This experiment is carried out to determine the percentage of calcium carbonate, CaCO3 in the toothpaste provided with the experimental technique known as back titration. A back titration is also known as indirect titration. A known mass of toothpaste is neutralised with a known concentration and volume of hydrochloric acid, HCl. The mixture is then further neutralised by a known concentration and volume of sodium hydroxide solution, NaOH to determine the number of mole of HCl that reacted with CaCO3 in the toothpaste. As the number of mole of CaCO3 is found through the mole ratio thus the mass of CaCO3 is known and the percentage of CaCO3 can be calculated. The percentage of CaCO3 from this experiment is about 20.14% which is close with the …show more content…
It accounts for about 20% to 30% of weight in the whole toothpaste. In common cases, a titrant may be titrated directly with a analyte. However, calcium carbonate is an insoluble salt so it is unsuitable to be titrated directly with acid for the neutralisation process. Hence as to determine the percentage of calcium carbonate in the toothpaste, a back titration method is required as toothpaste which contains calcium carbonate are generally insoluble thus it is difficult to be titrated as the end point of the titration maybe hard to be observed. Excess hydrochloric acid is added to the toothpaste to neutralise the calcium carbonate and the remaining unreacted hydrochloric acid is further titrated by sodium hydroxide solution to determine the volume of hydrochloric acid reacted with calcium carbonate in the …show more content…
This can also speed up the reaction between calcium carbonate and hydrochloric acid with a higher temperature as a higher temperature can let the particles to have a higher kinetic energy for more effective collisions to have a higher rate of reaction. From the calculation, the percentage if calcium carbonate in the toothpaste is found to be at about 20.14% which has a small difference from the literature value of 20%. This maybe cause by the errors when carried out the experiment such as parallax errors. For example, the position of eye level when taking reading or measuring which are not perpendicular to the scale of the instrument will result in the inaccuracy of the data. Next, the other error that might happen is the filter funnel is not inserted properly into the mouth of the conical flask causing some of the hydrochloric acid to be evaporated in the form of vapours. As some of the hydrochloric acid is evaporated, the concentration of the hydrochloric acid will be affected which will lead to the inaccuracy of the
Marwah Alabbad Post lab 10/21/15 Question 1: 1. Experiment 1: Number of trails NaOH concentration (M) Volume of HCl solution (mL) Initial volume of NaOH(mL) final volume of NaOH(mL) The volume of NaOH to titrate HCl (mL) Concentration of HCl (M) 1st 0.1023 25.0 10.05 36.12 26.07 0.085 2nd 0.1023 25.0 5.74 31.40 25.66 0.105 3rd 0.1023 25.0 9.84 35.52 25.68 0.105 First trail calculation: 0.02607L× (0.1023mole NaOH/1L)×(1 mol of HCL/1 mol of NaOH)×(1/0.025)= 0.085M of HCl
Exercise 1 1. Suppose a household product label says it contains sodium hydrogen carbonate (sodium bicarbonate). Using your results from Data Table 1 as a guide, how would you test this material for the presence of sodium bicarbonate? B BoldI ItalicsU Underline Bulleted list Numbered list Superscript Subscript33 Words
Enzymes are a form of protein that lowers activation energy and speeds up reactions as a catalyst. They are made by the stringing together of an abundant amount of amino acids and folded into a specific shape for chemical reactions. Turnip Peroxidase is the enzyme used in this lab and is derived from the vegetable. Enzymes are not used up or permanently altered by their environment Peroxidases are found in a range of organisms and function to break down alcohol (H2O2) and creates byproducts of oxygen and water. In this experiment, the reducing agent guaiacol is added with the substrate, hydrogen peroxide, to create water and oxygen.
Research Question: To investigate and compare how different temperature (5℃, 15℃, 25℃, 35℃, 45℃) can affect the concentration of carbon dioxide in soda water through titration with sodium hydroxide solution. Introduction: Carbon dioxide plays an important role in soft drinks. Soda water is manufactured by pumping carbon dioxide into water under high pressure. Carbon dioxide dissolves in water to form carbonic acid, which is the fizz we find in soft drinks. CO2 + H2O ⇌
Firstly, because the NaHCO3 compound was not stored in a sealed container, therefore dust particles could have changed the results, and making the product impure. Also, there are uncertainties associated with the instruments used in this experiment. This, if the products were measured slightly more than should be, this could have affected the concentrations of the solutions, and therefore causing a larger
These are particularly noteworthy in Mountain Dew, which, in two of the five trials, resulted in a mass gain of the teeth. This is also present in Trial 2 of the Pepsi. It was first assumed that the mass gain was due to the mass of the soft drink still on the marble chips, but after measuring them again the next day after allowing them to dry out, this theory was not supported. It is more likely that the initial mass of the marble chips were incorrectly noted, which resulted in incorrect data. Furthermore, both Coke and Fanta had a drastic difference in their highest and lowest percentage mass loss, with Coke’s greatest difference at 9.23% and Fanta’s at 8.18%.
