Chemistry Investigation on neutralisation reaction.
Chemistry Investigation on neutralisation reaction Plan Neutralisation is the reaction that occurs when an acid has its acidity, that is its hydrogen ions removed by, another chemical containing OH- hydroxide ions. Chemicals that can cancel out an acid in this way are: bases (metal oxides or hydroxides), alkalis (bases that dissolve), metals (e.g. magnesium) or metal carbonates (e.g. marble chips) All of these have a similar way of removing the hydrogen from the acids (they swap it or their metal atoms) but the reactions are quite different. They will all get quite hot if the acid is strong enough, but only the last two will make bubbles. Metals form hydrogen gas, carbonates make carbon dioxide. All of them will leave a neutral chemical after the reaction has finished, if all the acid has been used up. Titration is a technique used to calculate the concentrations or amounts of substances. In an acid base titration you may have an acid that you don't know the concentration of, and a base whose concentration you do know. The technique is to measure out accurately a volume of the alkali of unknown concentration into a flask, and fill up a burette with the acid. Add some indicator solution to the acid in the flask, so that when all the acid has reacted with the base, there will be a colour change. The burette is graduated. You then open the tap on the burette and let the acid
Chem MC analysis. In which of the following cases may it obtain a complete neutralization? (1)25.0 cm3 of 0.120 M sulphuric acid and 50cm3of 0.120M sodium hydroxide solution (2)50.cm3 of 0.5 M Sodium hydroxide and 0.025 moles of aqueous ammonium chlorid
Chemistry 4A Ivan Liu Chun Pok (12) Non-practical task Mc analysis Topics: Neutralization, Reaction between alkali and ammonium compound, Strength of Acid and Alkali, Reacting Masses, Volumetric Analysis In which of the following cases may it obtain a complete neutralization? (1)25.0 cm3 of 0.120 M sulphuric acid and 50cm3of 0.120M sodium hydroxide solution (2)50.cm3 of 0.5 M Sodium hydroxide and 0.025 moles of aqueous ammonium chloride (3)20.0cm3 of 0.100M phosphoric acid and 30.0cm3 of 0.200 M potassium hydroxide solution (4)Dissolve 0.2025g of solid sodium hydroxide in water and make up to 250cm3 of solution, then 25.0cm3 of this solution is added to 50.0cm3 of 1M hydrochloric acid A.2 B.1, 3 C.3, 4 D.1, 2, 4 Option 1: H2SO4(aq) + 2NaOH(aq)--> Na2SO4(aq) +H2O(l) Mole ratio of H2SO4 : NaOH = 1:2 ?Using the formula, Molarity of a solution M or mol dm-3 = Number of moles of solute (mol) / Volume of solution (dm3) ?Number of moles of solute (mol) = Molarity of a solution M or mol dm-3X Volume of solution (dm3) Number of moles of H2SO4 given: 0.12 X (25.0 /1000) = 0.003mol Number of moles of NaOH given: 0.12 X (50.0/1000) = 0.006mol Mole ratio of H2SO4 : NaOH = 0.003 /0.006 = 1:2 Therefore, option (1) is correct. Option 2 : NaOH(aq) + NH4Cl (aq)--> NaCl(aq)+ NH3(g) + H2O(l) Mole ratio of NaOH : NH4Cl
Anti-acids
Rates of Reaction in Antacids Aim I plan to investigate the effectiveness of indigestion tablet on HCl. I will take five different tablets containing a specific pH of alkali and discover how long it takes for the indigestion tablet to neutralise the acid. The test will tell me which is the fastest acting tablet is in the case of indigestion. Antacids Antacids are bases and most of them react on the excess build up of stomach acid and neutralises the acid. The most common of these bases are hydroxides, carbonates or bio carbonates. Neutralisation If you have a weak acid you can eliminate the acidic pH by combining it with a weak alkali. Neutralisation is a chemical reaction which is caused by combining both an acid and an alkali, this reaction causes both the solutions to cancel each other out, and this reaction leaves you with water and salt. Neutralisation is simply the combination of hydroxide ions OH and hydrogen ions H+ this process leaves you with the molecule H2O and forms salt. A Catalyst Catalyst are use to speed up a reaction without being used up. In some reaction like digestion catalyst are essentially needed to speed the process or digestion would take too long to gain nutrients. A catalyst breaks down a substance to provide a larger surface area to speed up the process. The Rate of Reaction The amount of time the reaction occurs. Magnesium +
equilibrium constant
