Question

A 300.0 mL buffer solution is 0.230 M in acetic acid and 0.230 M in sodium acetate.

Part A: What is the initial pH of this solution?

Part B: What is the pH after addition of 0.0050 mol of HCl?

Part C: What is the pH after addition of 0.0050 mol of NaOH?

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 {0.23/0.23}

= 4.745

Answer: 4.75

B)

mol of HCl added = 0.005 mol

CH3COO- will react with H+ to form CH3COOH

Before Reaction:

mol of CH3COO- = 0.23 M *0.3 L

mol of CH3COO- = 0.069 mol

mol of CH3COOH = 0.23 M *0.3 L

mol of CH3COOH = 0.069 mol

after reaction,

mol of CH3COO- = mol present initially - mol added

mol of CH3COO- = (0.069 - 0.005) mol

mol of CH3COO- = 0.064 mol

mol of CH3COOH = mol present initially + mol added

mol of CH3COOH = (0.069 + 0.005) mol

mol of CH3COOH = 0.074 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 {6.4*10^-2/7.4*10^-2}

= 4.682

Answer: 4.68

C)

mol of NaOH added = 0.005 mol

CH3COOH will react with OH- to form CH3COO-

Before Reaction:

mol of CH3COO- = 0.23 M *0.3 L

mol of CH3COO- = 0.069 mol

mol of CH3COOH = 0.23 M *0.3 L

mol of CH3COOH = 0.069 mol

after reaction,

mol of CH3COO- = mol present initially + mol added

mol of CH3COO- = (0.069 + 0.005) mol

mol of CH3COO- = 0.074 mol

mol of CH3COOH = mol present initially - mol added

mol of CH3COOH = (0.069 - 0.005) mol

mol of CH3COOH = 0.064 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 {7.4*10^-2/6.4*10^-2}

= 4.808

Answer: 4.81