Exercise 7 | The Properties of Acids and Bases

Benzoic acid C6H5COOH is a weak acid with a Ka = 6.3 x 10-5 M


1) Write the equation for the dissociation reaction of benzoic acid

2) Compare Ka and Kw and conclude

3) Calculate the concentration of all the species in a 0.250 M aqueous solution of benzoic acid

4) What is the pH of this solution?

1) Dissociation reaction:

C6H5COOH (aq) + H2O (l)  H3O+ (aq) + C6H5COO- (aq)

 

2) Ka = 6.3 x 10-5 M and Kw = 1.0 x 10-14 2

Ka >> Kw   ⇒   we can neglect the autoprotonation source of H3O+

[H3O+] depends only on the dissociation reaction

 

3) A t = 0, [C6H5COOH] = 0.250 M and [H3O+] = [C6H5COO-] = 0 M

At the equilibrium, [C6H5COOH] = 0.250 – X M and [H3O+] = [C6H5COO-] = X M


Ka = H3O+ C6H5COO-C6H5COOH = 6.3 x 10-5 M

⇒ at the equilibrium: X20.250 - X = 6.3 x 10-5

⇒ X2 + 6.3 x 10-5 X - 1.6 x 10-5 = 0

⇒ X = 3.9 x 10-3  or X = -4.0 x 10-3

⇒ The concentration cannot be negative ⇒ X = 3.9 x 10-3

⇒ at the equilibrium, [C6H5COOH] = 2.46 x 10-1 M and [H3O+] = [C6H5COO-] = 3.9 x 10-3 M

 

4) pH = -log [H3O+] = 2.4