Question

A 280.0 mL buffer solution is 0.250 M in acetic acid and 0.250 M in sodium acetate. For acetic acid, Ka=1.8×10−5.

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

Part B)  What is the pH after addition of 0.0150 mol of HCl?

Part C)  What is the pH after addition of 0.0150 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 {0.25/0.25}

= 4.745

Answer: 4.745

B)

mol of HCl added = 0.015 mol

CH3COO- will react with H+ to form CH3COOH

Before Reaction:

mol of CH3COO- = 0.25 M *0.28 L

mol of CH3COO- = 0.07 mol

mol of CH3COOH = 0.25 M *0.28 L

mol of CH3COOH = 0.07 mol

after reaction,

mol of CH3COO- = mol present initially - mol added

mol of CH3COO- = (0.07 - 0.015) mol

mol of CH3COO- = 0.055 mol

mol of CH3COOH = mol present initially + mol added

mol of CH3COOH = (0.07 + 0.015) mol

mol of CH3COOH = 0.085 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 {5.5*10^-2/8.5*10^-2}

= 4.556

Answer: 4.56

C)

mol of NaOH added = 0.015 mol

CH3COOH will react with OH- to form CH3COO-

Before Reaction:

mol of CH3COO- = 0.25 M *0.28 L

mol of CH3COO- = 0.07 mol

mol of CH3COOH = 0.25 M *0.28 L

mol of CH3COOH = 0.07 mol

after reaction,

mol of CH3COO- = mol present initially + mol added

mol of CH3COO- = (0.07 + 0.015) mol

mol of CH3COO- = 0.085 mol

mol of CH3COOH = mol present initially - mol added

mol of CH3COOH = (0.07 - 0.015) mol

mol of CH3COOH = 0.055 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 {8.5*10^-2/5.5*10^-2}

= 4.934

Answer: 4.93