Question

Calculate the pH of 1.00 L of the buffer 1.06 M CH₃COONa/1.09 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 0.060 mol NaOH:



(c) pH after further addition of 0.109 mol HCl:

Answer 1

a) Henderson Hasselbalch equation is

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

pKa of acetic acid = 4.75

[ A- ] = [ CH3COO- ] = 1.06M

[ HA ] = [ CH3COOH ] = 1.09M

pH = 4.75 + log (1.06/1.09)

= 4.75 - 0.012

= 4.74

b) Added NaOH react with CH3COOH

CH3COOH + NaOH -----> CH3COONa + H2O

Initial Concentration

[ CH3COOH ] = 1.09M

[ CH3COO- ] = 1.06M

after adding of NaOH

[CH3COOH] = 1.09M - 0.06M = 1.03M

[ CH3COO- ] = 1.06M + 0.06M = 1.12M

Therefore,

pH = 4.75 + log ( 1.12/1.03)

= 4.75 + 0.036

= 4.79

c) Added HCl react with CH3COO-

CH3COO- + HCl ------> CH3COOH + HCl

Before adding HCl

[ CH3COO- ] = 1.12M

[ CH3COOH ] = 1.03M

After adding HCl

[ CH3COOH ] = 1.03M + 0.109M = 1.139M

[ CH3COO- ] = 1.12M - 0.109M = 1.011M

Therefore,

pH = 4.75 + log( 1.011 /1.139)

=4.75 - 0.052

= 4.70