Experiment 13 Aim: To determine the equilibrium constant for esterification from ethanoic acid and propan-1-ol. Procedure: . 0.25 mole (equals to 14.3 cm3) of glacial ethanoic acid (density = 1.05 g cm-3) and 0.25 mole (equals to 18.8 cm3) of propan-1-ol (density = 0.8 g cm-3) was put into a clean, dry pear-shaped flask. It was mixed thoroughly. 2. 1.0 cm3 of the mixture was transferred by pipette to a 250 cm3 conical flask containing about 25cm3 deionized water and 2 drops of phenolphthalein indicator. It was titrated to end point with 0.50 M sodium hydroxide solution. The titre (V1 cm3)was recorded. 3. 8 drops of concentrated sulphuric(VI) acid was added to the remainder of the acid-alcohol solutions while continuously swirling the flask. Another 1.0 cm3 sample was titrated immediately. The titre (V2 cm3)was recorded. The difference between V1 and V2 Represents the volume to be subtracted from subsequent titration to correct for the amount of sulphuric (VI) acid present. 4. A few anti-bumping granules were added to the flask, and it was attached to a water-cooled reflux condenser. It was refluxed for 1 hour. The flask and its contents in an ice bath was cooled. 1.0 cm3 sample was removed from the flask for titration with the 0.50 M sodium hydroxide solution as before. The titre needed was recorded (V3)and was corrected for the sulphuric(VI) acid. 5. Refluxing was
To determine the concentration of limewater
To determine the concentration of limewater The aim of my experiment is to find the concentration of limewater solution provided. To do this I am provided with the following chemicals. Limewater: 250cm (1g dm ) Hydrochloric acid at concentration of 2 mol dm As you can see the hydrochloric acid is too concentrated for titration (i.e. one drop could change the colour of the solution.) therefore this acid will need to be diluted. Background knowledge: Limewater can be made by dissolved slacked lime in water to get a solution of calcium hydroxide (Ca (OH) 2 ) Quicklime + water --------> slaked lime CaO (s) + H2O (l) --------> CaCl2 (aq) + 2H2O (l) In this reaction it shows the calcium oxide reacts with water to produce calcium hydroxide. After adding more water to calcium hydroxide, it produces the saturated aqueous solution known as limewater. My experiment will be based on this neutralisation reaction between the limewater and hydrochloric acid. Ca(OH) 2 (aq) + HCl (aq) --------> CaCl2 (aq) + 2H2O (l) (back ground knowledge of limewater provided by AS chemistry 1 page...) Diluting hydrochloric acid: My first task for this experiment is to dilute the hydrochloric acid. To do this I first needed to work out how many moles I wanted for hydrochloric acid. For this part of my calculation I choose to use text book Calculations for chemistry to help me.
Investigation into the effect of acid/alkali strength on the heat change when acids and alkalis are mixed
Investigation into the effect of acid/alkali strength on the heat change when acids and alkalis are mixed Planning The main aim of this experiment is to investigate the heat change when acids and alkalis are mixed. The temperature of the acid and alkali that are going to be mixed will be taken prior to mixing, and after they have been mixed in a polystyrene cup. The maximum temperature rise will be noted as this will be the biggest heat change that has occurred. The highest temperature after mixing and the temperature prior to mixing will be subtracted to give the heat change. To ensure that this is a valid test the volume of the acid and alkali will be kept constant at 40cm3. The volume will be kept constant because if there were a differing volume of acid to alkali this would have an influence on the temperature rise as there is not the same amount of solution. The only variable in this experiment will be the strength of the acid and alkali. This will allow us to examine the manner in which the heat evolved differs for differing strengths. Comparisons can then be undertaken to see how concentration affects the heat change in set volumes of acid and alkalis. When the acid and alkali of the same volume are mixed, this will cause the process of neutralisation to occur. Neutralisation is the reaction between an acid and a base. It is the formation of a bond between
To determine the amount of ammonia in a sample of household cleaning product, 'cloudy ammonia', in the form of NH4OH through the process of volumetric analysis.
