Question

A buffer that contains 0.45 M of a base, B and 0.46 M of its conjugate acid BH⁺, has a pH of 8.19. What is the pH after 0.02 mol of NaOH are added to 0.84 L of the solution?

Answer 1

pH = 8.24

Explanation

Henderson-Hasselbalch equation is

pH = pKa + log([B]/[BH+])

pKa = pH - log([B]/[BH+])

pKa = 8.19 - log ( 0.45M/0.46M)

pKa = 8.19 + 0.0095

pKa = 8.20

Initial moles of BH+ = (0.46mol/1L)×0.84L = 0.3864 mol

Initial moles of B = ( 0.45mol/1L)× 0.84L = 0.3780mol

NaOH react with conjucate acid

BH+ + OH^- ------> B + H2O

After addition

No of moles of B = 0.3780 + 0.02 = 0.3980mol

[B] = (0.3980mol/0.84L)×1L = 0.4738M

No of moles of BH+ = 0.3864 - 0.02 = 0.3664

[BH+] = (0.3664mol/0.84L )×1L = 0.4362M

Applying Henderson -Hasselbalch equation

pH = 8.20 + log(0.4738M/0.4362M)

pH = 8.20 + 0.04

pH = 8.24