A group II metal hydroxide Essay

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). At the equivalence point moles of acid equal mole of base, and the solution contains only water and salt from the cation of the base and the anion of the acid. i.e. the vertical part of the curve in the graph. At that point, a tiny amount of alkali casuses a sudden, big change in pH. i.e. neutralised. Also shown in the graph are methyl orange and phenolpthalein. These two are both indicators that are often used for acid-base titrations. They each change colour at different pH ranges. For a strong acid to strong alkali titration, either one of those indicators can be used. However for a strong acid/weak alkali only methyl orange will be used due to pH changing rapidly across the range for methyl orange. That is from low to high pH i.e. red to yellow respectively pH (3.3 to 4.4), but not for phenolpthalein. Weak acid/strong alkali, phenolpthalein is used, the pH changes rapidly in an alkali range. From high to low pH, that is from pink to colourless pH(10-8.3) respectively but not for methyl orange. However for a weak acid/ weak alkali titrations there’s no sharp pH change, so neither can work. Therefore in this investigation, the titration will be between a 0.05 mol dm-3 of HCl with X(OH)2, using phenolphthalein. Dependant Variable: Is the volume of HCl to achieve a colour change that is from pink to colourless. The Controlled variables : 1. the same source of HCl 2. same concentration of HCl 3. Same source of X(OH)2 4. Same volume of X(OH)2 5. Same equipment, method, room temperature Controlled Variables How to control How to monitor 1. Same source of HCl Using the same batch of HCl or from the same brand will control this. If the concentration was not to be same throughout, then this will cause different ratios of the components of the solution, that might cause different volume of HCl to be obtained for the neutralization to occur. 2. Same concentration of HCl This will be controlled by using the same batch of HCl and from the same source i.e. the same brand. By using the same batch ensures that the reactant concentration is the same. If another batch were to be used causes the concentration to differ. This causes the HCl obtained to be different. 3.Same source of X(OH)2 Using the same batch of X(OH)2 or from the same brand will control this. If the concentration was not to be same throughout, then this will cause different ratios of the components of the solution that might cause different volume of HCl to be obtained for the neutralization to occur. 4. Same volume of X(OH)2 This will be controlled by using the same batch of X(OH)2 and from the same source i.e. the same brand. By using the same batch ensures that the reactant concentration is the same. If another batch were to be used causes the concentration to differ. This causes the HCl obtained to be different. 5. Same equipment, method, room temperature The method would be kept the same and the same set of equipment and brand will need to be used throughout. The room temperature will be kept throughout at 180C by using a water bath. If different equipment or brands were used then there would be a lot of anomalies in the experiment causing a huge amount of inaccuracy of measurement particularly. Results: Raw data results were collected by using 25.00 cm3 of X(OH)2 with phenolphthalein and the volume of HCl was obtained by the solution going from pink to colourless. The volume of HCl found in 50.0cm3 burette à ¯Ã‚ ¿Ã‚ ½ 0.05 cm3 Trial 1 Trial 2 Trial 3 Trial 4 Average 19.600 19.800 19.600 19.700 19.675 Qualitative results that occurred during the experiment: * Conical flask swirling not even between the trials * Difficult to judge ‘colourless’ solution change – subjective end point * Ability to measure 25cm3 * Filling of burette accurately with HCl – 0 point in right spot * Residual distilled water or solutions remain in conical flask i.e. diluted/interfered with subsequent solutions of X(OH)2 Average = trials (1+2+3+4)/4 Therefore: (19.6 + 19.8 + 19.6 + 19.7)/4 = 98.5/4 = 19.675 Due to the equation being 2HCl(aq) + X(OH)2 (aq) 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) X 2 = 34.020 58.330 Ca 40.080 (16.00 +1.01) X 2 = 34.020 74.100 Sr 87.620 (16.00 +1.01) X 2 = 34.020 121.640 Ba 137.340 (16.00 +1.01) X 2 = 34.020 171.360 To obtain the solubility’s