SRJC H2 Chem 2013 Prelim P2 Soln

1    1    2013 SRJC Prelim Paper H2 Chemistry P2 Suggested Solutions 1 Hard water, or water that contains high mineral content, such as calcium ions, pose serious problems in many industries. Other potential ions that could be present include zinc (through corrosion of pipings) and aluminium ions (whereby aluminium salts are added during the water purification process to start certain precipitation reactions). You are provided with a water sample that may contain the cations Ca 2+, Al3+ and Zn2+. The solution may also contain carbonate ions, CO 3 2–, and halide ions (either bromide or iodide). You are given only the following to conduct your test. Reagent Apparatus  aqueous ammonia  tests tubes  aqueous sodium hydroxide  funnel  dilute sulfuric acid  filter paper (a) Using the above only, write out a plan to separate each cation into three separate solutions. In your plan you should:  give a full description of the procedures you would use;  indicate the expected observations in each step. [5] Steps Test Observations 1 Add 1 cm 3 of the sample solution into a test tube. - 2 Add excess sodium hydroxide White ppt of Ca(OH) 2 will be produced. There may be slight dissolving of white ppt when NaOH is added in excess due to presence of A l3+ and Zn2+ ions. 3 Filter any precipitate being formed. White ppt of Ca(OH)2 will be collected as residue. Al3+ and Zn 2+ ions will be soluble and collected in the filtrate as Al(OH)4 – and Zn(OH)4 2- ions as a colourless solution. 4 Add dilute sulfuric acid Effervescence may be seen if CO3 2– is present. 5 Add aqueous ammonia in excess. White ppt of Al(OH)3 may be formed. White ppt of Zn(OH)2 may be formed, which is soluble in excess. 6 Filter any precipitate being formed. White ppt of Al(OH)3 will be collected as residue. Zn2+ ions will be soluble and collected in the filtrate as Zn(OH) 4 2- ions as a colourless solution. 1267

2    2    (b) A student proposed to use concentrated sulfuric acid instead of dilute sulfuric acid in (a). (i) Discuss how this change may affect the expected observations in (a). In the presence of a halide ion, concentrated sulfuric acid will act as an oxidising agent and react with the halide ions. This will interfere with the results as the coloured halogens produced may mask the white ppt of the metal hydroxides formed. (ii) Identify one potential safety hazard when the student make this change and state how you would minimise this risk. [4] Concentrated sulfuric acid is a corrosive acid. The student may adopt any one of the following : 1. Conduct the experiment in a fume hood 2. Wear safety goggles and gloves to prevent any contact with the eyes and skin. (c) You are also required to identify the anions present in the solution. (i) Propose two reagents to identify and distinguish between CO 3 2– and halide ion. For CO3 2–: dilute sulfuric acid. For halide ions: aqueous silver nitrate. (ii) Outline how you would use your chosen reagents, including conditions, to determine the anions present in the solution. In your plan you should:  give a full description of the of the order of adding the reagent;  quantity of reagent and apparatus used;  indicate the expected observations. [3] To 1 cm 3 of the sample in a test tube, add dilute sulfuric acid first, followed by the aqueous silver nitrate. No marks for switching sequence as the ppt produced will mask the CO 2 effervescence being formed. Observations: Effervescence will be produced if carbonate ions are present. Ppt will be produced if halides are present: 1. For bromide: cream ppt of AgBr 2. For iodide: yellow ppt of Ag I [Total : 12] H2O2 = 0.03 mol dm-3 1268

