Monoprotic (monobasic): An acid that has only one acidic hydrogen atom in its molecules; common examples are hydrochloric acid (HCl) and nitric acid (HNO3).Prediction and Scientific Background: Standard Molar Enthalpy of Neutralisation (/Hn,) is the enthalpy change per mole of water formed in the neutralisation of an acid by an alkali, (298K and 1 atm).* Unknown Acid (aq) + NaOH (aq) –> Salt (aq) + Water (l)* In aqueous solution, strong acids and bases are completely dissociated and Hneut is approximately equal to -57.9 kJ mol-1. This neutralization process corresponds to the reaction: H+ (aq) + OH- (aq) –> H2O (l)* For weak acids, this enthalpy change is less exothermic because some input of energy is required to dissociate the acid. Therefore the Hneut of weak acids and bases is more positive than -57kJmol-1.Hydrochloric acid (strong acid): As HCl is a strong acid; it is a good proton donor, with near to complete dissociation in water. HCl (aq) –> H+ (aq) + Cl- (aq). Hence a strong acid completely dissociates into ions.Ethanoic acid (weak acid): As CH3COOH is a strong acid; it is a poor proton donor. The dissociation in the water is equilibrium, with the equilibrium position well to the left-hand side of the equation.CH3COOH (aq) H+ (aq) + CH3COO- (aq). For every 250 molecules of ethanoic acid, only 1 molecule dissociates and only a small proportion of the potential acidic power is released as H+. A weak acid only partially dissociates into ions.Strong Acids: When an acid dissolves in water, a proton (hydrogen ion) is transferred to a water molecule to produce a hydroxonium ion and a negative ion depending on what acid you are starting from. In the general case:These reactions are all reversible, but in some cases, the acid is so good at giving away hydrogen ions that we can think of the reaction as being one-way. The acid is virtually 100% ionised. For example, when hydrogen chloride dissolves in water to make hydrochloric acid, so little of the reverse reaction happens that we can write:At any one time, virtually 100% of the hydrogen chloride will have reacted to produce hydroxonium ions and chloride ions. Hydrogen chloride is described as a strong acid. Thus, a strong acid is one which is virtually 100% ionised in solution. Other common strong acids include sulphuric acid and nitric acid.You may find the equation for the ionisation written in a simplified form:This shows the hydrogen chloride dissolved in the water splitting to give hydrogen ions in solution and chloride ions in solution. This version is often used in this work just to make things look easier. If you use it, remember that the water is actually involved, and that when you write H+(aq) what you really mean is a hydroxonium ion, H3O+.Weak Acids: A weak acid is one, which does not ionise fully when it is dissolved in water. Ethanoic acid is a typical weak acid. It reacts with water to produce hydroxonium ions and ethanoate ions, but the back reaction is more successful than the forward one. The ions react very easily to reform the acid and the water.At any one time, only about 1% of the ethanoic acid molecules have converted into ions. The rest remain as simple ethanoic acid molecules. Most organic acids are weak. Hydrogen fluoride (dissolving in water to produce hydrofluoric acid) is also a weak inorganic acid.Apparatus:* 3 – 100cm3 beakers* 50cm3 burette* Coffee cup calorimeter: This is a simple, inexpensive device used in many general chemistry labs. It is made of two nested and capped cups made of Styrofoam; making it a very good insulator. When a reaction occurs in such a calorimeter, minimal heat is lost to the surroundings. It the reaction involves heat, for example, nearly all the heat stays within the solution in the calorimeter, where it causes an easily measured temperature increase. In the calculations involving a coffee cup calorimeter, the negligible heat absorbed by the Styrofoam will have to be ignored. This works acceptably well when the maximum temperature is reached soon after the reaction is initiated.* Glass rod* Stand and clamp* Stopwatch* ThermometerProcedure: A watch glass will be placed on an electronic balance to provide an accurate reading of its weight; this reading however is not required and may create a series of problems. Therefore to deal with this, the balance will be set to zero, so that it would not record the weight of the watch glass and consequently would only record the weight of the given solute. The white anhydrous solute, of the unknown acid will slowly be placed onto the watch glass with a spatula, so that the reading would accurately match the specified amount required: -RMM = 135Moles = 1.00MGrams (1 litre) = 135 x 1.00= 135gGrams (50cm3) = 135 ? 