Question

A 260.0 mL buffer solution is 0.200 M in acetic acid and 0.200 M in sodium acetate

(For all answers express them using two decimal places)

A) What is the initial pH of this solution

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

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 {0.2/0.2}
= 4.745

Answer: 4.75

B)

mol of HCl added = 0.015 mol

CH3COO- will react with H+ to form CH3COOH

Before Reaction:
mol of CH3COO- = 0.2 M *0.26 L
mol of CH3COO- = 0.052 mol

mol of CH3COOH = 0.2 M *0.26 L
mol of CH3COOH = 0.052 mol

after reaction,
mol of CH3COO- = mol present initially - mol added
mol of CH3COO- = (0.052 - 0.015) mol
mol of CH3COO- = 0.037 mol

mol of CH3COOH = mol present initially + mol added
mol of CH3COOH = (0.052 + 0.015) mol
mol of CH3COOH = 0.067 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 {3.7*10^-2/6.7*10^-2}
= 4.487

Answer: 4.49

c)

mol of NaOH added = 0.015 mol

CH3COOH will react with OH- to form CH3COO-

Before Reaction:
mol of CH3COO- = 0.2 M *0.26 L
mol of CH3COO- = 0.052 mol

mol of CH3COOH = 0.2 M *0.26 L
mol of CH3COOH = 0.052 mol

after reaction,
mol of CH3COO- = mol present initially + mol added
mol of CH3COO- = (0.052 + 0.015) mol
mol of CH3COO- = 0.067 mol

mol of CH3COOH = mol present initially - mol added
mol of CH3COOH = (0.052 - 0.015) mol
mol of CH3COOH = 0.037 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 {6.7*10^-2/3.7*10^-2}
= 5.003

Answer: 5.00