Preparation of Ear Drops Practical – Titration of NaHCO3 Essay

IntroductionEardrops are preparations used for the treatment of various disorders of the ear.

They come in numerous dosages and consist of many different ingredients depending on what they have been recommended for1. Over time the composition of eardrops have improved and been more tailor made to cover a large range of specific ear problems. The most common problem involving ears is earwax. Earwax is produced to keep the ear clean and to protect it from infections. Sometimes the production of it either becomes too much.

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These earwax problems can cause various conditions such as dizziness, hearing loss, irritability as well as loss of balance2. When eardrops are administered they break up large compacted pieces of earwax into smaller manageable clumps. As mentioned the composition of eardrops vary depending on the specific ear problem. This experiment is based around the accuracy of a prepared eardrop.MethodThe method is devised into two separate sections:Preparation of eardrops1. Weigh out 5 grams of NaHCO32.

Add NaHCO3 to a volumetric flask then dissolve in 60ml of purified water.3. Measure 30ml of glycerol using a pipette.4. Add the glycerol to NaHCO3 solution5. Make volume of NaHCO3 solution to 100ml by using purified water. Shake once mixedTitration1. Take 5ml of eardrop and then add 20ml of water.

2. Place this in a conical flask.3.

Add 2-3 drops of methyl orange to be an indicator4. Then titrate eardrops against 0.1M from the burette.ResultsThe results obtained from the titration are as follows:Volume in cm3TrialTitre 1Titre 2Titre 3Titre 4Initial volume5050505050Final volume12.


20Total amount used37.5030.5030.6030.5030.

80From the results it can be clearly seen that the trial titration was overshot. The four titre results all fell within 0.30 cm3 of each over, showing accuracy. The average of the four titres came to be 30.60 cm3.To calculate the molarity of NaHCO3Using the formula Moles= Concentration/VolumeThe concentration is given at 0.

1 M and the volume obtained is 26.60 cm3.By rearranging this equation, molarity can be calculated:0.1 x 30.6 = 5.005(being the weight) x MolarityMolarity= 0.1 x 30.

6/5.005= 0.61Calculation of the number of grams of NaHCO3 in 100ml of prepared eardrops.In this experiment 5.005 grams of NaHCO3 was measured. To ensure that this is the correct amount used the molar mass of NaHCO3 is calculated and then multiplied by the moles to find out how much it is per litre.Molar of mass of NaHCO3 = 84 gramsmoles x molar mass- 0.61 x 84 = 51.

24 grams per litre.To calculate the amount per 100ml divide it by 1051.24/10= 5.124 grams per 100 mlThe calculated amount is close is to the measured amount but not spot on or as close as it could be. This can be because of various things, for one the level of certainty provided by the equipment in the lab. Accuracy of lab equipment such as beakers, pipettes and burettes in a college laboratory will not be as accurate as of that in a professional laboratory where precise equipment is needed. Another contributing factor wound be the use of a weighing boat, to say all of the powdered NaHCO3 was added to the beaker with no grains left in the dish would be incorrect as it is not easy to ensure that the entire 5.

005 grams entered the beaker. These minor details may have contributed to difference seen in the calculation and the amount actually weighed. Not withstanding differences may be caused by the accuracy of an individual when looking at markings on lab equipment.DiscussionThe mixture used in this experiment is just one of many mixtures used to make eardrops. Different eardrops require different amounts of ingredients, that where titrations come into importance.

Pharmacists use titrations to ensure that the amount they believe to be measuring is actually present. Especially with complex compositions of ear dropsBibliography1. http://www.earhelp.[Accessed on the 10th of March]2.[Accessed on the 10th of March]


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