Unit 2 Assignment A: Undertake titration and colorimetry to determine the concentration of
solutions.
I have been newly appointed as a technical assistant at chemical equip, a large chemical plant. As
part of my induction period, I have been asked to demonstrate skills in a range of practical
procedures and techniques. This will also allow me to progress in my role. Part of my role is to ensure
equipment and chemicals are calibrated, and safety checked. Regularly my company is required to
make and test standard solutions using titration and colorimetry procedures. I have been asked to
demonstrate my ability to carry out these techniques accurately. I have also been asked to create a
report including evidence of practical skills, along with an analysis of my technique and an evaluation
of my results including improvements.
Why do we calibrate equipment?
Calibration of equipment is required in order to ensure precision, standardisation, and repeatability,
which guarantees accurate and trustworthy results. If Your equipment is not calibrated this increases
the risk of systematic error and you receiving Inaccurate measurements and results.
Calibrating the Balance
1. First, I took the balance out of the box and cleaned it with a paper towel, in order to remove any
chemicals leftover from previous use.
2. I then switched the balance on and waited for 0.00 to display on the screen.
3. Next, I pressed the M button down until it displayed CAL on the screen.
4. I Then press the M button again until the screen flashed 100.00.
5. Then using tweezers, I carefully placed my 100g mass onto the center of the balance.
6. I then waited until the display showed PASS, as this meant my calibration was complete.
7. Finally, I removed the 100g weight of the scale.
Preparing the standard solution
Calculation of the mass of Sodium carbonate required.
Formula for Sodium carbonate = Na2C03
Concentration of Sodium carbonate = 0.05M
Volume = 250cm3
Conversion of Volume from cm3 to dm3 = 250/1000 = 0.25dm
Moles = Concentration x Volume
Moles = 0.05M x 0.25dm3 = 0.0125mol−1
Mr of Sodium Carbonate = (Na2CO3) = (23x 2) + (1 x 12) + (3 x 16) = 106
Mass of Sodium Carbonate = Mr x Moles
Mass = 106 x 0.0125 = 1.325g
solutions.
I have been newly appointed as a technical assistant at chemical equip, a large chemical plant. As
part of my induction period, I have been asked to demonstrate skills in a range of practical
procedures and techniques. This will also allow me to progress in my role. Part of my role is to ensure
equipment and chemicals are calibrated, and safety checked. Regularly my company is required to
make and test standard solutions using titration and colorimetry procedures. I have been asked to
demonstrate my ability to carry out these techniques accurately. I have also been asked to create a
report including evidence of practical skills, along with an analysis of my technique and an evaluation
of my results including improvements.
Why do we calibrate equipment?
Calibration of equipment is required in order to ensure precision, standardisation, and repeatability,
which guarantees accurate and trustworthy results. If Your equipment is not calibrated this increases
the risk of systematic error and you receiving Inaccurate measurements and results.
Calibrating the Balance
1. First, I took the balance out of the box and cleaned it with a paper towel, in order to remove any
chemicals leftover from previous use.
2. I then switched the balance on and waited for 0.00 to display on the screen.
3. Next, I pressed the M button down until it displayed CAL on the screen.
4. I Then press the M button again until the screen flashed 100.00.
5. Then using tweezers, I carefully placed my 100g mass onto the center of the balance.
6. I then waited until the display showed PASS, as this meant my calibration was complete.
7. Finally, I removed the 100g weight of the scale.
Preparing the standard solution
Calculation of the mass of Sodium carbonate required.
Formula for Sodium carbonate = Na2C03
Concentration of Sodium carbonate = 0.05M
Volume = 250cm3
Conversion of Volume from cm3 to dm3 = 250/1000 = 0.25dm
Moles = Concentration x Volume
Moles = 0.05M x 0.25dm3 = 0.0125mol−1
Mr of Sodium Carbonate = (Na2CO3) = (23x 2) + (1 x 12) + (3 x 16) = 106
Mass of Sodium Carbonate = Mr x Moles
Mass = 106 x 0.0125 = 1.325g
