Answer:
The buffer d has the best buffering capacity.
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
<em>x = 0,3 mol</em>
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
<em>x = 0,338 mol</em>
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
<em>x = 0,257 mol</em>
Thus, <em>buffer d needs more strong acid to change its pH. That means that have the best buffering capacity</em>
You can do the same process using strong base (Increasing pH in 1) and you will obtain the same results!
I hope it helps!
