While I am conducting the titration, I will put the conical flask on a white tile, so that I can see the colour change more easily.
It is suggested that any indicator is suitable when a strong alkali reacts with a strong acid, so I have chosen methyl orange for the titration.
Prediction
I have included calculations to show the approximate amount of acid needed to neutralise the alkali. From these calculations, you can see that if the concentration of the hydrochloric acid is 0.3 mol dm-3, there will only be approximately 2.5cm3 needed. Smaller measurements allow a larger percentage error, so in order to make the experiment more accurate, I will dilute the solution with 10 parts to one of water. When this is done, there will be approximately equal quantities of both reactants.
I predict, therefore, that in order to find a more accurate concentration of the lime water, I will use approximately 25 cm3 of the dilute hydrochloric acid.
Apparatus
50cm3 burette,
pipette,
volumetric flask,
conical flask,
retort stand,
boss clamp,
methyl orange indicator,
white tile,
measuring cylinder,
lime water,
hydrochloric acid,
distilled water
Plan
In order to work out the concentration, I plan to find the amount of hydrochloric acid needed to neutralise 25cm3 of the lime water. In order to do this I will need to titrate the acid into the solution with an indicator. When the indicator just changes colour, the solution will be neutral. I will carry out the titration as follows.
First, the hydrochloric acid must be diluted. It must be diluted 10 : 1, and I will make up enough to repeat the experiment several times. To conduct the experiment once, I will need 50cm3. to do this, I will use a volumetric flask. I will mix 5 cm3 of hydrochloric acid with 50 cm3 of distilled water. I will fill the burette with 50 cm3 of this solution. I will transfer 25 cm3 of lime water into the conical flask, using a measuring cylinder, and place it on the white tile below the burette. I will put a few drops of methyl orange in the conical flask. I will run the acid into the alkali until a change in colour is seen. At this point I will note the volume of acid used. In order to make the experiment more accurate, I will repeat this process, and take an average (mean) reading. From this, I will be able to find the exact concentration of the calcium hydroxide.
Before I perform the experiment, I must take care to rinse out all of the insides of the apparatus, for example the burette, pipette etc. This is top ensure that the apparatus is not contaminated with other chemicals, which could well affect the outcome of the experiment.
Risk assessment
The main risks involved with this experiment are due to the nature of the chemicals used. Hydrochloric acid is harmful to skin, and both are harmful to eyes. Safety goggles must be worn, and care must be taken to avoid contact with skin.
Results
The table above shows the volumes of hydrochloric acid that were titrated into the calcium hydroxide solution during the experiment, in order to neutralise the solution.
I have worked out the standard deviation of the results. It is 0.25 (2 significant figures).
Any result that is more than twice the standard deviation above or below the mean is an outlier and may be ignored as an anomaly. I am pleased to say that none of my results were more than 0.5cm3 away from the mean, so none need to be disregarded.
The acid I used was too concentrated to be appropriate for neutralising the calcium hydroxide solution. To use it at its original concentration would have provided a larger source for error. As I mentioned in my plan, I found it necessary to dilute the hydrochloric acid by 10 parts to one. Because the original concentration was given, I can work out the new concentration.
The original concentration was 0.3 mol dm-3, and as the dilution was one part hydrochloric acid to ten parts water. After the dilution, the new concentration was 0.03 mol dm-3. It is therefore possible to calculate the amount, in moles, in solution. From this I can calculate the amount, again in moles, of calcium hydroxide to react with it. If the volume of calcium hydroxide solution is known, the concentration of it can be found. I have included a calculation showing this. I worked out the concentration of the calcium hydroxide solution as 0.00089 (2 significant figures).
In conclusion, it is clear from the results, and the subsequent calculations that calcium hydroxide is reasonably insoluble, in comparison with other compounds. The results also show that the calculated concentration I found from the experiment was much lower than the approximate calculation that was given.
Evaluation
From the percentage error calculation, and by the deviation of the results, shown by the calculated value for standard deviation, (2.5) it is clear that there were sources of error in the experiment. I felt that although the apparatus provided a slight source of error during the experiment, the main source of error was my own judgement of when the indicator had changed colour. I felt that it was difficult to judge the same colour change every time. I think this was the main cause for the results deviation that can be seen from the results. This could not be included in the percentage error calculation, however, because of the difficulty in actually measuring this. It would be difficult to improve in a future experiment, but the experiment would have been improved if the human error factor was removed. A way of doing this could have been an electronic system involving a filter allowing only a certain colour of light, and an LDR, with a voltmeter.