Volumetric Analysis of Ammonia in Household Cleaning Product Aim: To determine the amount of ammonia in a sample of household cleaning product, 'cloudy ammonia', in the form of NH4OH through the process of volumetric analysis. Introduction: Neutralisation refers to the process whereby an acid reacts with a base in stoichiometric proportions to each other to form a salt and water. In this experiment, the neutralisation reaction can be summarized by the following equation: HCl (aq) + NH4OH (aq) NH4Cl (aq) + H2O (l) In this prac, the primary standard is Na2CO3. Primary standards are substances that possess certain properties (i.e. it is soluble), which enable it to be made up into a solution of a known concentration with distilled water to high degree of accuracy. A secondary standard on the other hand, is any solution that has an accurately known concentration. In this experiment, HCl acid is the secondary standard. The standardising of HCl can be summarized by the following equation: 2HCl (aq) + Na2CO3 (aq) 2NaCl(aq) + CO2 (g) + H2O (l) Since sodium carbonate is deliquescent, it is kept in a desiccator to prevent it from absorbing water vapour from the atmosphere. The equivalence point of a titration is the point where the reactants are present in stoichiometric proportions to each other. The end point of a titration is the stage at which the chosen
Biology Practical Investigation to find the lowest concentration of Copper Sulphate solution that will denature egg albumen.
Biology Practical Investigation to find the lowest concentration of Copper Sulphate solution that will denature egg albumen Introduction Proteins are organic compounds of large molecular mass. They are not truly soluble in water; instead they form colloidal suspensions. Proteins are made up of amino acids.. There are three main types of bonding. The first is a disulphide bond. This occurs between cysteine molecules in the same amino acid chain (intrachain) or between molecules in different chains (interchain). This is the strongest and last bond to break. The second bond is an ionic bond. At a suitable pH, an interaction may occur between ionised amino and carboxylic groups. This forms an ionic bond. These ionic bonds are weak and can be broken in an aqueous solution by changing the pH of the medium surrounding the polypeptide. The third bond is the hydrogen bond, which occurs between hydrogen and oxygen atoms within the polypeptide chain. Hydrogen bonds are not strong on their own, although a large number of them together makes them very strong. The primary structure is the linear sequence of amino acids stablised by peptide bonds. There are an endless number of different possible primary structure The secondary structure is the sequence of amino acids arranged as either an alpha helix of a beta pleated sheet stabilised by peptide s. bonds and hydrogen bonds. In an
To see how the concentration of acid, reacting with potassium carbonate, affects the rate of reaction
Aim: To see how the concentration of acid, reacting with potassium carbonate, affects the rate of reaction. Intro: This is the reaction I am using in my coursework: 2HCl + K2CO3 CO2 + 2KCL + H2O In order for substances to react together the particles in the substances must collide with each other and the collision must have enough energy. If there isn't enough energy, no reaction occurs. If there are lots of successful collisions then a lot of CO2 will be produced. The rate of a reaction depends on how many successful collisions there are in a given unit of time. A reaction can be made to go slower or faster by changing the concentration of a reactant. Acid particle Water molecule Potassium carbonate tablet 1 2 In dilute acid, there are not so many acid particles (see diagram 1). This means there is not much chance of acid particles hitting a potassium carbonate particle. In a more concentrated solution of acid, there are more acid particles (see diagram 2). There is now more chance of a successful collision occurring. Concentration is how much of a substance there is in a certain volume and is measured in Moles per litre of solution (M). The concentration of a solution is the amount of solute, in grams or Moles that is dissolved in a litre of solution. That is what my coursework is mainly about. I predict that on my
Analysis of Neutralisation of NaOH
Analysis The results of my investigation show that the larger the amount of hydrogen in the acid, the smaller the amount of acid needed to neutralise the alkali. This is because when an acid is added to an alkali each hydrogen ion in the acid joins an hydroxide ion in the alkali to form neutral water. Hydrogen ion from Hydroxide ion from Neutral water acid alkali The solution only becomes neutral if the amount of hydrogen ions and the amount of hydroxide ions are equal. If there were more hydrogen ions than hydroxide ions then once every hydroxide ion joined with 1 hydrogen ion there would be hydrogen ions left meaning the solution would become acidic. If there were more hydroxide ions than hydrogen ions then once every hydrogen ion joined with 1 hydroxide ion there would be hydroxide ions left meaning the solution would become alkaline. When an acid contains more hydrogen then there are a greater number of hydrogen ions per ml than in an acid containing less hydrogen so it will be stronger. For example in sulphuric acid (H2SO4) there would be more hydrogen ions than in the same amount of hydrochloric acid (HCl), twice as many because there are twice as many in the formula. The alkali in my experiment remained the same throughout (NaOH)
