Question

1. Calculate the pH of a solution of 0.2M acetic acid and 0.35M acetate ion. The pk

of acetic acid is 4.8.
pH = pk + log ([A] : [HA])
A.
5.10
B.
5.04
c.
5.25
D.
6.10
E.
6.00
of which

Answer 1

The correct answer is option C.

Step-by-step explanation:

The pH of the solution with weak acid and its conjugate base is given by the Henderson-Hasselbalch equation:

`pH=pK_a+\log[([A^-])/([HA])]`

Where:

`pK_a`= The negative logarithm of the dissociation constant of a weak acid

`[A^-]`= Concentration of conjugate base of a weak acid

`[HA]`= Concentration of weak acid

We are given a solution with acetic acid and acetate ion.

`HAc(aq)\rightleftharpoons H^+(aq)+Ac^-(aq)`

The concentration of acetic acid in a solution=
`[HAc]=0.2M`

The concentration of acetate ion in a solution =
`[Ac^-]=0.35M`

The pK_A of the acetic acid =
`pK_a=4.8`

The pH of the solution:

`pH=4.8+\log[(0.35 M)/(0.2M)]=5.04`

5.04 the pH of a solution of 0.2M acetic acid and 0.35M acetate ion.

Hence, the correct answer is option C.