This procedure will require a titration of an unknown concentration of limewater using a precisely known concentration of hydrochloric acid. Using equations the exact concentration of the limewater can be determined. This is useful as when an indicator such as phenolphthalein indicates the titration is complete, the known number of moles within the acid can be used to determine the concentration of the limewater.Equipment list:50.0 ml Burette1.00ml of HCl 2.00mol dm-3250cm3 Limewater of approximately 1.00g dm-399.0ml Distilled waterPhenolphthalein indicator100ml Measuring cylinder250ml Conical flaskWhite tilePipetteClampBossStandSafety:In regards to safety when working with any hydrochloric acid, at 2M it is a classed as an irritant so gloves must be worn throughout. Also regardless of the concentration, lab goggles must be worn. Calcium hydroxide is also an irritant so gloves must be worn.Method:Firstly add 25.0ml of the sample limewater to a conical flask.Place the flask on a white tile, to make the change in colour more noticeable.Add 5 drops of indicator to the flask any appropriate indicator could be used in this titration such as ?-naphtholbenzein or thymol blue, I am going to use phenolphthalein . This will be the indicator to tell me when the titration is complete.Fill the burette with the dilute HCl. I will talk about the dilution later.Firstly carry out a rough titration and record the hydrochloric acid level at which the indicator turns colourless. Record this result.Set up the experiment again and carry out a more precise titration. Repeating the titration is important as it removes the impact which abnormalities have on the experiment. Repeat this twice more and record these results.Preparation calculationsCa(OH)2 + 2 HCl –>CaCl2 + 2H2OThis shows that one mole of limewater reacts with two moles of hydrochloric acid.The limewater is approximately 1.00g dm-3 this is 0.0135mol dm-3 as the molecular weight of calcium hydroxide is 74g dm-3 ? mr = mol dm-31.00 ? 74 = 0.0135mol dm-3The concentration of the hydrochloric acid is at first too much, so we dilute the acid.This is quite simple, first I need to calculate the desired concentration, then the dilution factor.It makes sense to have a concentration quite close to 0.0135mol dm-3 as this would mean I would theoretically need just over double the volume of Ca(OH)2 than HCl as one mole of Ca(OH)2 reacts with two moles of HCl.I will try and get 0.02mol dm-3 as it is close to 0.0135mol dm-3 yet will be easier to measure.Dilution:The equation for finding the dilution factor of something is:Quoted from: = Molarity V = Volume con = Referring to the concentrate dil = Referring to the dilution(M con) (V con) = (M dil) (V dil)(2.00) (V con) = (0.02) (100)(2.00) (V con) = 2.00?2.00(V con) = 1.00I want 100ml as I will put 25.0ml in the conical flask each of the four times I carry out the experiment. So I will use 1.00ml in my dilution with 99.0ml of distilled water as this is not involved in the reaction it is ok. This is a ratio of 1:99.Equipment for dilution:Glass pipettes (100ml 10.0ml)Using a glass pipette add 100ml of distilled water to a measuring cylinder.Remove 1.00ml of distilled water using the smaller pipette. (Using two pipettes and removing 1.00ml is more accurate than using many decreasing in size pipettes to add the same volume.)Using the smaller glass pipette add 1.00ml of HCl to the measuring cylinder.Put a bung on and gently shake to mix.The dilution of the acid seems of no use but it means it is possible to increase the degree of accuracy of the test as each drop of HCl has a smaller and more measurable effect in the reaction. A large concentration would complete the reaction too quickly.After carrying out the titration I got these results:* 34.6* 34.5* 34.45* 34.5Handling of results:Firstly using these values I intend to find the average number of moles within the volume of HCl I used in the titration. When calculating an average I get:(34.6 + 34.5 + 34.45 + 34.5) ? 4138.05 ? 4 = 34.5 (3s.f)HClNumber of moles = concentration (mol dm-3) x average volume (?).N = C x VN = 0.02 ? 0.0345N = 0.00069 molesCa(OH)2Number of moles = 0.000345 because of the ratio 2:1. 2 HCl to every Ca(OH)2.Concentration (mol dm-3) = number of moles ? volume (?)Concentration = 0.000345 ? 0.025Concentration = 0.0138mol dm-3This is the concentration in moles per decimetre cubed.In grams per decimetre cubed it is:g dm-3 ? mr = mol dm-3g dm-3 = 0.0138 x 74g dm-3 = 1.02 g dm-3 (to 3 s.f)Evaluation:The concentration of the limewater sample to my calculations is 1.02g dm-3 this is sensible as the value I was asked to prove was close was 1.00g dm-3, I still do not know the concentration exactly which the lab prepared but it is likely my answer though I think precise, is out. I think that though my results where fairly accurate but if I where asked for the worst, the first titration and third titration were most imprecise. The first one was my roughest titration so possibly the least precise. The titration in this result possibly continued past the end point and more acid was going when the titration was already over. This is simply human error as I wasn’t quite quick enough at stopping the acid from leaving the burette as I should have been.When doing my third titration I may have only put 4 drops in the flask because I thought I saw 2 drops go in, in one. This left the flask with less indicator and a slightly paler colour. This would have needed less acid to make the change in colour because the change would be less dramatic and happen easier.Apart from these errors my results were quite accurate because I did use the equipment I chose to use well. The glass pipettes are very accurate to a very fine margin of error. The burette is also a very precise piece of equipment.Overall my experiment worked well as I got a reasonable outcome and used all the equipment well, any mistakes I had made would have been lessened by the repetition of my experiment.


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