we want to prepare an acetate buffer of pH 6.420 from a 0.664 M acetic acid solution pKa =4.752 and a 2.50 M KOH solution. If we have 975 ml of the acetic acid solution, how many milliliters of the KOH solution do we need to add to make a buffer of pH 6.42?

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 =

i think it's c. i'm not sure but i think so.

answeridentity;