20= 6.75gNext a measuring cylinder will be filled with an accurate 50cm3 of distilled water. The unknown acid solute will then be poured into a clean 100cm3 beaker (labelled U). Subsequently the distilled water will be gently poured into the beaker, making sure that any remaining solute on the watch glass will be rinsed gently into the beaker. To ensure the solute had fully dissolved, a thin glass rod could be used to grind and mix any large residual pieces – this would ensure that results were not inaccurate, as the acid would be prepared incorrectly.A 100cm3 beaker (labelled NaOH) will be filled with approximately 75cm3 of sodium hydroxide (1.00M). A burette will be rinsed off with approximately 10cm3 of the sodium hydroxide, which will then be run into the sink. This consequently guarantees that each solution will not become diluted by excess water, or otherwise cause inaccurate results to arise due to the addition of any foreign chemicals from previous experiments. Moreover the beakers will be labelled according to their content, so the different colourless solutions can be clearly identified; reducing the probability of mistakes, which may cause serious problems and/or queries to arise.A stand and clamp will be acquired to which the burette will be placed vertically. The sodium hydroxide must be steadily poured into the burette until the meniscus of the solution is in line with the measurement 0 – making sure the measurements and results are as precise as possible. Subsequently the hydroxide will be run into a clean coffee cup, making sure that practically 50cm3 is contained within this calorimeter. An appropriate calorimeter for the investigation will be plastic cup, as it would sufficiently limit the amount of heat loss during the chemical reaction of neutralisation, as the material acts as an insulator, thus providing more reliable and accurate results leading to a defined inference. Afterwards the burette will be firstly rinsed off with tap water and then with approximately 10cm3 of unknown acid, which will consequently be run into the sink (to remove any excess sodium hydroxide).Similarly the unknown acid will be steadily poured into the burette until the meniscus of the solution is in line with the measurement 0. Then 50cm3 of the sodium hydroxide will be run into a clean 100cm3 glass beaker. At this point, two comparable thermometers will be used to test the temperature of each solution for 3 minutes; a stopwatch will be used to record the temperature (used effectively to gain accurate recordings) at 0.5-minute intervals (every 30 seconds). When the temperature is efficiently recorded the unknown acid will be added to the sodium hydroxide. Afterwards a lid consisting of a small hole (in the centre) will immediately be placed securely on the cup to ensure a minimal amount of heat is lost (due to evaporation or convection), however in this gap of approximately 10-20 seconds a significant amount of heat could be lost or gained thus causing a change (major or minor) that may result in inaccurate recordings developing. When the initial drop of the acid will be applied to the sodium hydroxide the thermometer (that fits through the specific hole) will be used to stir the mixture and additionally record the temperature of the mixture, with the aid of the stopwatch at regular 0.5-minute intervals (every 30 seconds) for a period of five minutes. This experiment will be repeated a further two times and due to this a mean value can be obtained for the temperature change.Safety procedures: Safety glasses and gloves will be worn for health and safety reasons, and therefore as the chemicals are hazardous a risk assessment has been constructed in order to reduce the possibility of accidents or future problems. When chemicals have been used they must be sealed to confine any toxic gases being released. Furthermore chemicals may have to be placed within a fume cupboard to rid of any noxious fumes. Below is a risk assessment concerning the chemicals used within the investigation, along with information regarding ‘potential’ dangers, as well as precautions that may need to be executed when dealing with an unknown acid:Chemical and Structural FormulaRisk AssessmentToxicologyAction (Personal Protection)Sodium Hydroxide (NaOH)Harmful/IrritantVery corrosive and causes severe burns. May cause serious permanent eye damage. Very harmful by ingestion. Harmful by skin contact or by inhalation of dust.Safety glasses, Neoprene or PVC gloves, adequate ventilation.Unknown AcidCould be harmful, an Irritant, highly flammable or even toxicCould causes burns, be toxic by inhalation, ingestion and through