Question

Calculate the pH of 1.00 L of the buffer 1.03 M CH₃COONa/0.97 M CH₃COOH before and after the addition of the following species. (Assume there is no change in volume.)

(a) pH of starting buffer:

(b) pH after addition of 0.080 mol NaOH:

(c) pH after further addition of 0.144 mol HCl:

Answer 1

A) pH of starting buffer

Henderson- Hasselbalch equation is

pH = pKa + log([A^-]/[HA])

where,

pKa = pKa of weak acid , 4.74

A^- = conjucate base , CH3COO^-

HA = weak acid , CH3COOH

substituting the values

pH = 4.74 + log(1.03M/0.97M)

pH = 4.74 + 0.03

  pH = 4.77

B) pH after addition of 0.080mol NaOH

Initial moles of CH3COOH = 0.97mol

Initial moles of CH3COO^-= 1.03mol

moles of NaOH added = 0.080mol

NaOH react with weak acid CH3COOH

CH3COOH + OH^- -----------> CH3COO^- + H2O

Affter addition of NaOH

moles of CH3COOH = 0.97mol - 0.080mol = 0.89mol

moles of CH3COO^- = 1.03 mol + 0.080mol = 1.11mol

[CH3COOH] = 0.8900M

[CH3COO^-] = 1.110M

Applying Henderson - Hasselbalch equation

pH = pKa + log( 1.110M/0.8900M)

pH = 4.74 + 0.10

pH = 4.84

C) pH after adfition of 0.144mol of HCl

HCl react with conjucate base CH3COO^-

CH3COO^- + H⁺ --------> CH3COOH

After addition of HCl

moles of CH3COOH  = 0.97mol + 0.144mol = 1.114mol

moles of CH3COO^- = 1.03mol - 0.144mol = 0.886 mol

[CH3COOH] = 1.114M

[CH3COO^-] = 0.886M

Applying Henderson-Hasselbalch equation

pH = 4.74 + log( 0.8860M/1.114M)

pH = 4.74 -0.10

  pH = 4.64