1. Benzoic acid, C6H5COOH is a weak acid. When dissolved in water the following reaction takes place: C6H5COOH(aq) + H2O(aq)λ C6H5COO-(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 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 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 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) c) Addition of HCl leads to and increase in [Cl-(aq)] This means that [Ag+(aq) ][Cl-(aq)] is greater than Ksp And AgCl will precipitate from the solution i. e. solubility has decreased. (2 marks) (Marks available: 4)