Objective: To determine the extent of solvolysis of ammonium borate in water by calorimetry.
Title : Solvolysis of the Salt of A Weak Acid and A Weak Base Objective: To determine the extent of solvolysis of ammonium borate in water by calorimetry. Results: Part I: 25ml of 1.75M NaOH + 100ml of 0.5M HCl Initial temperature of NaOH = 27.0 Initial temperature of HCl = 26.0 Table 1: Time(seconds) Temperature of mixture() Time(seconds) Temperature of mixture() 0.0 26.5 360.0 29.5 30.0 29.0 420.0 29.5 60.0 30.0 480.0 29.5 90.0 30.0 540.0 29.5 120. 30.0 600.0 29.5 150.0 29.5 660.0 29.5 180.0 29.5 720.0 29.5 210.0 29.5 780.0 29.5 240.0 29.5 840.0 29.5 270.0 29.5 900.0 29.5 300.0 29.5 From the graph of Temperature of mixture against Time, ΔT1 = 3.5°C Part II: 25ml of 1.75M NaOH + 100ml of 0.5M H3BO3 Initial temperature of NaOH = 25 Initial temperature of H3BO3 = 25.0 Table 2: Time(seconds) Temperature of mixture() Time(seconds) Temperature of mixture() 0.0 25.0 360.0 28.0 30.0 28.5 420.0 28.0 60.0 28.0 480.0 28.0 90.0 28.0 540.0 28.0 120. 28.0 600.0 28.0 150.0 28.0 660.0 28.0 180.0 28.0 720.0 28.0 210.0 28.0 780.0 28.0 240.0 28.0 840.0 28.0 270.0 28.0 900.0 28.0 300.0 28.0 From the graph of Temperature of mixture against Time, ΔT2 = 3.5°C Part III: 25ml of 1.75M Ammonia solution + 100ml of 0.5M HCl Initial temperature of Ammonia solution = 25.0 Initial temperature
An Environmental Case Study - Acid Rain: Causes, Effects and Solutions
An Environmental Case Study - Acid Rain: Causes, Effects and Solutions Acid rain is the broad term used to describe rainfall, snow, fog or sleet which has a higher level of acidity than that of natural, unpolluted rain. The acidity of a substance is measured by its pH. The pH of deionised water is pH 7, or neutral. Anything above 7 is classed as being an alkaline or base and anything below pH 7 is acidic. The pH of normal rain is 5.6 due to the reaction of water and carbon dioxide in the atmosphere to form a mildly acidic carbonic acid. The pH of acid rain ranges between pH 5-3 with most acidic deposition ranging between pH 5.0-4.3. Acid rain is formed when gases such as sulphur dioxide and nitrogen oxides. These gases are by products of human activity and are released during industrial processes, most notably the burning of fossil fuels to produce electricity. Smaller quantities are released by the burning of fuels for cars, but these emissions have been greatly reduced due to the introduction of three-way catalytic convertors which reduce the nitrogen oxides released in exhaust fumes. By far, the most common cause of acid rain is sulphur dioxide which makes up 70% of acid rain formation with the various oxides of nitrogen being responsible for the remaining 30%. Around 70Tg(S) per year SO2 is released through human activity, 7-8Tg(S) per year through volcanic emissions and
Deriving a Solubility Curve
Deriving a Solubility Curve- Practical Report Aim: - To observe the change in solubility of an ionic compound, potassium nitrate, as the temperature is increased. - To plot the solubility curve of an ionic compound, potassium nitrate, on the basis of experimental data. Apparatus: As per prac sheet Method: As per prac sheet Safety: See risk assessment sheet Results: Solubility Table Mass of solute (KNO3) Mass of solvent (H2O) Solubility (g of KNO3/100g of H2O) Temperature (° C) (at first sign of crystallisation) 6.5g 5ml 30g 81°C 6.5g 8ml 81.25g 54°C 6.5g 1ml 59.1g 47°C 6.5g 4ml 46.43g 27°C 6.5g 7ml 38.24g 20°C 6.5g 20ml 32.50g 20°C 6.5g 23ml 28.26g 3°C Find solubility curve attached to the end of the report. Calculations: To find solubility, the mass of solute (KNO3) was divided by the mass of solvent (H2O), and then multiplied by 100. 6.5 × 100 = 130 5 6.5 × 100 = 59.1 1 6.5 × 100 = 38.24 7 6.5 × 100 = 28.26 23 Questions: Refer to prac sheet for questions. ) From the graph of the solubility curve, it can be determined that: a) the solubility of potassium nitrate at 50°C is approximately 74g/ 100g of H2O. b) the temperature at which the solubility of potassium nitrate is 80g/100g of H2O, is approximately 53.5°C. c) the maximum amount of potassium nitrate that would dissolve in 100g of water at 60°C is
