Finding the concentration of an acid sample
Finding the concentration of an acid sample Concentration is the amount of a substance in a given volume1. There is different concentration in all solids, liquids and gases. And here lies the aim of my coursework. "to find the accurate concentration of a given sample of acid solution" The solution of acid I have been given is Sulphuric Acid (H2SO4 herein). We have been told that the concentration lies in between 0.05 and 0.15 mol dm-3. To find the concentration accurately I will titrate the solution using a known indicator with a known endpoint and Sodium Carbonate (Na2CO3 herein) using the following equation: H2SO4(aq) + Na2CO3(aq) › Na2SO4(aq) + H2O(l)+ CO2(g) Background Theory A titration is when and acid is run into a base or an alkali with an indicator added. This indicator will have a known end point. So once you have measured how much of the acid or base required to make the solution reach the endpoint. From there we can measure the concentration. So what I need to decide is which indicator I will use. There can be strong and weak acids, and the same for alkalis and bases. The Bronsted-Lowry theory2 states that acids are H+ donors and that bases are a H+ receiver. A strong acid is defined as having a strong tendency to donate a H+. H2SO4 is a well known acid with a strong base. While the sodium Carbonate is the opposite as it is relatively weak. So now that I
Composition of Hydrates
Composition of Hydrates Purpose: The purpose of this lab was to determine the percentage of water in a hydrate and to calculate the coefficient of the water of hydration of a hydrate. Data Table: Mass of Crucible & Cover 22.03 g Mass of Crucible & Cover & Hydrate 24.45 g Mass of Crucible & Cover & Anhydrous Salt 23.71 g Analysis: Hydrate Mass (Before Heating): 24.45-22.03 = 2.42 g Anhydrous Salt Left (After Heating): 23.71-22.03 = 1.68 g Water Lost: 24.45-23.71 = .74 g Molar Mass of Hydrate used: CuSO4?5H20 = 63.55+32.07+64+10.10+80 = 249.72 g/mole Molar Mass of Anhydrous Salt: CuSO4 = 63.55+32.07+64 = 159.62 g/mole Molar Mass of all the Waters in Compound: 5H2O = 80+10.10 = 90.10 g/mole Theoretical Percentage of Water in Hydrate: 90.10 * 100 = 36.98% 249.72 Experimental Percentage of Water in Hydrate: .74 * 100 = 30.58% 2.42 Percent Error in the Lab: 36.08 - 30.58 = 15.24% 36.08 Questions: ) Regardless of the mass of hydrate, the percentage of water lost stays the same. But, the value is not always the theoretical value, often close; it may vary, depending on the time you have kept the hydrate over the fire. 2) If I used 5 grams of hydrate, I would expect to lose around 36.98% of water. That's about 1.849 grams. Conclusion: The purpose of this lab was to determine the percentage of water in a hydrate and to calculate the coefficient of the water
Redox Titration Aims/Objective : To determine the molar concentration of the given KMnO4 solution.
Chemistry TAS Report . Experiment Number : 16 2. Date : 30/04/2008 3. Title : Redox Titration 4. Aims/Objective : To determine the molar concentration of the given KMnO4 solution. 5. Introduction / Theory: 6. Relevant Equations/Chemical Reactions Involved : Half-equation : MnO4-(aq) + 8H+(aq) + 5e- › Mn2+(aq) + 4H2O(l) C2O42-(aq) › 2CO2(g) + 2e- Overall equation : 2MnO4-(aq) + 16H+(aq) + 5C2O42-(aq) › 2Mn2+(aq) + 8H2O(l) + 10CO2(g) 7. Chemicals : Oxalic dehydrate acid crystal 1.58 g KMnO4 solution 80 cm3 1 M dilute sulphuric acid 80 cm3 8. Apparatus and equipment : 250 cm3 conical flask 3 burette 250 cm3 beaker 2 filter funnel 25.0 cm3 pipette pipette filler safety spectacle white tile wash bottle 250 cm3 volumetric flask 00 cm3 beaker 2 dropper heat-proof mat Bunsen burner 25 cm3 measuring cylinder 9. Procedure : . 1.58 g of oxalic acid crystal was weighted accurately which was then dissolved in distilled water. 2. The solution was transferred to volumetric flask and made up to 250 cm3 (the mark) by distilled water. 3. 25 cm3 of oxalic acid was pipetted in a conical flask. About 25 cm3 of 1 M dilute sulphuric acid was added to it so that the solution was acidified. 4. The mixture was heated to about 60 oC by Bunsen burner. 5. The hot mixture was titrated with KMnO4 solution until the colour changed
Determination of the equilibrium constant for the reaction
