A group II metal hydroxide Essay Example for Free

A group II metal hydroxide probeTo 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 HCLTheory1.Titrations are the reaction between an acid solution with an floor. In this reaction (called neutralization), the acid donates a proton (H+) to the root (base). When the two solutions are combined, the products made are flavor and water.For example2HCl(aq) + X(OH)2 (aq) XCl2 (aq) + 2H2O (l)This shows the oneness of the products i. e. salt organism XCl2 and water.So titration therefore helps to find the denseness for a solution of unknown submerging. This involves the controlled addition of a hackneyed solution of known.Indicators are used to determine, at what stage has the solution r each(prenominal)ed the equivalence point(inflextion point). This means at which, does the procedure of moles base added equals the number of moles of acid present. i.e. pH 7Titration of a strong Acid with a voiceless BaseA s shown in the graph, the pH goes up slowly from the start of the tiration to near the equivalence point. i.e (the startle 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 channelize in pH. i.e. neutralised.Also shown in the graph are methyl radical chromatic and phenolpthalein. These two are both indexs that are often used for acid-base titrations. They each change emblazon at divergent pH ranges.For a strong acid to strong alkali titration, either one of those indicators place be used.However for a strong acid/weak alkali only methyl orangish allow for be used due(p) to pH changing rapidly across the range for methyl orange. That is from low to high pH i.e. red to sensationalistic respectively pH (3.3 to 4.4), but not for phenolpthalein. timid acid/strong alkali, phenolpthalein is used, the pH changes rapidly in an alkali range. From high to low pH, that is from pink to colorless 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.Therefore in this investigation, the titration ordain be between a 0.05 mol dm-3 of HCl with X(OH)2, employ phenolphthalein.Dependant VariableIs the mint of HCl to achieve a colour change that is from pink to colourless.The Controlled variables 1. the same source of HCl2. same tightness of HCl3. Same source of X(OH)24. Same mickle of X(OH)25. Same equipment, method, board temperatureControlled VariablesHow to controlHow to monitor1. Same source of HClUsing 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 drift different ratios of the components of the solution, that great power cause different volume of HCl t o 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 go overs 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 big businessman 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 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 diff erent.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 cause a huge amount of in true statement 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* conic flask swirling not even between the trials* Difficult to judge colourless solution change inseparable end point* Ability to measure 25cm3* Filling of burette accurately with HCl 0 point in right spot* Residual distilled water or solutions rema in in conelike flask i.e. reduce/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 equating 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 in a higher placeMoles of acid is the number of moles= concentration X volumei.e. the volume will be used from the reasonableTherefore=0.05mol/dm3 x 19.675 cm3=19.6 cm3 / snow0 = 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 100 cm3It is assumed that there are four lots of 25 cm3= 4 x 0.00049= 0.00196 molesThe side by side(p) series of results will be used to calculate solubility of each compound by their smokestack in 100 cm3The total Mr has been reckon in the table below for each co mpound.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 determine 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 doubtfulness in measurementUncertainty due to pipet 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 cm3Assumed due to measurable 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 likely to be Ca(OH)2 as the calculated solubility is closest to the literature look on 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 rest 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 of opinion. Showing that it cannot be X(OH)2 solution.This is similarly 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 erroneousness of Ca(OH)2 = (0.170 0.145)/0.170 X 100= (0.025/0.170) X 100= 14.705% passim the experiment there were systematic errors and random errors that were met.Uncertainties/limitationsErrorType of errorQuantity of error business relationship for errorImprovementsMeasurement in buretteSystematic error+/- 0.05cm3Equipment limitation, this is because the line where each of the reading might not be precise.Different manufacturer should be used with bigeminal trials in order to plus 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 no t 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 error non quantifiableHuman observation subjective measurement. This is because even though a white cover 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 elaborateness 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 er rorNot quantifiableChange solution concentrations due to differences in evaporation rate in the surrounding air.Controlled lab environmentCalibration error in buretteSystematic errorNot quantifiable0 line incorrectly markedDivisions on burette inexactUse 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 determine 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 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 sho uld 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 meliorate 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-%20 Acid-Base%20Reactions%20and%20Titration.pdf