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  3. PEARSON (PEARSON)
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  5. Applied Science 2016 NQF
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  7. Unit 19 - Practical Chemical Analysis

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applied science - unit 19 A part 3 - sodium bicarbonate - distinction level
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this is part 3 of the unit 19 A . it was graded as a distinction by my teacher Investigation 3 : Determining the Sodium Bicarbonate Content of Bottled Water

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  • May 14, 2024
  • 12
  • 2023/2024
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Investigation 3 : Determining the Sodium Bicarbonate Content of Bottled Water

Background : Various chemicals, including sodium bicarbonate, are present in bottled water. You'll
be able to ascertain the amount of sodium bicarbonate present in bottled water thanks to this
investigation. You will first use sodium carbonate to standardize a hydrochloric acid solution before
dilution the acid to the proper concentration. When you want to know how much sodium
bicarbonate is in the bottled water, you'll titrate it using dilute hydrochloric acid.

Equipment

Part 1: Conical flask 100ml , 10ml pipette and pump

Part 2:White tile , Distilled water, 250cm Volumetric Flash, pH Meter , Conical flask 250cm3 ,
Burette , 25cm Pipette and Pump.



Healthy and safety

• Lab coat to be worn with safety glasses

• hazard warning and safety sign to be read before any work

• Bags or coat must be kept under bench or completely out of the way

• Spills or accident must be told to a teacher. Rinse skin instantly if any chemical touches

• Broken glass must be cautiously place in the yellow glass bin provide

Chemical

0.1Mol of Hydrochloric Acid – the hazard associated with this chemical are it can cause chemical
burns to your skin and eye, they are highly reactive and can cause explosion, they are also an
environmental hazard. By following the health and safety this risk can be minimized

Phenolphthalein- the hazard associated with this chemical is that is can cause irritation to the skin
and eyes, are highly toxic if digested and are an environmental hazard. This risk can be minimized by
following the health and safety

0.2Mol of sodium bicarbonate -the hazard associated with this chemical – it can cause eye and skin
irritation, it can cause respiratory problems, the are highly reactive so it can cause explosion plus
they are environmental hazard. By following the health and safety this risk can be minimized

, Part 1 : Standardise hydrochloric acid with sodium carbonate.

Method

To begin the experiment, we will need to transfer a 10 cm3 aliquot of the hydrochloric acid solution
into a 100 cm3 conical flask. This is done by carefully measuring out the 10 cm3 of solution and
transferring it into the conical flask using a pipette. It is important to be precise in this step as any
error in the volume of the solution added will affect the accuracy of the results. Next, a few drops of
phenolphthalein indicator are added to the hydrochloric acid solution in the conical flask.
Phenolphthalein is a commonly used indicator in acid-base titrations, which changes from colourless
to pink as the pH of the solution increases. We will then titrate the hydrochloric acid solution with
the sodium carbonate solution of known concentration, which is 0.2 mol/dm3. The sodium
carbonate solution is added gradually to the hydrochloric acid solution, while stirring the mixture,
until the solution just changes from colourless to pink. This colour change marks the endpoint of the
titration. At this point, we will record the volume of the sodium carbonate solution used to reach the
endpoint of the titration. This volume is known as the titre, and it will be recorded in Results table 1.
We will then repeat steps 1 to 3 until we obtain concordant results, which are readings within 0.1
cm3 of each other. This helps to increase the precision and accuracy of our measurements.

Experiment Rough 1 2 3 Average

Initial volume (cm3) 0.00 0.00 4.00 8.00

Final volume (cm3) 7.00 4.00 8.00 12.00

Titre volume (cm3) 7.00 4.00 4.00 4.00 4.00



Calculation

Na2CO3(aq) + 2HCl(aq) →+ 2NaCl(aq) CO2(g) + H2O(l)



Step 1: moles of Na2CO3(aq) used

Average titre = 4.00

Moles = conc x vol

Moles = 0.2 x ( 4.) = 0.0008



Step 2 : moles of HCL used

0.0008 x 2 = 0.0016



Step 3 : conc of HCL

Moles / vol = c/ 0.01 dm = 0.0016/ 0.01 = 0.16

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