The preparation, analysis, and reactions of an ethanedioate complex of iron
Experiment 1 Title: The preparation, analysis, and reactions of an ethanedioate complex of iron Date: 5-2-2007 and 12-2-2007 Objective: * To prepare oxalate complex of iron from the reaction with ammonium iron (II) sulphate * To deduce the oxidation states of iron in the complex * To determine the amount of iron and oxalate in the oxalate complex of iron by titration. * To investigate the molecular formula of the complex formed by preparation, analysis and the reaction of an ethanedioate (oxalate) complex of iron * To deduce the chemical properties of the complex of iron from the reactions with dilute sodium hydroxide, ammonium thiocyanate solution, and ammonium thiocyanate solution in the presence of dilute sulphuric acid respectively, compared to the reaction of iron(III) chloride and the reagents above. Introduction Iron is nearly always determined by reduction to the dipositive state followed by titration with manganate(VII) or dichromate(VI). However oxalate would interfere and must be determined first by titration with permanganate. After titration, any iron present will be Fe(III) then reduced by tin(II) chloride and hydrochloric acid, and the Fe(II) determined with dichromate. Materials and Methods Procedure for prepartion 10.53g of ammonium iron(II) sulphate, (NH4)2Fe(SO4)2?6H2O , was weighed using a rough balance and put into a 400cm3 beaker,
To find out the amount of dissolved oxygen in a water sample
Experiment 9 Aim To find out the amount of dissolved oxygen in a water sample. Procedure . A water sample was collected by a 250cm3 volumetric flask. The flask was filled completely with water without trapping any air bubbles. 2. 1 cm3 of manganese(II) sulphate solution was added to the sample using a pipette. The solution was discharged well below the surface. 3. 1 cm3 of alkaline potassium iodide solution was introduced similarly. It was sure that no air becomes entrapped. The bottle was inverted to distribute the precipitate uniformly. 4. When the precipitate has settled at least 3 cm below the stopper, 1 cm3 of concentrated sulphuric(VI) acid was introduced well below the surface. The stopper was replaced and carefully mixed until the precipitate disappears. A magnetic stirrer was used. 5. The mixture was allowed to stand for 5 min and then 100 cm3 of the acidified sample was withdrawn into a 250 cm3 conical flask. 6. 0.0125M sodium thiosulphate was titrated until the iodine color becomes faint. Then 1 cm3 of starch solution was added, and continue adding the thiosulphate solution until the blue color disappears. Result Titration Result First try Second try Initial (cm3) 3.3 1 Final (cm3) 1 7 Net (cm3) 7.7 6 The first try encountered error, and only the result of the second try was used. Calculation I2+2S2O32- --> 2I-+S4O62- 2Mn(OH)3(s) + 2I-(aq)
MASS OF LITHIUM
DETERMINING THE MASS OF LITHIUM METHOD 1 RESULTS Mass of lithium used 0.12g Amount of hydrogen gas produced 182cm³ TREATMENT OR RESULTS 2Li(s) + 2H2O(l) --> 2LiOH(aq) + H2(g) Moles of hydrogen gas collected Moles of H2 volume / 24000(cm³) 82 / 24000 0.007583333 moles Moles of H2 0.0076 moles Lithium that reacted H2 : Li : 2 0.0076 : 2 × 0.0076 0.0076 : 0.0152 Moles of Li 0.0152 moles Relative atomic mass of lithium R.A.M mass / moles 0.12 / 0.0152 7.894736842 R.A.M of Li 7.8947 METHOD 2 RESULTS Titration of aqueous LiOH with 0.100 mol dm-3 HCl. Start (cm³) End (cm³) Titre (cm³) 0 34.90 34.90 0 34.80 34.80 0 34.90 34.90 Average result (34.80 + 34.90) / 2 = 69.7 / 2 = 34.85 TREATMENT OF RESULTS LiOH(aq) + HCl(aq) --> LiCl(aq) + H2O(l) Moles of HCl used in titration Moles of HCl concentration × (volume / 1000) 0.100 × (34.85 / 1000) 0.003485 moles Moles of HCl 0.0035 moles LiOH used in titration HCl : LiOH : 1 0.0035 : 0.0035 Moles of LiOH 0.0035 moles Number of moles of LiOH present in 100cm³ of solution from method 1 25cm³ of LiOH is pipette each time. There's 100cm³ of LiOH, so it will be: 0.003485 × 4 = 0.01394 Moles present in 100cm³ of LiOH 0.0139 moles Relative atomic mass of Lithium R.A.M mass / moles 0.12 / 0.0139 8.633093525 R.A.M of Li 8.633 HAZARD OF CHEMICALS IN THIS
