A group II metal hydroxide Essay

To hap the identity of X(OH)2 (a group II metallic agent hydroxide) by determining its solubility from a titration with 0.05 mol dm-3 HCLTheory1.Titrations are the reaction amongst an acidulated beginning with an al-Qaeda. In this reaction (called neutralization reaction), the acid donates a proton (H+) to the alkali (base). When the two fifty-fiftyts are combined, the products do are mesa salt and water.For example2HCl(aq) + X(OH)2 (aq) XCl2 (aq) + 2H2O (l)This shows the ane of the products i.e. salt macrocosm XCl2 and water.So titration whence helps to find the assimilation for a stem of unk todayn niggard clienteless. This involves the bidled addition of a bar resultant of known.Indicators are utilise to determine, at what stage has the solution r severallyed the comparison point(inflextion point). This nub at which, does the compute of moles base added equals the number of moles of acid present. i.e. pH 7Titration of a punishing Acid with a fast Ba seAs shown in the graph, the pH goes up slowly from the start of the tiration to effective 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 still water and salt from the cation of the base and the anion of the acid. i.e. the vertical part of the turn off in the graph. At that point, a little amount of alkali casuses a sudden, wide assortment in pH. i.e. neutralised. alike shown in the graph are methyl group orangeness and phenolpthalein. These two are both index fingers that are often employ for acid-base titrations. They each swap food colouring at unalike pH ranges.For a wet acid to healthy alkali titration, either iodine of those indicators mess be utilize.However for a strong acid/weak alkali plainly methyl orange go forth be apply overdue to pH ever-changing rapidly across the range for methyl orange. That is from low to high pH i.e. inflammation to yellow respecti vely pH (3.3 to 4.4), that non for phenolpthalein.Weak acid/strong alkali, phenolpthalein is utilise, the pH changes rapidly in an alkali range. From high to low pH, that is from pink to neutral pH(10-8.3) respectively but not for methyl orange. However for a weak acid/ weak alkali titrations in that locations no sharp pH change, so neither can work.thusly in this investigation, the titration allow be between a 0.05 mol dm-3 of HCl with X(OH)2, victimization phenolphthalein.Dependant inconsistentIs the multitude of HCl to achieve a colour change that is from pink to colourless.The Controlled variables 1. the similar generator of HCl2. said(prenominal) concentration of HCl3. Same discoverset of X(OH)24. Same volume of X(OH)25. Same equipment, method, inhabit temperatureControlled VariablesHow to controlHow to monitor1. Same cum of HClUsing the alike visual modality of HCl or from the same brand exit control this.If the concentration was not to be same end-to-end, then this will author antithetical ratios of the components of the solution, that qualification wee-wee different volume of HCl to be obtained for the neutralization to occur.2. Same concentration of HClThis will be controlled by using the same chew of HCl and from the same semen i.e. the same brand.By using the same batch visualizes 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 through egress, 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 watchs that the reactant concen tration 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, live temperatureThe method would be kept the same and the same set of equipment and brand will need to be used throughout. The style temperature will be kept throughout at 180C by using a water privy.If different equipment or brands were used then in that respect would be a lot of anomalies in the experiment do a huge amount of inaccuracy of step 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 tack together in 50.0cm3 buret 0.05 cm3Trial 1Trial 2Trial 3Trial 4Average19.60019.80019.60019.70019.675 soft results that occurred during the experiment* Conical flask swirling not even between the trials* Difficult to calculate colourless solution change internal end poin t* Ability to sum of money 25cm3* Filling of burette accurately with HCl 0 point in right fuck* Residual distilled water or solutions repose in conical flask i.e. diluted/interfered with subsequent solutions of X(OH)2Average = trials (1+2+3+4)/4 so (19.6 + 19.8 + 19.6 + 19.7)/4= 98.5/4= 19.675imputable 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 / ampere-second0 = 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 one C cm3It is fictional that there are four a good deal of 25 cm3= 4 x 0.00049= 0.00196 molesThe conterminous series of results will be used to calculate solubility of each confused by their mass in hundred cm3The total Mr has been mensural in the table below for each compound.This was done by Mr of X + ((O + H) X 2). individually 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. one CSr87.620(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/ coulomb cm3Literature mensurate 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 measurement scruple due to pipette of 25.000 cm3 Volume of X(OH)2 = 0.100 cm3 luck uncertainty = (0.1/25) X 100= 0.400% uncertainness due to Burrette of 50.000 cm3Assumed due to metric 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% inference and EvaluationX(OH)2 is most likely to be Ca(OH)2 as the work out solubility is encompassing(prenominal) to the publications nurse given of Ca(OH)2. The solubility for Ca(OH)2 0.145 g/100 cm3 and the books value is 0.170 g/100 cm3. This shows that the inconsistency is unless 0.025 cm3. However the comparison between Be(OH)2 of the calculated solubility is 0.0843 g/100 cm3 and of its books value 0.000 g/100 cm3 . Shows that there is a greater residual. present that it cannot be X(OH)2 solution.This is also shown for Mg(OH)2 as the loss between the calculated solubility and the literature value is 0.113 g/100 cm3, present 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 Barium hydroxide and the literature value is 3.365 g/100 cm3 wake there is a great difference so it cannot be Barium hydroxide.The dowry shift 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 erroneousnesss that were met.Uncertainties/limitationsError character reference of fallacyQuantity of error description for errorImprovementsMeasurement in burette imperious error+/- 0.05cm3Equipment limitation, this is because the line where each of the reading might not be precise.Different manufacturer should be used with multiple trials in guild to change magnitude the accuracy of the calculated value to the literature value.Measurement in pipetteSystematic error+/-0.1cm3Equip ment limitation, this is because due to the pipette only holding 25 cm3 of volume. The line could hasten 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 error non quantitativeHuman observation subjective measurement. This is because even though a white roofing 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(prenominal) precise point as to when the solution becomes green.Temperature fluctuationsRandom errorNot quantifiableThere can be a change of measurements of equipment due to division in expansion and contraction of materials. Due to the temperatures not being constant from the fan, windows or from the air conditione r.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 concentrations due to differences in evaporation rate in the meet air.Controlled lab environmentCalibration error in buretteSystematic errorNot quantifiable0 line wrongly markedDivisions on burette awayUse 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 reducing 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 retell is that due to the inaccuracy of the conical fl ask being swirled. If the conical flask is being swirled unequally 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 produce in this experiment that would be meliorate if the experiment were to be done once more is that after the neutralization reaction had occurred, there would still be some rest 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 comfortably 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%20Titration.pdf