7_Acid_base - < Acid_base/AL90P1_3a-b 3(a Explain why...

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<<< Acid_base/AL90P1_3a-b 3. (a) Explain why phenolphthalein turns pink in a solution of sodium carbonate, but remain colourless in a solution of sodium hydrogencarbonate. (2 marks) 3. (b) Describe what would be observed when an excess of dilute ammonia solution is added to (i) aluminium sulphate solution; (ii) aluminium sulphate solution to which an excess of ammonium chloride has been added. Account for any differences in observation. (4 marks) <<< Acid_base/AL90P2_3b 3. (b) The pH of a 0.02 M ethanedioic acid solution is 1.8. Calculate the concentrations of each species present in the solution at 25 o C. [Dissociation constants of ethanedioic acid at 25 o C are: K 1 = 6.5 x 10 -2 mol dm -3 K 2 = 6.1 x 10 -5 mol dm -3 (7 marks) <<< Acid_base/AL91P13b 3. (b) Write equation(s) to describe the reaction of (i) ZnCl 2 (s) with water (i i) NaHCO 3 (s) with water (2 marks) <<< Acid_base/AL91P2_2c 2. (c) A weak base MOH has an ionization constant K b = 2.0 × 10 -5 mol dm -3 , where K b = [M + ] [OH - ]/[MOH]. Solution S is made up of 0.10 mol of MOH in 1.0 dm 3 , and a second solution T has 0.10 mol of MOH and 0.50 mol of MCl in 1.0 dm 3 . (i) Calculate the pH of solutions S and T. (i i) 0.01 mol of a strong acid HX is added separately to the solution S and T. The new pH value for solution S is 10.26; calculate that for solution T. What conclusion can you draw from this result? (K w = 1.0 × 10 -14 mol 2 dm -6 ) (8 marks) <<< Acid_base/AL92P2_3c 3. (c) (i) Explain the difference between the equivalence point and the end point of a titration. (ii) 40.0 cm 3 of an aqueous solution of a weak acid, HA(aq), was titrated with a strong base, MOH(aq), at 298 K. The initial pH, before the addition of base, was 2.70. At the equivalence point of the titration, the pH was 8.90. Calculate (I) the initial concentration of the acid. (II) the volume of the base added to reach the equivalence point.
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(III) the concentration of the base. (The dissociation constant of the weak acid and the ionic product of water at 298 K are
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7_Acid_base - < Acid_base/AL90P1_3a-b 3(a Explain why...

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