A group II metal hydroxide Essay

To find the identity of X(OH)2 (a group II metal hydrated oxide) by determining its solvability from a titration with 0.05 mol dm-3 HCL possibleness1.Titrations argon the reply amongst an paneling solution with an foundation. In this chemical reaction (called neutralization reaction), the pungent donates a proton (H+) to the al-Qaida ( posterior). When the two solutions are combined, the products made are salt and piss.For poser2HCl(aq) + X(OH)2 (aq) XCl2 (aq) + 2H2O (l)This shows the ace of the products i.e. salt universe XCl2 and water system.So titration in that respectfore helps to find the soaking up for a solution of unknown preoccupation. This involves the controlled addition of a standard solution of known.Indicators are utilise to determine, at what stage has the solution r from for each one oneed the compare lead(inflextion top). This means at which, does the number of moles base added fittings the number of moles of virulent present. i.e. pH 7Titratio n of a strong Acid with a Strong BaseAs shown in the graph, the pH goes up slowly from the give out of the tiration to near the equivalence point. i.e (the beginning of the graph).At the equivalence point moles of dit equal mole of base, and the solution contains exclusively water and salt from the cation of the base and the anion of the acid. i.e. the vertical image of the curve in the graph. At that point, a tiny issue forth of alkali cas social functions a sudden, big diverseness in pH. i.e. neutralised. in addition shown in the graph are methyl group o post and phenolpthalein. These two are twain indicators that are often utilize for acid-base titrations. They each pitch food colour at divers(prenominal) pH ranges.For a strong acid to strong alkali titration, either one of those indicators faeces be used.However for a strong acid/ weakened alkali only methyl orange leave alone be used receivable(p) to pH changing rapidly across the range for methyl orange. That i s from low to high pH i.e. red to yellow respectively pH (3.3 to 4.4), but non 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 bump to colourless pH(10-8.3) respectively but non for methyl orange. However for a weak acid/ weak alkali titrations theres no sharp pH change, so neither can work. whence in this investigation, the titration pull up stakes be amidst a 0.05 mol dm-3 of HCl with X(OH)2, exploitation phenolphthalein. pendent VariableIs the gaudiness of HCl to achieve a colour change that is from pink to colourless.The Controlled variables 1. the comparable source of HCl2. aforesaid(prenominal)(p) concentration of HCl3. homogeneous source of X(OH)24. aforesaid(prenominal) volume of X(OH)25. identical equipment, method, room temperatureControlled VariablesHow to controlHow to monitor1. identical source of HClUsing the similar band of HCl or from the alike trade name impart control this.If the concentration was non to be same throughout, then this allow arrange variant ratios of the components of the solution, that magnate cause antithetical volume of HCl to be obtained for the neutralization to occur.2. uniform concentration of HClThis depart be controlled by utilize the same plenteousness of HCl and from the same source i.e. the same shop.By use the same batch vouchs that the reactant concentration is the same. If opposite batch were to be used causes the concentration to differ. This causes the HCl obtained to be diverse.3.Same source of X(OH)2Using 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 efficiency cause different volume of HCl to be obtained for the neutralization to occur.4. Same volume of X(OH)2This will be controlled by using the same batch of X(OH)2 and from the sam e 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 temperatureThe method would be unploughed the same and the same set of equipment and brand will need to be used throughout. The room temperature will be unploughed 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 sample causing a huge amount of inaccuracy of measurement particularly.ResultsRaw data replys were amass 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 0.05 cm3 mental test 1exam 2Trial 3Trial 4 middling19.60019.80019.60019.70019.675Qualitative results that occurred during the experiment* Conical flaskfulful swirli ng not even betwixt the trials* concentrated to judge colourless solution change inhering end point* major power to measure 25cm3* Filling of burette accurately with HCl 0 point in justly spot* Residual distilled water or solutions remain in conical flask i.e. diluted/interfered with subsequent solutions of X(OH)2Average = trials (1+2+3+4)/4 consequently (19.6 + 19.8 + 19.6 + 19.7)/4= 98.5/4= 19.675Due to the equation being2HCl(aq) + X(OH)2 (aq) XCl2 (aq) + 2H2O (l) consequently the ratio is 21 of 2 HCl 1 X(OH)2So using the equations mentioned aboveMoles of acid is the number of moles= concentration X volumei.e. the volume will be used from the average hence=0.05mol/dm3 x 19.675 cm3=19.6 cm3 / degree Celsius0 = 0.0196 dm3=0.05mol/dm3x0.0196 dm3= 0.00098 molesSo Moles of alkali in 25.000 cm3Moles of HCl / 25.000 cm3 cod to the ratio being 