Experiment 4: Back Titration (Formal Report)
The objective of the experiment is to determine the percentage by mass of calcium carbonate(CaCO3) in toothpaste through back titration. To determine the amount of CaCO3, a known mass of toothpaste reacts with a standard acid solution (HCl) of known concentration and known volume. The solution containing the toothpaste and excess acid is then back titrated with a standard base solution (NaOH). By obtaining the volume of NaOH used for back titration, we are able to find the volume of HCl which reacted away. From the volume of HCl that reacted away, we are thus able to find the mass of CaCO3 present and hence calculate the average percentage by mass of CaCO3 in toothpaste. My results for the experiment is that the toothpaste contains an average of 20.9% of CaCO3. As my results fall within the acceptable range of 18% to 22%, I conclude that my experiment is a success.
Toothpaste contains active ingredients which help to fight cavities and reduce the risks of gum disease. (toothpaste)Toothpaste contains approximately 20% of calcium carbonate.
Calcium carbonate is mainly naturally found in chalks and limestone. It can be used in home products such as toothpaste and shoe polish.(wassup) Hydrochloric acid (HCl) was discovered by the alchemist Jabir ibn Hayyan around the year 800. It is a highly corrosive acid and it also constitutes majority of gastric acid. (HCL) Sodium Hydroxide (NaOH) is discovered by Humphrey Day in year 1807. It is a highly corrosive base, has a pH of 13 and is very soluble in water. NaOH can be used in removal of paint and in the manufacturing of plastics and textiles. (naoh)
The experiment is conducted to find the percentage of mass of CaCO3 in toothpaste.
Toothpaste mainly contains fluoride, abrasives such as calcium carbonate and flavours.(toothpaste) The percentage of mass of CaCO3 to the toothpaste should be between 18% to 22%.
Back titration is used to find the concentration of an analyte that is insoluble in water but soluble in acid. To perform back titration, react the analyte with excess acid (or base) and then back titrate the acid that is unreacted with a base (or acid). For this experiment, add a weighed portion of toothpaste to a known excess 0.1672 M standard HCl solution. After the reaction is completed, the resulting solution is back titrated against a 0.0797 M of standard NaOH solution.
Chemical reactions that are involved:
CaCO3 + 2 HCl ? CaCl2 + H2O + CO2
NaOH + HCl (excess) ? NaCl + H2O
To derive the percentage of CaCO3 in toothpaste, firstly find the amount of NaOH required for back titration and then calculate the volume of excess HCl using mole ratio. After finding the volume of excess HCl, subtract the value from the original volume of HCl to find the volume of HCl that reacted away. From the volume of HCl that reacted away, we can calculate the mass of CaCO3. To find the percentage of CaCO3 in toothpaste, it is mass of CaCO3 divided by the mass of toothpaste, multiplied by 100%. Hence, we are able to derive the percentage of CaO3 in toothpaste.
10.00mLof 0.1672M standard HCl was pipetted into a clean conical flask.
0.1000g to 0.2000g of toothpaste was removed from the container with a glass rod and the weight was recorded on the datasheet.
The glass rod with the weighed toothpaste from step 1 was placed in the conical flask and the toothpaste was dislodged completely from the glass rod.
About 10.00mL of deionised water was added to wash down the acid from the glass rod.
A funnel was inserted into a flask and it was heated gently over a hot plate until the reaction was completed. It had taken about 3 to 5 minutes.
The funnel and the side wall of the flask was rinsed with small amount of deionised water.
The flask was allowed to cool to room temperature.
1 to 2 drops of methyl orange indicator was added and the excess HCl was back titrated with the 0.0797M NaOH. The indicator had changed colour from red (or pink) to orange. A yellow colour at the end point had indicated that it is over titrated.
The above steps were repeated twice.
