VCE mid‐year exams Chemistry By Christopher Dwyer In 1992 VCAB, the then controlling body of Year 12 assessment, introduced a June examination in Chemistry. As 2012 VCE students you will, at 9.00 AM on Wednesday June 13, sit the twenty first and last June examination in Chemistry. Consider two questions from that 1992 June exam show that key principles of volumetric analysis have not changed. Q1. In an acid‐base titration to determine the exact concentration of an approximately 1.1 M NaOH solution, 0.0900 M HCl is in a 50.0 mL burette. The NaOH(aq) is added to the titration flask using a pipette. The most appropriate volume for the pipette is A. 2.00 mL, B. 5.00 mL, C. 10.00 mL, D. 20.00 mL. Q2. 40.0 g NaOH was dissolved in water and the solution was left exposed to the air for several days. During this time several 50 mL portions of the solution were analysed for hydroxide ion concentration which was shown to decrease over time. Explain why the c(OH‐) in the solution decreased over time. Well before the days of digital camera, the 1992 exam included a stoichiometry question based on black and white photography. Q3. When a photographic plate containing 0.50 g of silver bromide, AgBr, was developed, 35 per cent of the AgBr was reduced to Ag. The unreacted AgBr was removed by placing the plate in 100 mL of 0.200 M Na2S2O3 solution. The reaction occurring is represented by the equation AgBr(s) + 2S2O32‐(aq) → Ag(S2O3)23‐(aq) + Br‐(aq) (a) Calculate the n(Na2S2O3) which reacted with AgBr. (b) Calculate the concentration of the Na2S2O3(aq) solution after the reaction with AgBr. The current Chemistry study design has been in effect since 2008, so the 2008‐2011 VCAA examinations, and assessment reports, provide you with examples not only of recent stoichiometry related questions but also student performance on those questions. Often performance on stoichiometry is a good indicator of overall exam performance. Do not be put off by an ‘unfamiliar’ or seemingly detailed context; the fundamentals inevitably remain the same. Question 4, in Section B of the 2011 examination provides a detailed description of the method by which the % P2O5 in a fertiliser sample is determined. It proved to be a challenging question. The description contains up to 8 items of data, of which only two and the given molar masses need to be used to ‘Calculate the percentage, by mass, of P2O5, in the fertiliser.’ Often thinking about the ‘final step’ in the calculation, in this case ‘dividing the m(P2O5) by the m(fertiliser) and multiplying by 100’ will help you choose the appropriate data at the right stage in your solution. The contrast in student performance on this question, and the perhaps more familiar context Question 5, was quite marked. Make sure you attempt both questions before checking out the assessment report. An important part of volumetric analysis is the choice of indicator in an acid‐base titration. The equivalence point of a reaction is the exact point where the amounts of reactants added match the mole ratio indicated in the equation for the reaction. The pH at the equivalence point determines the choice of indicator and that pH depends on the products of the reaction. In the titration of CH3COOH(aq) with NaOH(aq), the pH at the equivalence point is 8.8, not because NaOH is in excess – which it cannot be at the equivalence point – but because CH3COO‐(aq) produced in the titration reaction CH3COOH(aq) + NaOH(aq) → CH3COO‐(aq) + Na+(aq) + H2O(l) makes the solution slightly basic. The indicator chosen is ideally one that includes 8.8 in its recognised colour change range, i.e. phenolphthalein (8.3‐10.0). Make use of the Data Book. A common misconception is that because CH3COOH(aq) is a weak acid, then less NaOH(aq) would react with it than with the same number of mole of HCl(aq). Acid strength not a factor in titration with a strong base. This point has caught out many students since 1992, and also did so on Question 11 in Section A in 2011. There are a number of instances where ‘dilution’ is encountered in VCE chemistry, most commonly in volumetric analysis and quantitative analyses using AAS or UV‐Vis spectroscopy. The reasons for dilution are usually clearly linked with the ‘range’ of acceptable data. In a titration for example, experiments are usually designed for a titre around 20 mL. Should a situation arise where a titration required the refilling of a burette before the endpoint is reached, then dilution of the sample being analysed is a relatively simple solution. However it would not be a random dilution, it would be by a factor that would bring the titre close to 20.0 mL. Similarly, in spectroscopic analysis, if the absorbance of the ‘unknown’ falls outside the range of the calibration curve, then the solution could be diluted ‘by a factor that would bring its absorbance inside the range of the calibration curve’. AAS and UV‐Vis provide an interesting comparison with respect to light sources. Both depend on the ‘absorption’ of energy as electrons move to higher energy levels. In AAS, in an analysis for Ni content, a Ni lamp is used, because only Ni atoms in the flame will absorb wavelengths emitted by that light source. In UV‐Vis however, a key part of the analysis is the determination, via an absorption spectrum, of a wavelength that will be strongly be absorbed by the active ingredient but not by other species in the solution. Spectroscopy is a very powerful tool in modern chemistry. An empirical formula and the peak with the highest m/z ratio on a mass spectrum may allow you to determine a molecular formula, say C3H6O2. Knowledge of organic chemical pathways suggests that the compound may be a carboxylic acid or an ester. The IR spectrum of the compound, showing a peak in the absorption band 1670‐
1750 cm‐1 but no broad absorption band at 2500‐3300 cm‐1 suggests that the compound is an ester, either CH3COOCH3, or HCOOCH2CH3. Three signals on the high resolution 1H NMR spectrum, including a singlet, a triplet and a quartet indicates the compound is ethyl methanoate. Can you suggest a reaction pathway for the production of ethyl methanoate from ethene and methane? From the ester perspective, on Friday April 15, Qantas undertook Australia’s first commercial flight using sustainably derived biofuel. Biofuels, usually methyl or ethyl esters of fatty acids are produced by condensation reactions, as are large biomolecules and proteins. The production of proteins in our cells goes back to DNA. Make sure you check back overall all dot points in the key knowledge for both areas of study. The assessment reports for the 2008‐2011 VCAA exams are recommended reading after you have attempted the questions on the exams. Prepare thoroughly. Don’t forget to practice ‘reading time’. How well you use that time in the exam sets the scene for the following 90 minutes. Answers to questions. 1. A – on the basis of an approximately 20 mL titre – do the calculations. 2. The same reason we do not use NaOH as a primary standard, reacts with atmospheric CO2. 3. (a) 3.5x10‐3 mol, (b) 0.165 M.