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 =

answer is: c. the experiment shows that the red substance experienced a chemical change.

this is chemical change (chemical reaction), because two new substances are formed (metalig liquid and gas), the atoms are rearranged and the reaction is followed by an energy change.

this is example of chemical decomposition.

chemical decomposition (analysis) is the separation of a single chemical compound into its two or more elemental parts or to simpler compounds.  

chemical decomposition is opposite of chemical synthesis.