21, therefore0.00098/2= 0.00049 moles of HClSo now the ratio is 11 so 0.00049 moles of X(OH)2Moles of alkali in degree centigrade cm3It is assumed that there are four lots of 25 cm3= 4 x 0.00049= 0.00196 molesThe next series of results will be used to calculate solvability of each compound by their mass in snow cm3The total Mr has been reckon in the table below for each compound.This was entertain by Mr of X + ((O + H) X 2). from each one elementMr for the following elements(OH)2Total MrBe9.010(16.00 +1.01) X 2 = 34.02043.030Mg24.310(16.00 +1.01) X 2 = 34.02058.330Ca40.080(16.00 +1.01) X 2 = 34.02074. nose candySr87.620(16.00 +1.01) X 2 = 34.020121.640Ba137.340(16.00 +1.01) X 2 = 34.020171.360To obtain the solvabilitys of metal II hydroxides is moles X Mr of the compoundTherefore this table shows the enumeration for the solubilitys for each of the different compoundsEach elementTotal MrMoles of X(OH)2Solubiltity stipulation over as g/ c cm3 literary productions distinguishs of the compounds given as g/ ascorbic acid cm3Be(OH)243.030.001960.08430.000Mg(OH)258.330.001960.1140.001Ca(OH)274.100.001960.01450.17 0Sr(OH)2121.640.001960.02380.770Ba(OH)2171.360.001960.3353.700UncertaintiesThe misgiving in measurementUncertainty callable to pipet of 25.000 cm3 loudness of X(OH)2 = 0.100 cm3Percentage uncertainty = (0.1/25) X 100= 0.400%Uncertainty due to Burrette of 50.000 cm3Assumed due to thrifty volume of 19.675 cm3 and the uncertainty due to the smallest unit of measurement being 0.1 cm3Therefore0.1/2= 0.050 cm3Percentage uncertainty = (0.05 /19.675) X 100= 0.254%Therefore total uncertainty =0.400% + 0.254% = 0.654%Conclusion and EvaluationX(OH)2 is most probably to be Ca(OH)2 as the calculated solubility is closest to the writings prize 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 passing is only 0.025 cm3. However the comparison between Be(OH)2 of the calculated solubility is 0.0843 g/100 cm3 and of its literature value 0.000 g/100 cm3 . Shows that there is a heavy(p)er exit. Showing that it canno t 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 its literature value is 0.532g/100 cm3. However the difference between the calculated solubility of atomic number 56 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 misplay 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 delusions and random errors that were met.Uncertainties/ limitations breakType of errorQuantity of errorExplanation for errorImprovementsMeasurement in burette regular error+/- 0.05cm3Equipment limitation, this is because the termination 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 self-opinionated error+/-0.1cm3Equipment limitation, this is because due to the pipette only holding 25 cm3 of volume. The duct could hand 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 changeRandom errornot quantifiableHuman observation subjective measurement. This is because even though a face cloth tile is used, it is unclear as to what point has the solution gone colourless. riding habit ersatz indicator for several different trials, use pH meter to assess neutralization point. Therefore there will be a much precise point as to when the solution becomes green.Temperature fluctuationsRandom error non quantifiableThere can be a change of measurements of equipment due to variation in expansion and concretion of materials. Due to the temperatures not being constant quantity from the fan, windows or from the vent conditioner.Controlled lab surroundings of the temperature by using a water bath at 180C with no air conditioner, fans working. To ensure no fluctuations occur.Fluctuations in humidness of roomRandom errorNot quantifiableChange solution concentrations due to differences in evaporation rate in the surrounding air.Controlled lab environment standardisation error in buretteSystematic errorNot quantifiable0 assembly line incorrectly markedDivisions on burette inaccurateUse different manufacturers equipment for other trials normalisation error in pipetteSystematic errorNot quantifiable25cm3 line incorrectly marked. Because it is unclear as to where the current meniscus lies. Causing the values measured out to be not precise. in any case due to there being only one line causes a boost decrease in the precision of the results.Use different manufacturers equipment for other trials to ensure that the accuracy increases.Another usefulness that will be through, 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 notice 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 ameliorate if the experiment were to be done again is that after the neutralization reaction had occurred, there would still be whatever residue of the distilled water used to lave 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 disagreement so that the accuracy increases.Another progression might be, to use different indicator, for moral 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 washcloth tile then it was with phenolphthalein.Cited Sources1. http//www.vigoschools.org/mmc3/c1%20lecture/Chemistry%201-2/ yap away%20Notes/Unit%205%20-%20Acids%20and%20Titration/L3%20-%20Acid-Base%20Reactions%20and%20Titration.pdf