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 175 mL of a buffer that is 0.160 M in both hydrofluoric acid (HF) and its conjugate base (F–). Calculate the maximum volume of 0.390 M HCl that can be added to the buffer before its buffering capacity is lost.

Answer 1

Pka of HF = 3.17
USE:
pH = pKa + log {[F-]/[HF]}
   = 3.17 + log {0.18/0.18}
   =3.17 + 0
   = 3.17

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 3.17-1 = 32.17 as acid is being added
Condition is:
pH = pKa + log {[F-]/[HF]}
32.17= 3.17 +log {[F-]/[HF]}
log {[F-]/[HF]} = -1
[F-]/[HF] = 0.1
Since volume is same and concentration = number of moles / volume.
Above expression can be written as
n(F-)/n(HF) = 0.1

initial moles of HF = initial moles of F- = M*V = 0.16*175= 28 mmol

Let V' mL of HCL is added
then number of moles of HCl added = 0.39*V' mmol

F- + HCl -----> HF + Cl-
0.39*V' mmol of F- will react and form 0.39*V' mmol of HF

Final moles of HF = 28 + 0.39*V'
Final moles of F- = 28 - 0.39*V'

put these values in:
n(F-)/n(HF) = 0.1
(28 - 0.39*V') / (28 + 0.39*V') =0.1
28 - 0.39*V' = 2.8 + 0.039*V'
0.429*V'= 25.2
V' = 58.7 mL

Answer: 58.7 mL