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Essay Unit 2- Laboratory Techniques & their Application
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- PEARSON (PEARSON)
Essay Unit 2- Laboratory Techniques & their Application
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TitrationSince January I have been doing experiments and titration is one of them. I did different
titrations, with hydrochloric acid (HCl), which was involved in all of them, sodium hydroxide and
sodium carbonate, using indicator and pH meter. Some of the instruments I used, I needed to
calibrate to make sure that my equipment was working properly and ensure good readings. I
calibrated the volumetric equipment, weight balance and pH meter.
First, I used water as standard for calibration of volumetric equipment. I used an empty beaker,
measuring cylinder, burette, and pipette. I zero the balance, then I place the beaker on the
balance and recorded its mass. Then I took 25 cm3 of distilled water from a pipette and put it in
the beaker and then recorded its new mass. And to calculate the amount of water delivered I
just make the difference between the two values. I discarded the water and dried the beaker
with a paper towel. In contrast if I did not want to go through the calculations, I could zero the
balance, place the empty beaker and then zero the balance, then add the amount of water I
want, and the balance would give me just the amount of water, which was different for each
equipment. I followed the same steps for burette, 100 cm3 beaker and for the 25 cm3
measuring cylinder. I did the calibration in a temperature of 23 degrees measured with mercury
thermometer for all of them.
This table shows my results.
Pipette Burette 100 cm3 25 cm3 measuring
beaker cylinder
Mass of beaker plus 85.19 85.22 79.93 84.61
25 cm3 water (g)
Mass of empty 60.22 60.22 60.22 60.22
beaker (g)
Mass of 25 cm3 24.97 25.00 19.71 24.61
water (g)
Volume= mass ÷ density
Volume= 24.97÷0.997538
Volume=25.03 cm3
,Volume= mass ÷ density
Volume= 25.00 ÷ 0.997538
Volume= 25.06 cm3
Volume= mass ÷ density
Volume= 24.61 ÷ 0.997538
Volume= 24.67 cm3
0.997538 is the density of water at 23 degrees Celsius. Water has a lot of different densities for
each degree Celsius.
As shown in my results, the first two equipment, the pipette and the burette were the ones
who gave me accurate results. This is the purpose of doing calibration, to ensure that each
equipment that you choose will give you a result close to the true values. This is the reason I
used a burette in all following titrations I will be talking about. I believe that If I had repeated
the calibration, I could get 25 cm3 for the measuring cylinder, as the value I got was close to it.
Titration 1
Part 1: Preparation of a Standard Solution
Equipment:
- Distilled water
- 250 cm3 beaker
- 250 cm3 volumetric flask with stopper
- Pipette
- Spatula
- Filter funnel
- Solid sodium carbonate
- Glass rod
- Specimen boat
Method:
1. First of all, I set up the whole equipment. I calibrated my weigh balance.
2. I zero the balance, placed the specimen boat on the balance and took the reading.
, 3. Then I put the anhydrous sodium carbonate with spatula on the specimen boat and took
the reading.
4. Then I transferred the anhydrous sodium carbonate to a large beaker, accurately and
precisely.
5. After I added 150 cm3 of distilled water to the same beaker and stir with a glass rod until
sodium carbonate is completely dissolved.
6. After it all dissolved properly, I transferred all the solution to a 250 cm 3 volumetric flask
with a funnel. I rinsed the beaker into the funnel to remove the rests and put in the
volumetric flask. And then I rinsed the funnel as well to remove the rests.
7. I completed the solution until 250 cm3 incredibly careful with pipette with distilled
water, then I put the stopper firmly and I shook it so they could combine (mix the
contents).
8. Then I let it rest for 24 hours, after cleaning everything.
Results:
Mass of specimen boat + Na2CO3 (g) 2.36
Mass of specimen boat (g) 1.00
Mass of Na2CO3 (g) 1.36
Molecular mass
Na2CO3= (23×2) + 12 + (16×3)
Na2CO3=106 g/mol
Number of moles
n= m ÷ Mr
n= 1.36 ÷ 106
n= 0.01283019 moles
Concentration
Concentration= moles ÷ (volume÷1000)
Concentration= 0.01283019 ÷ (250÷1000)
Concentration= 0.05132076 mol/dm3
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