Question

Acetic acid has a Ka of 1.8*10^-5. Three acetic acid/ acetate buffer solutions, A,B, and C, wer made using varying concentrations: 1. [acetic acid] ten times greater than [acetate] 2. [acetate] ten times greater than [acetic acid] 3. [acetate] = [acetic acid] Match each buffer to the expected pH pH = 3.74 pH= 4.74 pH = 5.74

Answer 1

Step-by-step explanation:

It is given that
`K_(a)` for acetic acid is
`1.8\tiemes 10^(-5)`. And, its
`pK_(a)` value will be calculated as follows.

`pK_(a) = -log_(10) K_(a)`

=
`-log_(10) (1.8 * 10^(-5)`

= 4.74

And, according to the Henderson-Hasselbalch equation,

pH =
`pK_(a) + log ([Salt])/([Acid])`

(1). When [Acetic acid] is ten times greater than [acetate]

This means that
`\frac{[\text{Acetate}]}{[\text{Acetic acid}]} = (1)/(10)`

So, pH =
`pK_(a) + log \frac{[Acetate]}{[\text{Acetic Acid}]}`

=
`4.74 + log (1)/(10)`

= 3.74

(2). When [Acetate] ten times greater than [Acetic acid]

This means that
`\frac{[Acetate]}{\text{Acetic acid}}` =
`(10)/(1)`

pH =
`4.74 + log_(10) 10`

= 5.74

(3). When [acetate] = [acetic acid]

This means that
`\frac{\text{Acetate}}{\text{Acetic acid}}` = 1

pH =
`4.74 + log_(10) 1`

= 4.74