2 Iodo To the titr a illus (a) (b ) ometry is a a sample o oxidising a ated again s strated belo Iodine e iodine c a KI (aq) to (i) Su Iod int be (ii) Su in m In t I- + (iii) Dra ) 250 cm 3 concentr The con c via the io Run C w 0.0300 m of reactio The rat e triiodide triiodide The con c technique of hydrogen agent oxidis st standard ow. Re Re exists as a an only b e o form a da ggest a rea dine do hav termolecula tween wate ggest with making iodi the presenc + I2 ⇌ I3 - aw the struc of hydrog e rations in a centration o odometric tit was conduc mol dm-3 wh on is indepe e of reacti o formed ove produced w centration o used to ana n peroxide, ses the iodi d thiosulfate eaction 1 : H eaction 2 : I3 black soli d e prepared rk reddish-b ason for the e favourabl ar VDW i er, thus iod an equation ne more so ce of excess - cture of I3 - en peroxid e series of ex of the triiodi tration. cted with c o hile varying endent to hy on can be m er time and was 240 x 1 of hydrogen alyse the co excess but de to triiod i solution u s H2O2 + 3I- + 3 - + 2S2O3 2- d at stand a by dissol v brown solut insolubility le interactio is incom p ine does no n, how the a oluble. s KI, iodine formed and e was rea c xperiment n ide liberate oncentration the concen ydrogen ion measured b it was foun 0-3 mol dm- peroxide u oncentration t known a m ide ion. Th e sing starch 2H+  I3 - + -  3I - + S4 ard conditi o ving iodine tion. y of iodine in on with wate patible wi t ot dissolve i addition of e dissolve d d suggest its cted with a c named Run d at respec n of hydro g ntration of io ns. by the inc r d out that th -3 in Run C. used in run D n of an oxid mount of io d e triiodide i o indicator. + 2H2O 4O6 2- ons. An a q in excess n water. er molecule th the h y n water. excess pota ue to comp s shape. linear cidified iodi d C. ctive times c en peroxide odide. It is k ease in th e he maximum D was 0.020 ising agent dide is use d on solution The reacti o queous sol u potassium es as the ex ydrogen b assium iodi lex formatio de ions of can be dete e held con s known that t e concentr a m concentr 00 mol dm- 3  3  . d which is then ons are ution of iodide, xtensive onding de aids on: [4] various ermined stant at the rate ation of ration of 3. 1269

4    4    (i) Using the graph above and relevant information, determine the order of reaction with respect to hydrogen peroxide, iodide and hydrogen ion. Rate of reaction is independent to change in [H+]  zero order w.r.t. H+ Taking Run C only, since [H 2O2] is constant at 0.30 and order of reaction is zero wrt H+, the varied [I -] causes the product of [I3 - ] to change. Mentioned in the question the maximum concentration of [ I3 -] obtained is 240 x 10 -3 mol dm -3, using half-life method from the graph (show clear working on graph) Consistent half-life of 180 s was observed  1st order reaction wrt to I- 1270

5    5    Taking the initial rate for Run C and Run D where [I -] are constant while [H2O2] is varied. Let rate = k [H2O2]y[I-] From Run C, initial rate = 3 450 10 8.33 1060    mol dm-3 s-1 From Run D, initial rate = 3 470 10 5.83 10120    mol dm-3 s-1 Comparing Run C and Run D: y runC y runD 4y 4y rate k[I ] [0.03] rate k[I ] [0.02] 8.33 10 k[I ] [0.03] 5.83 10 k[I ] [0.02] y0 . 8 y1          Therefore, order of reaction w.r.t. hydrogen peroxide = 1 Rate = k[H2O2][I-] (not required by question) (ii) In run C , the excess hydrogen peroxide was titrated with 10.00 cm3 of 0.10 mol dm-3 of acidified potassium manganate(VII). Determine the total amount of triiodide formed. Amt of H2O2 initial = (250/1000) x 0.03 = 7.5 x 10-3 mol From the reaction b/w H2O2 and KMnO4: 5H2O2  2MnO4 - Amt of MnO4 - = (10/1000) x 0.10 = 1 x 10-3 mol Amt of H2O2 remaining = 5/2 x 1 x 10-3 = 2.5 x 10-3 mol Amt of H2O2 reacted = 7.5 x 10-3 – 2.5 x 10-3 = 5.0 x 10-3 mol   H2O2 + 3I- + 2H+  I3 - + 2H2O Amt of I3 - produced = 5.0 x 10-3 m