Question

Benzoic acid is a weak monoprotic acid occurring in most of berries. The acid dissociation constant of benzoic acid is 6.3 x 10-5. Find pH and percent of dissociation of 0.001 M solution of benzoic acid.

Answer 1

Given:

The values given in the question are as follows;

Benzoic acid dissociation constant Ka = 6.3 x 10-5.

Concentration of benzoic acid = 0.001M.

Solution:

Step 1 - Determine [H+]

Benzoic acid dissociates in water as:

C6H5COOH      H+ + C6H5COO-

Initial concentration 0.001M 0 0

Change concentration 0.001 - X + X   + X

Equilibrium concentration 0.001 - X X X

The formula for Ka = [H+] [C6H5COO-] / [C6H5COOH]

Substituting the values;

6.3 x 10-5 = [X][X] / [0.001 -X]

0.000063 = [X]2 / [0.001 - X]

0.000063 [0.001 - X] = [X]2

0.000000063 - 0.000063X = X2

X2 + 0.000063X -  0.000000063 = 0

Solving the above quadratic equation, X = 0.0002215 or X = -0.0002845.

Since X represents the concentration of ions in solution, it cannot be negative. Hence X =  0.0002215. That is [H+] =  0.0002215M

Step 2 - pH calculation:

pH = - log [H+]

= - log(0.0002215) = - (-3.6546) = 3.6546 = 3.65

Hence the pH is 3.65.

Step 3 - Percentage dissociation:

We know that at equilibrium [H+] =  [C6H5COO-] = 0.0002215M; Also the initial [C6H5COOH] = 0.001M

Percentage dissociation = ([H+] / [C6H5COOH]) x 100 = (0.0002215M / 0.001M) x 100 = 0.2215 x 100 = 22.15%

Hence the percentage dissociation of 0.001M benzoic acid is 22.15%.