Question

1. Calculate the pH of the buffer of a solution of 0.02M acetic acid (pKa = 4.76) and 0.04 M acetate

2. Based on your answer to the previous question, how much of a 0.5 M solution of HCl must be added to 200 mL of the above mentioned solution to lower the pH to 2.50? Ignore volume changes.

Answer 1

1)

use:

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

= 4.76+ log {0.04/0.02}

= 5.06

Answer: 5.06

2)

Let volume of HCl be V mL

mol of HCl added = 0.5*V mmol

Before adding HCl

Before Reaction:

mol of CH3COONa = 0.04 M *200.0 mL

mol of CH3COONa = 8 mmol

mol of CH3COOH = 0.02 M *200.0 mL

mol of CH3COOH = 4 mmol

0.5*V HCl will react with 0.5*V of CH3COONa to form extra 0.5*V of CH3COOH

After adding HCl

mol of CH3COOH = 4+0.5*V mmol

mol of CH3COONa = 8-0.5*V mmol

use:

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

2.5 = 4.76+log {[CH3COONa]/[CH3COOH]}

log {[CH3COONa]/[CH3COOH]} = -2.26

[CH3COONa]/[CH3COOH] = 0.005495

So,

(8-0.5*V)/(4+0.5*V) = 0.005495

8-0.5*V = 0.022 + 0.002748*V

(0.5+0.002748)*V = 8 - 0.022

0.502748*V = 7.978

V = 15.9 mL

Answer: 15.9 mL