Chemistry TAS Report . Experiment Number : 10 2. Date : 18/01/2008 3. Title : Determination of the equilibrium constant for the reaction : Fe3+(aq) + SCN-(aq) <=> FeSCN2+(aq) 4. Aims/Objective : To determine the equilibrium constant for the reaction : Fe3+(aq) + SCN-(aq) <=> FeSCN2+(aq) 5. Introduction / Theory: In this experiment, the equilibrium constant for the formation of a complex ion, FeSCN2+(aq), is determined. Complex ions, thiocyanatoiron(III) ions, are formed from iron(III) ions and thiocyanate ions in aqueous solution : Fe3+(aq) + SCN-(aq) <=> FeSCN2+(aq) (1) The equilibrium constant for this reaction is: Kc = [FeSCN2+(aq)] (2) [Fe3+(aq)][SCN-(aq)] The product complex ion is the only one of the three species which has an appreciable color (blood-red). 6. Relevant Equations/Chemical Reactions Involved : Fe3+(aq) + SCN-(aq) <=> FeSCN2+(aq) 7. Chemicals : 0.002 M KSCN(aq) 50 cm3 0.2 M Fe(NO3)3(aq) 20 cm3 8. Apparatus and equipment : Boiling tube 5 Dropper 2 0 cm3 measuring cylinder Lamp 25 cm3 measuring cylinder Wash bottle Safety spectacle 250 cm3 beaker 9. Procedure : . 0.2 M Fe(NO3)3(aq) was used and 10 cm3 of 0.08 M, 0.032 M, 0.0128 M, 0.00512 M Fe(NO3)3(aq) were prepared respectively. 2. The solutions were added by using 10 cm3 measuring cylinder to
Indigestion Tablet
Chemistry Lab Report The Best Indigestion Tablet To: Miss Lina Subeh By: Dara Masri Abstract: In this experiment, we are to find which antacid (indigestion tablet), Rennie or Novagel, is the most effective. This is done using titration method using a burette filled with Sodium Hydroxide, in order to measure the amount used to neutralize or lower the pH of the tablet powder mixed with Hydrochloric Acid. This will be indicated by a change in color using Methyl Orange Indicator after adding Sodium Hydroxide. The more the amount of Sodium Hydroxide used, the more it will show that the type of that indigestion tablet is not effective, because we will conclude that the Hydrochloric Acid was not neutralized or the pH of it was lowered, so more Sodium Hydroxide was needed to do that. After doing this experiment, we found out that Rennie needed less Sodium Hydroxide to neutralize its acidity than Novagel, and that meant that Rennie is more effective than Novagel. Key Words: Indigestion Rennie Novagel Sodium Hydroxide Methyl Orange Indicator Titration Experiment Table of Contents: Abstract ............................................................................... 2 Aim ...................................................................................... 4 Hypothesis ............................................................................ 4 Literature
Experiment investigating hydrogen bonding in different chemicals.
Tsuen Wan Public Ho Chuen Yiu Memorial College Form 6 Chemistry Practical Experiment 8: Hydrogen Bonding Date of experiment: 27-1-2011 Objective: A. To discover the existence of hydrogen bonds between ethanol molecules B. To measure the strength of hydrogen bond formed between ethanol molecules C. To investigate the formation of hydrogen bonds between molecules of ethyl ethanoate and trichloromethane D. To measure the strength of hydrogen bonds formed between molecules of ethyl ethanoate and trichloromethane Introduction: A hydrogen bond is the attractive interaction of a hydrogen atom with an electronegative atom, such as nitrogen, oxygen or fluorine, that comes from another molecule or chemical group. The hydrogen must be covalently bonded to another electronegative atom to create the bond. These bonds can occur between molecules, or within different parts of a single molecule. The hydrogen bond (5 to 30 kJ/mol) is stronger than a van der Waals interaction, but weaker than covalent or ionic bonds. This type of bond occurs in both inorganic molecules such as water and organic molecules such as DNA. Intermolecular hydrogen bonding is responsible for the high boiling point of water (100 °C) compared to the other group 16 hydrides that have no hydrogen bonds. Intramolecular hydrogen bonding is partly responsible for the secondary, tertiary, and quaternary structures
Determination of the concentration of limewater solution in g dm-3 as accurately as possible against a standard solution of HCl acid
PLAN The aim of my investigation is to determine the concentration of limewater solution in g dm-3 as accurately as possible against a standard solution of HCl acid. BACKGROUND INFORMATION ABOUT LIMEWATER Limewater solution is a clear coloured saturated Ca(OH)2 produced when calcium carbonate (limestone) is decomposed to form calcium oxide CaO. Water is then added to form CaO (quicklime), to produce slaked lime. Again, excess water is added to form Ca(OH)2 , limewater. Limewater, which is an alkali is used principally in medicine as an antacid as a neutraliser for acidic poisoning or treatment of burns. Limewater as an alkali would have a pH scale of 9-14. I am going to set up my equipment in the form of an acid/base titration since it is a way of measuring quantities of reactants and can be very useful in determining an unknown concentration or following the progress of a reaction which is related to my investigation. The reaction is between an acid (HCl) and alkali, (Ca (OH)2) and therefore it's a neutralisation reaction. In order to determine the concentration of limewater in an experimental way, it is very important to use the most appropriate equipment available. This is why I have chosen to use a pipette, volumetric flask and a burette, as they have an accuracy of +-0.005 cm3, which should produce a suitable accuracy for my experiment. The apparatus I am going to
