 -1.5 mmole)

pH = 3.475 + log 76.5 mmole/98.5 mmol = 3.475 -0.11 = 3.365= 3.37