A group II metal hydroxide Essay Example for Free

A congregation II metal hydroxide EssayTo picture the identity of X(OH)2 (a group II metal hydroxide) by determining its solvability from a titration with 0.05 counterspy dm-3 HCLTheory1.Titrations are the reception between an venomous solution with an alkali. In this reaction (called neutralisation), the acid donates a proton (H+) to the alkali (base). When the two solutions are combined, the products made are salt and water.For example2HCl(aq) + X(OH)2 (aq) XCl2 (aq) + 2H2O (l)This shows the one of the products i. e. salt macrocosm XCl2 and water.So titration at that placefore helps to find the concentration for a solution of unknown concentration. This involves the controlled addition of a standard solution of known.Indicators are employmentd to determine, at what stage has the solution r severally(prenominal)ed the equivalence point(inflextion point). This mode at which, does the number of moles base added equals the number of moles of acid present. i.e. pH 7Titrat ion of a strong Acid with a Strong BaseAs 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 however 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 measuring stick of alkali casuses a sudden, big swop 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 smorgasbord 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 volition 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 theres no sharp pH change, so neither can work. thereof in this investigation, the titration give be between a 0.05 mol dm-3 of HCl with X(OH)2, using phenolphthalein.Dependant VariableIs the volume of HCl to achieve a colour change that is from pink to colourless.The Controlled variables 1. the resembling stemma of HCl2. like concentration of HCl3. Same source of X(OH)24. Same volume of X(OH)25. Same equipment, method, room temperatureControlled VariablesHow to controlHow to monitor1. Same source of HClUsing the same people 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 differe nt volume of HCl to be obtained for the neutralization to occur.2. Same concentration of HClThis 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 take cares 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)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 might 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 same source i.e. the same brand.By using the same batch get winds that the reactant concentration is the same. If another batch were to be used causes the concentration to differ. This causes the HCl obtaine d to be different.5. Same equipment, method, room temperatureThe 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.ResultsRaw 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 0.05 cm3Trial 1Trial 2Trial 3Trial 4Average19.60019.80019.60019.70019.675Qualitative results that occurred during the experiment* Conical flask swirling not tied(p) 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 solu tions remain in conic flask i.e. diluted/interfered with subsequent solutions of X(OH)2Average = trials (1+2+3+4)/4Therefore (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)Therefore 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 averageTherefore=0.05mol/dm3 x 19.675 cm3=19.6 cm3 / grand piano = 0.0196 dm3=0.05mol/dm3x0.0196 dm3= 0.00098 molesSo Moles of alkali in 25.000 cm3Moles of HCl / 25.000 cm3due 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 ampere-second 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 solubility of each compound by their mass in 100 cm3The total Mr has been calculated in the submit below for each compound .This was done by Mr of X + ((O + H) X 2).Each 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 = 34.02074.100Sr87.620(16.00 +1.01) X 2 = 34.020121.640Ba137.340(16.00 +1.01) X 2 = 34.020171.360To obtain the solubilitys of metal II hydroxides is moles X Mr of the compoundTherefore this table shows the calculation for the solubilitys for each of the different compoundsEach elementTotal MrMoles of X(OH)2Solubiltity given as g/100 cm3Literature entertains of the compounds given as g/100 cm3Be(OH)243.030.001960.08430.000Mg(OH)258.330.001960.1140.001Ca(OH)274.100.001960.01450.170Sr(OH)2121.640.001960.02380.770Ba(OH)2171.360.001960.3353.700UncertaintiesThe uncertainty in measurementUncertainty due to pipette of 25.000 cm3 Volume of X(OH)2 = 0.100 cm3Percentage uncertainty = (0.1/25) X 100= 0.400%Uncertainty due to Burrette of 50.000 cm3 sham due to measured volume of 19.675 cm3 and th e 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 likely to be Ca(OH)2 as the calculated solubility is closest to the literature nurse 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 its 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 its literature value is 0.532g/100 cm3. However the differe nce 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 magisterial errors and random errors that were met.Uncertainties/limitationsErrorType of errorQuantity of errorExplanation for errorImprovementsMeasurement in buretteSystematic error+/- 0.05cm3Equipment limitation, this is because the landmark where each of the discipline might not be precise.Different manufacturer should be used with multiple trials in order to attach the accuracy of the calculated value to the literature value.Measurement in pipetteSystematic error+/-0.1cm3Equipment 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 shou ld be used with multiple trials in order to increase the accuracy of the calculated value to the literature value.Point of colour change hit-or-miss errorNot quantifiableHuman observation subjective measurement. This is because plain 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 fluctuationsRandom errorNot quantifiableThere 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 roomRandom errorNot quantifiableChange solution conce ntrations due to differences in evaporation rate in the surrounding air.Controlled lab environmentCalibration error in buretteSystematic errorNot quantifiable0 line incorrectly markedDivisions on burette inaccurateUse different manufacturers equipment for other trialsCalibration error in pipetteSystematic errorNot quantifiable25cm3 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 manufacturers equipment for other trials to ensure that the accuracy increases.another(prenominal) feeler 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 accur acy 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 Sources1. http//www.vigoschools.org/mmc3/c1%20lecture/Chemistry%201-2/Lecture%20Notes/Unit%205%20-%20Acids%20and%20Titration/L3%20-%20Acid-Base%20Reactions%20and%20Titra tion.pdf