Question

Calculate the pH of 1.00 L of a buffer 0.98 M CH3COONa/0.92 M CH3COOH before and after the addition of the following species

A. pH of starting buffer:

B. pH after addition if 0.040 mol NaOH:

C. pH after further addition if 0.101 mol HCl

Answer 1

A)

Ka = 1.8*10^-5

pKa = - log (Ka)

= - log(1.8*10^-5)

= 4.745

use:

pH = pKa + log {[conjugate base]/[acid]}

= 4.745+ log {0.98/0.92}

= 4.772

Answer: 4.77

B)

mol of NaOH added = 0.04 mol

CH3COOH will react with OH- to form CH3COO-

Before Reaction:

mol of CH3COO- = 0.98 M *1.0 L

mol of CH3COO- = 0.98 mol

mol of CH3COOH = 0.92 M *1.0 L

mol of CH3COOH = 0.92 mol

after reaction,

mol of CH3COO- = mol present initially + mol added

mol of CH3COO- = (0.98 + 0.04) mol

mol of CH3COO- = 1.02 mol

mol of CH3COOH = mol present initially - mol added

mol of CH3COOH = (0.92 - 0.04) mol

mol of CH3COOH = 0.88 mol

Ka = 1.8*10^-5

pKa = - log (Ka)

= - log(1.8*10^-5)

= 4.745

since volume is both in numerator and denominator, we can use mol instead of concentration

use:

pH = pKa + log {[conjugate base]/[acid]}

= 4.745+ log {1.02/0.88}

= 4.809

Answer: 4.81

C)

mol of HCl added = 0.101 mol

CH3COO- will react with H+ to form CH3COOH

Before Reaction:

mol of CH3COO- = 0.98 M *1.0 L

mol of CH3COO- = 0.98 mol

mol of CH3COOH = 0.92 M *1.0 L

mol of CH3COOH = 0.92 mol

after reaction,

mol of CH3COO- = mol present initially - mol added

mol of CH3COO- = (0.98 - 0.101) mol

mol of CH3COO- = 0.879 mol

mol of CH3COOH = mol present initially + mol added

mol of CH3COOH = (0.92 + 0.101) mol

mol of CH3COOH = 1.021 mol

Ka = 1.8*10^-5

pKa = - log (Ka)

= - log(1.8*10^-5)

= 4.745

since volume is both in numerator and denominator, we can use mol instead of concentration

use:

pH = pKa + log {[conjugate base]/[acid]}

= 4.745+ log {0.879/1.021}

= 4.68

Answer: 4.68