# Exam-style Questions: Ionic Equilibria

1. Benzoic acid, C6H5COOH is a weak acid. When dissolved in water the following
reaction takes place:

C6H5COOH(aq) + H2O(aq)λC6H5COO-(aq) + H3O+(aq)

a) Explain what is meant by the term 'weak acid'.

(1 mark)

b) A solution is made of benzoic acid at 25°C. The acid dissociation
constant for benzoic acid is 6.3 x 10-5 moldm-3. Write
down an expression for Ka.

(1 mark)

c) Calculate the pH of a 0.005M solution of benzoic acid.

(1 mark)

(Marks available: 3)

Answer outline and marking scheme for question: 1

a) A weak acid is one that does not completely dissociate when dissolved
in aqueous solution.

(1 mark)

b) Ka = [C6H5COO-][ H3O+]

[C6H5COOH]

(1 mark)

c) From equation we know that [C6H5COO-]
= [ H3O+]. We can replace these concentrations in the
expression of Ka with x2.

Take [C6H5COOH] = 0.005M (as degree of dissociation
is small there will be a negligible effect on this concentration).

Ka = x2 / 0.005 = 6.3 x 10-5 (1)

x2 = 0.005 x (6.3 x 10-5) = 3.2 x 10-7

x = 5.7 x10-4 (1)

[H+] = 5.7 x 10-4 moldm-3

pH = -lg[H+]

pH of solution = -lg(5.7 x 10-4) = 3.2

(1 mark)

(Marks available: 3)

2. Calculate the pH of the following solutions.

a) 0.1M HCl(aq).

(2 marks)

b) 0.1M Be(OH)2(aq)

(2 marks)

(Marks available: 4)

Answer outline and marking scheme for question: 2

a) Acid is fully dissociated - [H+] = 0.1 moldm-3

pH = -lg[H+]

pH = -lg[0.1] = 1

(2 marks)

b) Base is fully dissociated - [OH-] = 2 x 0.1 = 0.2 moldm-3+

pOH = -lg[OH-] = 0.7

pH + pOH = 14

pH = 14 -0.7 = 13.3

(2 marks)

(Marks available: 4)

3. Precipitation reactions are an important method of identifying the inorganic
ions in a solution.

For example silver nitrate is used to identify halides in solution.

a) Write down the expression for the solubility product, Ksp, for silver
chloride, AgCl.

(1 mark)

b) Ksp for AgCl is 1.8 x 10-10 mol2dm-6. Calculate
the concentration of Ag+(aq) in a saturated solution of
AgCl.

(1 mark)

c) What effect will the addition of HCl have on the solubility of AgCl?

(2 marks)

(Marks available: 4)

Answer outline and marking scheme for question: 3

a) Ksp = [Ag+(aq) ][Cl-(aq)]

(1 mark)

b) From equation [Ag+(aq) ] = [Cl-(aq)]

Ksp = [Ag+(aq) ]2 = 1.8 x 10-10

[Ag+(aq) ] = 1.3 x 10-5 moldm-3

(1 mark) 