Question

Buffer capacity is a measure of a buffer solution\'s resistance to changes in pH as strong acid or base is added. Suppose that you have 115 mL of a buffer that is 0.180 M in both acetic acid (CH3COOH) and its conjugate base (CH3COO–). Calculate the maximum volume of 0.100 M HCl that can be added to the buffer before its buffering capacity is lost.

Answer 1

Pka of CH3COOH = 4.75
USE:
pH = pKa + log {[CH3COO-]/[CH3COOH]}
   = 4.75 + log {0.18/0.18}
   =4.75 + 0
   = 4.75

Buffer capacity is expressed as the amount of strong acid or base, that must be added to 1 liter of the solution to change its pH by one unit.

SO pH has to become 4.75-1 = 3.75 as acid is being added
Condition is:
pH = pKa + log {[CH3COO-]/[CH3COOH]}
3.75= 4.75 +log {[CH3COO-]/[CH3COOH]}
log {[CH3COO-]/[CH3COOH]} = -1
[CH3COO-]/[CH3COOH] = 0.1
Since volume is same and concentration = number of moles / volume.
Above expression can be written as
n(CH3COO-)/n(CH3COOH) = 0.1

initial moles of CH3COOH = initial moles of CH3COO- = M*V = 0.18*115= 20.7 mmol

Let V' mL of HCL is added
then number of moles of HCl added = 0.1*V' mmol

CH3COO- + HCl -----> CH3COOH + Cl-
0.1*V' mmol of CH3COO- will react and form 0.1*V' mmol of CH3COOH

Final moles of CH3COOH = 20.7 + 0.1*V'
Final moles of CH3COO- = 20.7 - 0.1*V'

put these values in:
n(CH3COO-)/n(CH3COOH) = 0.1
(20.7 - 0.1*V') / (20.7 + 0.1*V') =0.1
20.7 - 0.1*V' = 2.07 + 0.01*V'
0.11*V'=18.63
V' = 169

So upto 169 mL its fine
Answer: 169 mL