skin absorption. Could be probable human carcinogen. Could cause damage to kidneys. Could cause allergic reactions. Could cause sensitisation. Could cause heritable genetic damage. Could be very destructive to mucous membranes, upper respiratory tract as well as eyes and skin.Safety glasses, Neoprene or PVC gloves, adequate ventilation.Calculations: A concentration of 1.00M NaOH and 1.00M of the unknown acid will be used within the experiment as it provides a simple 1:1 ratio enabling calculations to be constructed without difficulty, thus limiting errors in the analysis. Furthermore these moderately dilute concentrations are able to safely and appropriately handled under standard conditions.Results Table: Temperatures will be measured to the one decimal place. 0 will be the temperature of the reactants (NaOH and unknown acid).Time (minutes – 1d.p.)0.00.51.01.52.02.53.04.04.55.0Temperature (oC – 1d.p.)In order to demonstrate how the molar enthalpy change of neutralisation would be calculated, temperatures and numerical values could be practically invented, but the molarity and volumes will remain the same.Results: When 50cm3 of 1.00M NaOH at 25.5oC was added to 50cm3 of the 1.00M unknown acid at 25.5oC; in the coffee cup calorimeter, the temperature increased to a 32.4 oCAssumptions:* Because solutions are relatively dilute, it can be assumed that their specific heat capacities are close to that of water, 4.18 J/g/oC* The density of the solutions = 1.0 g/cm3 (actual density of NaOH at 25oC is 1.04 g/cm3)* There are no heat lost to the coffee cup itself or to the surrounding air* The acid is completely neutralised* Heat evolved = Amount of Substance x Specific Heat Capacity x Temperature ChangeQ = mcTAnalysis: Two particular factors have to be calculated – the system’s heat capacity and the temperature change.Specific Heat = Heat Capacity/ Mass (g)By rearranging the formula: Heat capacity = Mass (g) x Specific HeatThe mass in this equation refers to the total grams of the combined solutions, but volumes are known. So the densities have to be used to calculate mass.Density = Mass/VolumeSolution: Because the densities for the solution are both 1.00 g/cm3 the cm3 are numerically the same as the grams of the solutions. Each solution mixed has a mass of 50.0g, therefore its total mass = 100.0g. We can now calculate the heat capacity of the final solution at the instant of mixing and immediately before any reaction occurs to increase its temperature. For this solution, the specific heat is 4.18 J/g/oC, so:Heat Capacity = Mass x Specific Heat= 100.0g x 4.18 J/g/oC= 418 J/oCThus the heat capacity of the solution in the calorimeter is 418 J/oC. The reaction increases the system’s temperature by t = 32.4 – 25.5oC = 6.9 oC, so the energy evolved by the reactions can be calculated from:Heat Capacity x t = Heat Energy (Evolved)orHeat Evolved = 418 J/oC x 6.9 oC= 2.9 x 103 JThe temperature change will be obtained from a graph similar to thisNote: Four graphs will be drawn, three to display the individual temperature change for each experiment and another graph to illustrate the mean temperature change. The graphs will bare similarity to this:This is the heat for the specific mixture prepared, but joules per mole of unknown acid are required. To calculate the number of moles of the unknown acid the equation: Molarity (M) = Moles Solute1000cm3 SolutionThe molarity of the acid is 1.00M, so in 50cm3 of the unknown acid solution is 0.0500mol.50.0cm3 of the unknown acid x 1.00 mol = 0.005mol1000cm3The neutralisation of 0.0500mol of acid released 2.9 x 103 J. So, the heat released is:Energy Evolved = 2.9 x 103 J = 58 x 103 J/mol = -58 kJ/mol (exothermic)0.0500molEnthalpy: The term used for the total energy of a system ‘when it is at constant pressure’ is called the enthalpy of the system, symbolised by H. Enthalpy is a state function so when a system reacts at constant pressure and absorbs or evolves energy, it experiences an enthalpy change, H, defined by:H = H final – H initialH final is the enthalpy of the system in its final state and H initial is the enthalpy of the system in its initial state. H is a state function. Its value depends solely on the difference between the initial and final states and not on the mechanism by which the system undergoes this change. For a chemical reaction, the initial state refers to the reactants and the final state to the products, so for a chemical reaction the equation for H can be rewritten as follows: H = H products – H reactantsIf a change is exothermic, the system loses heat and its final enthalpy must be less than its initial enthalpy.Thus: H final
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