Results and Calculations
Results 1st 2nd
Mass of toothpaste (g) 0.1462 0.1463
Initial burette reading (mL) 0.00 0.00
Final burette reading (mL) 13.00 12.90
Volume of NaOH used (mL) 13.00 – 0.00 = 13.00 12.90 – 0.00 = 12.90
Using the mole ratio, (M_NaOH x V_NaOH)/(M_HCl x V_(HCl (excess)) )= 1/1
Volume of excess HCl (mL) 0.0797 x 13 = 0.1672 x V_HCl
V_HCl=6.1968 (5 sf)
0.0797 x 12.90 = 0.1672 x V_HCl
V_HCl= 6.1491 (5 sf)
Volume of HCl reacted away (mL) 10.00 – 6.1968 = 3.8032 10.00 – 6.1491 = 3.8509
Using the mole ratio, (moles of ?HCl?_reacted)/(M_HCl x V_(HCl reacted) )= 1/1
Moles of ?HCl?_reacted (moles) Moles of ?HCl?_reacted
= 0.1672 x 3.8032/1000
= 6.3590 x ?10?^(-4) Moles of ?HCl?_reacted
= 0.1672 x 3.8509/1000
= 6.4387 x ?10?^(-4)
From equation: CaCO3 + 2 HCl ? CaCl2 + H2O + CO2
(moles of ?HCl?_reacted)/(moles of Ca?CO?_3 )=2/1
Moles of CaCO3 (moles) (6.3590 x ?10?^(-4))/2=3.1795 x ?10?^(-4) (6.4387 x ?10?^(-4))/2=3.2193 x ?10?^(-4)
Mass of CaCO3 (g) 3.1795 x ?10?^(-4) x 100.1
=0.031827 (5 sf) 3.2193 x ?10?^(-4) x 100.1
= 0.032225 (5 sf)
% CaCO3 in the toothpaste 0.031827/0.1462 x100%=21.769% (5 sf) 0.032225/0.1463 x100%
= 22.027% (5 sf)
Average % of CaCO3 in toothpaste sample (21.769+22.027)/2=21.9% (3 sf)
Firstly, the mixture of toothpaste with water and HCl is heated. This is because when CaCO3 reacts with HCl, carbon dioxide is one of the products. Carbon dioxide (CO2) can increase the acidity of the solution, thus through heating the solution, CO2 will evaporate and will have lesser effect on the acidity of the solution. After reacting HCl with CaCO3, there is unreacted HCl due to the excess amount of HCl added initially. The excess amount of HCl is titrated with NaOH till the resulting solution is neutralised. To know that the solution is neutralised, I added 1 to 2 drops of methyl orange indicator. The initial colour of the indicator was red as the solution was acidic but the colour changed to orange, which indicated the end point and that the solution was neutralised. Upon reaching the end point, I recorded down the volume of NaOH titrated. To calculate the percentage by mass of CaCO3 in toothpaste, I firstly use the mole ratio of NaOH and HCl (excess) to find the volume of excess HCl. As the initial volume of HCl used is 10.00mL by subtracting the volume of excess HCl from 10.00mL I am able to find the volume of HCl that reacted away. From there, I am able to find the number of moles of HCl that reacted away. Using the mole ratio of HCl to CaCO3, I am able to find the number of moles of CaCO3. The number of moles of CaCO3 multiplied by the molecular mass of CaCO3, which is 100.1, will give the mass of CaCO3. Hence, by dividing the calculated mass of CaCO3 by the mass of toothpaste, the percentage of CaCO3 in toothpaste is determined. I then repeated the experiment once to calculate the average percentage of CaCO3 in the toothpaste sample.
From the results, the average percentage of CaCO3 in toothpaste is 21.9%. My 2 readings of 21.8% and 22.0% differ by 0.2% could be due to the judgement error on the colour change of the indicator. Also, the difference in reading could be due to the different durations in heating the solution HCl and toothpaste. I had to stop heating when the first droplet drips into the conical flask, however the droplet dripped out at different timings. Due to the difference in heating time, 1 solution may be slight more acidic than the other. This will thus affect the volume of NaOH and the percentage by mass of CaCO3 in toothpaste. However, I feel that my experiment is still considered a success as my first 2 readings and average percentage falls within the acceptable range of 18% to 22%.
Back titration definition
Different uses of calcium carbonate
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