Then the balloon is lifted up so that the baking soda runs into bottle to react with the vinegar inside. Immediately the balloon is inflated by the carbon dioxide formed. The baking soda is kept constant when the experiment is repeated for another different amount of vinegar. Results/Findings When sodium bicarbonate and vinegar mix,
Using the Law of Definite Proportions, the mass of this product was used to determine the number of moles of copper and chlorine in the sample, which led to being able to determine the
Experiment 2: Distillation and Purification of Liquids Angela Kaiser 100125701 ELL 308 September 19th, 2015 Introduction and Experimental: The purpose of this experiment was to determine the ratio of dichloromethane (DCM) to cyclohexane in a DCM/cyclohexane solution by carrying out a fractional distillation. The temperature and volume of distillate were measured periodically to determine the volume both components in the solution. The experiment was performed as written in “Experiment 2: Distillation and Purification of Liquids” from the Chemistry 2050 Lab Manual for Organic Chemistry Part 1, Fall 2015. Results and Observations:
If only one reactant is increased, then the chemical reaction will only produce a certain amount of products after the limiting reagent is used up, and in this experiment, the most mass the reaction could produce was 0.4 grams. Although we kept adding calcium chloride, not adding sodium hydroxide in the same proportions will not yield more product, which is the main goal in conducting this lab. We should have seen a plateau at 0.4 grams to show that the limiting reagent inhibited further Ca(OH)2 production, but we made several mistakes in our experiment, which made the data unusable to conclude. Once again, the data is polluted, so these number are not accurate, but it is the data our group has to work with. The theoretical yield should have been more than the actual yield, and the percentages should have been less than 100.
Upon cooling, it was shaken until no bubbles were formed. 20 mL of each brand of soft drinks was titrated with NaOH solution. 3 drops of phenolphthalein was used as an indicator if it has already completely reacted. The acidity can then be calculated referring on the known concentration and volume of base; and the known volume of acid.
AIM To design an investigation to study the kinetics of a reaction of your choice RESEARCH QUESTION With respect to hydrochloric acid (HCl), what is the order of reaction in the reaction between HCl and calcium carbonate (CaCO3) determined by changing the concentration of HCl and measuring the volume of carbon dioxide gas (CO2) collected in 30 seconds whilst keeping the mass of the powdered CaCO3 constant and the temperature of the reaction system at 25oC? BACKGROUND INFORMATION Calcium carbonate (CaCO3) is a chemical compound that is commonly found in rocks such as chalk, limestone, marble and travertine in all parts of the world. It also used as a form of medicine as a dietary supplement for a person with insufficient calcium intake because calcium is needed by the body for healthy bones, muscles, nervous system, and heart. CaCO3 is also used as an antacid to relieve
They sell their specific brand of toothpaste as the best and then use the rhetorical appeals of ethos, logos, and pathos along with color schemes and word placement to convince you of their claims. Colgate instills people with an urgency to take care of their teeth and a confidence that their toothpaste will do the job better than any other
Introduction The goal of the experiment is to examine how the rate of reaction between Hydrochloric acid and Sodium thiosulphate is affected by altering the concentrations. The concentration of Sodium thiosulfate will be altered by adding deionised water and decreasing the amount of Sodium thiosulphate. Once the Sodium thiosulphate has been tested several times. The effect of concentration on the rate of reaction can be examined in this experiment.
The mass of vinegar used during the experiment was 4.108 grams. It was determined that there were .003129 moles of CH3COOH in the vinegar sample. Using this information and the molar mass of CH3COOH, which was 60.05 g/mol, the mass of acetic acid in the vinegar was calculated: 4.Vinegar is a 5% aqueous solution of acetic acid. Since the mass of acetic acid within the vinegar was calculated as .18789 g in step 3, the percent of CH3COOH was calculated using the following equation: To calculate the percent error, the experimental value of 4.5% acetic acid in vinegar was subtracted by the theoretical value of 5% and divided by 5% to yield a percent error of 8.54%. The following is a copy of the calculations done using decimals: 5.The equivalence point of the titration curve measured in step 1 was 25.25 mL of NaOH.