VJC H2 CHEM P3 ANS FINAL Prelim

 VJC 2011 9647/03/PRELIM/11 [Turn over 1 Victoria Junior College 2011 H2 Chemistry Prelim Exam 9647/3 Suggested Answers 1 Use of the Data Booklet is relevant to this question. Cyanidin is a natural pH indicator that can be found in many red berries, including grapes, blackberry and cherry. Aqueous soluti ons of cyanidin change colour with pH, appearing red at pH < 3, violet at pH 7-8, and blue at pH > 11. In addition, cyanidins are antioxidants that can scavenge free radicals and counteract ageing caused by oxidative damage in our body tissues and cells. Cyanidin Cyanidin (CyH) is a weak monoprotic acid . In one experiment, 25.0 cm 3 of cyanidin was titrated against 0.10 mol dm–3 of dilute NaOH at 25°C. The following titration curve was obtained using a data logger. × 26.20 Volume of NaOH/ cm3 A B pH

 VJC 2011 9647/03/PRELIM/11 [Turn over 2 (a)(i) Explain what is meant by a weak monoprotic acid, with reference to CyH. [1] A weak monoprotic acid is an acid that dissociates partially to donate only one proton per acid molecule. i.e. CyH Cy + H+ (ii) Calculate the molar concentration of the cyanidin solution. [1] Since n CyH = nNaOH, Conc. of cyanidin = 0.25 20.2610.0  = 0.105 mol dm3 (iii) Given that the solution of cyanidin is onl y 0.8% dissociated into ions, calculate the value of Ka for cyanidin, stating clearly its units. [2] [H+] = [Cy] = 0.008 x 0.105 = 8.40 x 10-4 mol dm3 Ka = ][ ]][[ CyH HCy  K a = 105.0 )1040.8( 24 = 6.72 x 10-6 mol dm3 [OR 6.77 x 10 -6 if dissociation of CyH is considered] (iv) Explain, with the aid of an appropriate equation, why the pH at end-point B is greater than 7. [2] At end-point B, only the conjugate base of the weak acid, Cy is present. Cy + H2O CyH + OH Cy hydrolyses in water to produce excess OH ions. Hence, pH > 7. (v) Calculate the pH value at end-point B. Hence, state the colour change observed at the end-point. [4] Conc. of salt = 20.260.25 20.2610.0   = 0.0512 mol dm3 Kb = ][ ]][[   Cy OHCyH = a w K K = 6 14 1072.6 10    = 1.49 x 109 mol dm3 9 22 1049.10512.0 ][ ][ ][      OH Cy OH [OH -] = 8.73 x 106 pOH = - log 10 [OH-] = 5.06 pH = 14 – pOH = 8.94 Colour change at end-point: violet to blue.

 VJC 2011 9647/03/PRELIM/11 [Turn over 3 (b) An aqueous solution of cyanidin CyH and Cy- can act as a buffer. (i) Copy the titration curve onto your answer script and label I – the buffer region II – the point corresponding to maximum buffer capacity by indicating the pH and volume of NaOH added. [1] (ii) Explain, with the help of equations, how an aqueous mixture of CyH + and Cy can control pH when relatively small amounts of acid or base is added to the solution. [3] When a small amount of acid is added, Cy-(aq) + H+(aq)  CyH(aq) Large reservoir of Cy- ions from the salt in the buffer remove the additional H+ ions and the pH of the solution remains almost constant. When a small amount of base is added, CyH(aq) + OH-(aq)  Cy-(aq) + H2O(l) Large reservoir of CyH molecules in the buffer remove the additional OH - ions and the pH of the solution remains almost constant. (iii) In a sample of cherry juice at 25°C, the concentration ratio of Cy - to CyH was found to be 3 to 1. Using the Ka value determined above, calculate the pH value of the cherry juice. [1] pH = pKa + lg [ Cy H] ]-[ Cy = - lg (6.72 x 10 -6) + lg 1 3 = 5.65 pH 26.20 Volume of NaOH/cm3 A X B  buffer region X  maximum buffer capacity × 13.10 5.17

