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

Answer:

The correct answer is option C.

Explanation:

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

$pH=pK_a+\log[\frac{[A^-]}{[HA]}]$

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:

$pH=4.8+\log[\frac{0.35 M}{0.2M}]=5.04$

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

Hence, the correct answer is option C.