Answer:
253.5mL must be added
Explanation:
To find pH of a buffer we need to use H-H equation:
pH = pKa + log [Acetate] / [Acetic acid]
Where pH is pH we want: 6.42; pKa is 4.752 and [] could be taken as moles of each species.
6.42 = 4.752 + log [Acetate] / [Acetic acid]
1.668 = log [Acetate] / [Acetic acid]
46.5586 = [Acetate] / [Acetic acid] (1)
And moles of acetate buffer = Initial moles of acetic acid are:
975mL = 0.975L * (0.664mol / L) = 0.6474 moles acetate buffer =
0.6474 = [Acetate] + [Acetic acid] (2)
Replacing (2) in (1):
46.5586 = 0.6474 - [Acetic Acid] / [Acetic acid]
46.5586 [Acetic acid] = 0.6474 - [Acetic Acid]
47.5586 [Acetic acid] = 0.6474
[Acetic acid] = 0.01361 moles
And [Acetate]: 0.6474 - 0.01361moles
[Acetate] = 0.63379moles
Now, in the beginning we have just Acetic acid that is converted in acetate with the use of KOH:
Acetic acid + KOH → Acetate
That means the moles of KOH added are the moles of acetate we need. That means we need to add 0.63379 moles of KOH that comes from a 2.50M. The volume is:
0.63379 moles KOH * (1L / 2.50 moles) = 0.2535L =