Mole Ratios in a Chemical Reaction
Renee Buettel Period D4 Ms. Parziale /7/08 Lab #10: Mole Ratios in a Chemical Reaction Paul Bergin Abstract The main objective of this lab experiment was to balance the given chemical equation and to find the correct mole-to-mole ratio of it. The theoretical balanced equation was Pb(NO3)2(aq) + K2Cr2O7(aq) --> PbCr2O7(s) + 2KNO3(aq). In addition, the ratio of moles was one to one.and the correct mole-to-mole ration was one to one. The experimental results matched this ratio. The theory that was proven was that balancing equations give the correct mole ratio of a chemical equation. Introduction In a chemical equation, there are two sides. The chemicals on the tail end are called the reactants and the chemicals on the other side are called products. An example of this given by Coefficients (2008) is 2H2 + O2 --> 2H2O. In this example, "2H2 + 02" is the reactants and "2H2O" is the product. Also, "-->" is the sign for "yield." The big 2s in front of H2 and H2O are called coefficients. In this case, the first 2 indicates that there are 2 molecules of H2, which also means that there are 4 atoms of hydrogen in the reactant part of the equation. The other 2 signifies that there are 2 molecules of H2O as the product. This means that every molecule of H2O that contains 2 atoms of hydrogen contains 4 hydrogen atoms and 2 oxygen atoms. According to Chemistry Formulas (2005), the
Estimated heat distribution by convection in water
Estimated heat distribution by convection in water Introduction This report assesses the distribution of heat by convection in water to estimate the heat conductivity of water. The transfer of heat from a heating coil to a fluid is conduction but the heat transfer within the fluid is convection. This is basically fluid flow of particles arising from nature, heat, chemical or kinetics. The distribution of heat is assessed with various factors introduced. In this case a magnetic stirrer and a motor. This report presents an estimate of the effect of free and forced convention on the distribution of heat in water. Experimental method The apparatus were arranged as shown in fig. 1. A beaker of five litre capacity was places on a motor, four litres (4L) of cold water was put in a beaker. A heating coil and three thermometers were placed at various depths in the beaker of water and their various distances from the base of the beaker were recorded. Power was supplied to the motor and heating coil and at intervals of four minutes each; the temperatures on all three thermometers were read simultaneously. After four successful readings, the electricity supply was disconnected and the ambient temperature was recorded. This same procedure was repeated twice, the first with a magnetic stirrer and the next time without the magnetic stirrer but the motor operating. Distance from base
An experiment to show the relationship between shape and diffusion rate
An experiment to show the relationship between shape and diffusion rate Aim To see whether there is a relationship between the surface area and the diffusion rate Hypothesis I predict that the smaller blocks of agar will turn clear, or diffuse first, as it has a smaller surface area. This is because there is less surface area and volume for the sulphuric acid to diffuse into. Apparatus * Three sizes of agar, 20x20x20mm, 20x20x10mm, 20x20x5mm * 240ml of sulphuric acid [80ml per beaker] * 3 100ml beakers * Tile used for placing the agar * Tissue to wipe off the sulphuric acid off the agar * 3 scalpels * Ruler, measurable in mm * Stop clock * Calculator Method . First, cut three pieces of sulphuric acid in the following sizes 20x20x20mm, 20x20x10mm, 20x20x5mm, as accurately as possible 2. Next, fill the three beakers with 80ml of sulphuric acid each 3. Then, prepare the stop clock, and make sure it is has been reset 4. After, place the three blocks of agar into the sulphuric all at the same time, as well as starting the stop clock once the agar is in the sulphuric acid. 5. Carefully stir the three beakers using the scalpels. 6. Watch until one of the blocks have gone completely clear. 7. Once one of the blocks have gone completely clear, stop the stop clock and take out the three blocks of agar and place on the tissue, and wipe off the excess sulphuric