 VJC 2011 9647/03/PRELIM/11 [Turn over 4 (iv) Sam tried to dilute the same cherry juice sample by adding 100 cm 3 of deionized water at 25°C. Predict the pH of the resulting sample. [1] Dilution has no effect on [ Cy H] ]-[ Cy and Ka. Thus, pH remains the same at 5.65. (c) Cyanidin can be synthesized via a number of steps from either of the two molecules shown below: Molecule P Molecule Q (i) Which molecule P or Q will have a higher boiling point? Explain your answer. [2] P will have a higher boiling point as more energy will be required to overcome the stronger inter-molecular hydrogen bonding in P. On the other hand, due to the proximity of the –OH groups in molecule Q, intra- molecular hydrogen bonding predomin ates. As such, less energy is required to break the weaker dipole-dipole interactions in Q. (ii) Draw the structures of all the possible organic products when P and Q react separately with chlorine in trichloromethane. [2] Products formed from P: Products formed from Q: [Total: 20]

 VJC 2011 9647/03/PRELIM/11 [Turn over 5 2 (a) The elements in Period 3 range from metals on the left of the Periodic Table to non-metals on the right. By describing one physical property of the element and one chemical property of its chloride, explain why phosphorus can be regarded as a non-metal. Write equation(s) where relevant. Physical property of phosphorus: Phosphorus does not conduct electrici ty due to absence of delocalised electrons in its simple molecular structure. (OR It has low melting point due to weak dispersion forces to be overcome in its simple molecular solid structure.) Chemical property of PCl 3 or PCl5: It undergoes hydrolysis to form an acidic solution due to its covalent nature. PCl 3 + 3H2O → H3PO3 + 3HCl or PCl5 + 4H2O → H3PO4 + 5HCl [2] (b)(i) The element aluminium and its compounds have some properties characteristic of metals, and some of non-metals. Aluminium hydroxide, for example, is known to be amphoteric. Explain the meaning of the word in italics. An amphoteric compound is one which reacts with both acids and bases. Aluminium sulfate and calcium oxide are sometimes added to water supplies to co- precipitate suspended solids and bacteria. A small amount of aluminium-containing ions remains in solution and its presence in drinking water may contribute to the mental illness known as Alzheimer’s disease. (b)(ii) Write a balanced equation for the reaction that occurs when aluminium sulfate and calcium oxide is added to water, given that aluminium hydroxide is one of the products formed. Al2(SO4)3 + 3CaO + 3H2O  2Al(OH)3 + 3CaSO4 (b)(iii) Explain why adding too much calcium oxide would increase the probability of contracting Alzheimer’s disease. Write equations for all reactions that occur. CaO dissolves slightly to form an alkaline solution of Ca(OH)2. CaO + H2O  Ca(OH)2 OH- thus formed reacts with Al(OH)3 to form soluble Al(OH)4 -. Al(OH)3 + OH-  Al(OH)4 - [4]

 VJC 2011 9647/03/PRELIM/11 [Turn over 6 (c) Explain each of the following as fully as you can. Justify your answers with relevant data from the Data Booklet. Write balanced equations, including state symbols, for any reaction that occurs. (i) HC l  can be prepared by adding concentrated sulfuric acid to solid sodium chloride. However the yield of H I is very low when concent rated sulfuric acid is added to solid sodium iodide. NaCl (s) + H2SO4(l)  NaHSO4 (s) + HCl (g) Na I (s) + H2SO4(l)  NaHSO4 (s) + HI (g) 8 H I (g) + H2SO4(l)  4I2 (g) + H2S (g) + 4H2O (l) C l2 + 2e 2 Cl- Eθ = +1.36V I2 + 2e 2 I- E θ = +0.54V I- is a stronger reducing agent compared to C l- (as shown by the less positive Eθ value hence able to reduce H2SO4 to H2S. Hence very little of HI is left behind. [3] (ii) When a hot glass rod is plunged into a gas jar of gaseous H I, some violet vapour is seen. On repeating the experiment with HBr, no change is observed. Hot glass rod able to decompose HI to H2 an