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Aim. To find the identity of X(OH)2 (a group II metal hydroxide) by determining its solubility from a titration with 0.05 mol dm-3 HCL

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Introduction

To find the identity of X(OH)2 (a group II metal hydroxide) by determining it's solubility from a titration with 0.05 mol dm-3 HCL Theory:1. Titrations are the reaction between an acid solution with an alkali. In this reaction (called neutralization), the acid donates a proton (H+) to the alkali (base). When the two solutions are combined, the products made are salt and water. For example: 2HCl(aq) + X(OH)2 (aq) XCl2 (aq) + 2H2O (l) This shows the one of the products i.e. salt being XCl2 and water. So titration therefore helps to find the concentration for a solution of unknown concentration. This involves the controlled addition of a standard solution of known. Indicators are used to determine, at what stage has the solution reached the 'equivalence point'(inflextion point). This means at which, does the number of moles base added equals the number of moles of acid present. i.e. pH 7 Titration of a strong Acid with a Strong Base: As shown in the graph, the pH goes up slowly from the start of the tiration to near the equivalence point. i.e (the beginning of the graph). ...read more.

Middle

XCl2 (aq) + 2H2O (l) Therefore the ratio is 2:1 of 2 HCl : 1 X(OH)2 So using the equations mentioned above: Moles of acid is the number of moles= concentration X volume i.e. the volume will be used from the average Therefore: =0.05mol/dm3 x 19.675 cm3 =19.6 cm3 / 1000 = 0.0196 dm3 =0.05mol/dm3x0.0196 dm3 = 0.00098 moles So Moles of alkali in 25.000 cm3 Moles of HCl / 25.000 cm3 due to the ratio being 2:1, therefore 0.00098/2= 0.00049 moles of HCl So now the ratio is 1:1 so 0.00049 moles of X(OH)2 Moles of alkali in 100 cm3 It is assumed that there are four lots of 25 cm3 = 4 x 0.00049 = 0.00196 moles The next series of results will be used to calculate solubility of each compound by their mass in 100 cm3 The total Mr has been calculated in the table below for each compound. This was done by : Mr of X + ((O + H) X 2). Each element Mr for the following elements (OH)2 Total Mr Be 9.010 (16.00 +1.01) X 2 = 34.020 43.030 Mg 24.310 (16.00 +1.01) ...read more.

Conclusion

Use different manufacturer's equipment for other trials to ensure that the accuracy increases. Another improvement that will be done, if the experiment were to be repeated is that due to the inaccuracy of the conical flask being swirled. If the conical flask is being swirled unevenly there is a chance of inaccurate results of when the colourless solution occurs. Therefore a stirring rod should be used to increase the accuracy of the swirls of the reaction in the conical flask. Another limitation that arouse in this experiment that would be improved if the experiment were to be done again is that after the neutralization reaction had occurred, there would still be some residue of the distilled water used to rinse out the equipment. This can be improved by increase the number of repeats of rinse. This would ensure that more of the diluted solution would have been removed. Also the trials can also increase, to 10 repeats so that there is more variance so that the accuracy increases. Another improvement might be, to use different indicator, for example methyl orange. Due to the colour change would be from red to yellow would make it easier for the pH 7 to be more easily recognized against a white tile then it was with phenolphthalein. Cited Sources: 1. http://www.vigoschools.org/~mmc3/c1%20lecture/Chemistry%201-2/Lecture%20Notes/Unit%205%20-%20Acids%20and%20Titration/L3%20-%20Acid-Base%20Reactions%20and%20Titration.pdf 2. ...read more.

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