Answer:
The buffer d has the best buffering capacity.
Step-by-step explanation:
It is possible to obtain the pH of a buffer using Henderson-Hasselbalch formula:
pH = pka + log₁₀ [A⁻]/[HA]
For CH₃COOH/CH₃COONa buffer:
pH = 4,8 + log₁₀ [CH₃COONa]/[CH₃COOH]
a. pH of this buffer is:
pH = 4,8 + log₁₀ [0,4]/[0,2]
pH = 5,1
As buffering capacity can be thought of as the amount of strong acid that must be added to a buffered solution to change its pH by 1:
For a pH of 4,1:
4,1 = 4,8 + log₁₀ [0,4-x]/[0,2+x]
Where x are the moles of strong acid added.
0,200 = [0,4-x]/[0,2+x]
0,0400 + 0,2x = 0,4 - x
x = 0,3 mol
d. pH of this buffer is:
pH = 4,8 + log₁₀ [0,4]/[0,6]
pH = 4,62
For a pH of 3,62:
3,62 = 4,8 + log₁₀ [0,4-x]/[0,6+x]
Where x are the moles of strong acid added.
0,066 = [0,4-x]/[0,6+x]
0,0396 + 0,066x = 0,4 - x
x = 0,338 mol
e. pH of this buffer is:
pH = 4,8 + log₁₀ [0,3]/[0,6]
pH = 4,5
For a pH of 3,5:
3,5 = 4,8 + log₁₀ [0,3-x]/[0,6+x]
Where x are the moles of strong acid added.
0,050 = [0,3-x]/[0,6+x]
0,030 + 0,05x = 0,3 - x
x = 0,257 mol
Thus, buffer d needs more strong acid to change its pH. That means that have the best buffering capacity
You can do the same process using strong base (Increasing pH in 1) and you will obtain the same results!
I hope it helps!