of metal II hydroxides is moles X Mr of the compound Therefore this table shows the calculation for the solubility’s for each of the different compounds Each element Total Mr Moles of X(OH)2 Solubiltity given as g/100 cm3 Literature values of the compounds given as g/100 cm3 Be(OH)2 43.03 0.00196 0.0843 0.000 Mg(OH)2 58.33 0.00196 0.114 0.001 Ca(OH)2 74.10 0.00196 0.0145 0.170 Sr(OH)2 121.64 0.00196 0.0238 0.770 Ba(OH)2 171.36 0.00196 0.335 3.700 Uncertainties: The uncertainty in measurement: Uncertainty due to pipette of 25.000 cm3 : Volume of X(OH)2 = à ¯Ã‚ ¿Ã‚ ½ 0.100 cm3 Percentage uncertainty = (0.1/25) X 100 = 0.400% Uncertainty due to Burrette of 50.000 cm3: Assumed due to measured volume of 19.675 cm3 and the uncertainty due to the smallest unit of measurement being 0.1 cm3 Therefore 0.1/2= à ¯Ã‚ ¿Ã‚ ½ 0.050 cm3 Percentage uncertainty = (0.05 /19.675) X 100 = 0.254% Therefore total uncertainty = 0.400% + 0.254% = 0.654% Conclusion and Evaluation: X(OH)2 is most likely to be Ca(OH)2 as the calculated solubility is closest to the literature value given of Ca(OH)2. The solubility for Ca(OH)2 0.145 g/100 cm3 and the literature value is 0.170 g/100 cm3. This shows that the difference is only 0.025 cm3. However the comparison between Be(OH)2 of the calculated solubility is 0.0843 g/100 cm3 and of it’s literature value 0.000 g/100 cm3 . Shows that there is a greater difference. Showing that it cannot be X(OH)2 solution. This is also shown for Mg(OH)2 as the difference between the calculated solubility and the literature value is 0.113 g/100 cm3, showing that it still has a greater difference than Calcium hydroxide does. The difference between Sr(OH)2 and it’s literature value is 0.532g/100 cm3. However the difference between the calculated solubility of Barium hydroxide and the literature value is 3.365 g/100 cm3 showing there is a great difference so it cannot be Barium hydroxide. The percentage error of Ca(OH)2 = [(0.170 – 0.145)/0.170] X 100 = (0.025/0.170) X 100 = 14.705% Throughout the experiment there were systematic errors and random errors that were met. Uncertainties/limitations Error Type of error Quantity of error Explanation for error Improvements Measurement in burette Systematic error +/- 0.05cm3 Equipment limitation, this is because the line where each of the reading might not be precise. Different manufacturer should be used with multiple trials in order to increase the accuracy of the calculated value to the literature value. Measurement in pipette Systematic error +/-0.1cm3 Equipment limitation, this is because due to the pipette only holding 25 cm3 of volume. The line could have been where the actual reading might not be Causing the result to not be precise. Different manufacturer should be used with multiple trials in order to increase the accuracy of the calculated value to the literature value. Point of colour change Random error Not quantifiable Human observation – subjective measurement. This is because even though a white tile is used, it is unclear as to what point has the solution gone colourless. Use alternative indicator for several different trials, use pH meter to assess neutralization point. Therefore there will be a more precise point as to when the solution becomes green. Temperature fluctuations Random error Not quantifiable There can be a change of measurements of equipment due to variation in expansion and contraction of materials. Due to the temperatures not being constant from the fan, windows or from the air conditioner. Controlled lab environment of the temperature by using a water bath at 180C with no air conditioner, fans working. To ensure no fluctuations occur. Fluctuations in humidity of room Random error Not quantifiable Change solution concentrations due to differences in evaporation rate in the surrounding air. Controlled lab environment Calibration error in burette Systematic error Not quantifiable 0 line incorrectly marked Divisions on burette inaccurate Use different manufacturer’s equipment for other trials Calibration error in pipette Systematic error Not quantifiable 25cm3 line incorrectly marked. Because it is unclear as to where the true meniscus lies. Causing the values measured out to be not precise. Also due to there being only one line causes a further decrease in the precision of the results. 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