Question

A buffer solution that is 0.10 M sodium acetate and 0.20 M acetic acid is prepared. Calculate the initial pH of this solution The K, for CH3COOH is 1.8 x 10-5 M. As usual, report pH to 2 decimal places. Answer: A buffered solution resists a change in pH. Calculate the pH when 21.2 mL of 0.031 M HCl is added to 100.0 mL of the above buffer. Answer:

Answer 1

1)

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 {0.1/0.2}

= 4.444

Answer: 4.44

2)

mol of HCl added = 0.031M *21.2 mL = 0.6572 mmol

CH3COO- will react with H+ to form CH3COOH

Before Reaction:

mol of CH3COO- = 0.1 M *100.0 mL

mol of CH3COO- = 10 mmol

mol of CH3COOH = 0.2 M *100.0 mL

mol of CH3COOH = 20 mmol

after reaction,

mol of CH3COO- = mol present initially - mol added

mmol of CH3COO- = (10 - 0.6572) mmol

mol of CH3COO- = 9.3428 mmol

mol of CH3COOH = mol present initially + mol added

mol of CH3COOH = (20 + 0.6572) mmol

mol of CH3COOH = 20.6572 mmol

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 {9.343/20.66}

= 4.4

Answer: 4.40