Redox titration of copper evaluation
Redox Titration of Sodium Thiosulphate against Copper (II) Sulphate My results from this titration can be seen in the table below: Rough st 2nd 3rd Final Reading/cm3 29.30 25.25 30.40 28.80 Initial reading/ cm3 05.05 0.80 6.20 4.55 Titre/ cm3 24.25 24.45 24.20 24.25 The concordant results I will be using in my calculation are the titres from my 2nd and 3rd titration as there are within 0.01 cm3 of one another. To calculate the average titre I will use the following method: Average Titre = (24.20 + 25.25) 2 = 24.225 cm3 In this titration there are two half equations that are involved, the first is when the copper (II) sulphate is added to the potassium iodide: 2Cu2+ + 4I- --> 2CuI + I2 The second half equation takes place when the sodium thiosulphate is titrated into the solution containing the Cu2+ and iodine: I2 + 2S2O32- --> S4O62- + 2I- As I need to find the concentration of the Cu2+ I have to look at both half equations to find the ratio of Cu2+ to S2O32-, which is 1:1. This is because in each step of the reaction there are 2 moles of each. This will mean that the number of moles of Cu2+ will be the same as the number of moles of S2O32-: n = c x v n of S2O32- = 0.102827763 x 24.225 1000 = 2.491002571 x 10-3 mol n of Cu2+ = 2.491002571 x 10-3 mol Now that I have the number of moles of Cu2+ I can
Chemistry Iodine Clock
AIM To investigate how the rate of reaction varies using the Iodine Clock experiment with changes in temperature, different concentrations of substrates and the use of transition metals. BACKGROUND THEORY The iodine clock reaction was first studied by Augustus Harcourt and William Esson. They studied the reaction between iodide and hydrogen peroxide (1) In the experiment Potassium Peroxodisulphate (K2S2O8) is used instead of Hydrogen Peroxide, in which the Peroxodisulphate ions react with the iodide ions from the Potassium Iodide solution to form Sulphate ions and iodine. It can be represented by this half equation S2O82-(aq) + 2I-(aq) 2SO42-(aq) + I2(aq) The reaction involves the oxidation of the iodide ions to iodine molecules. Sodium thiosulphate increases the time taken for the iodide ions to turn to iodine. All the Sodium thiosulphate has to be used up before the colour of the starch to change. Sodium thiosulphate changes the iodine to iodide ions, without sodium thiosulphate the reaction would be too fast for getting reliable results. The reaction between the Sodium thiosulphate and the iodine can be shown as: 2S2O32-(aq) + I2(aq) S4O62-(aq) + 2I-(aq) Rates of reaction The rate of reaction is a measure of how fast a reaction occurs. If the reaction fast then the rate of reaction is high but if the reaction is slow then the rate of reaction is
Determination of DO & BOD in a water sample
Determination of DO & BOD in a water sample Objective Using volumetric analysis, the amount of dissolved oxygen (DO) in a water sample can be found. On standing another sample for 5 days, the difference between these two values is defined as biochemical oxygen demand in 5 days (BOD5). The aquatic life needs certain amount of DO for survival, whereas the value of BOD reflects the water quality. Principle In alkaline solution, dissolved oxygen will oxidize manganese(II) into manganese(III): 4Mn2+(aq) + 8OH-(aq) + O2(aq) + 2H2O(l) › 4Mn(OH)3(s) The amount of DO can be found by titrating the iodine produced from potassium iodide by manganese(III) with sodium thiosulphate: 2Mn(OH)3(s) + 2I-(aq) + 6H+(aq) › 2Mn2+(aq) + I2(aq) + 6H2O(l) I2(aq) + 2S2O32-(aq) › 2I-(aq) + S4O62-(aq) Chemicals water sample, MnSO4, conc.H2SO4, alkaline KI, standard Na2S2O3, starch solution Apparatus titration apparatus, pipette, 250cm3 volumetric flask x2, beaker, droppers, magnetic stirrer, white tile Procedure 1.> Collect a water sample with two 250cm3 volumetric flasks. Remember to fill the flasks completely with water without trapping any air bubbles. 2.> To one of the volumetric flask, add about 1 cm3 of MnSO4 solution well below the surface with a dropper. 3.> Similarly, introduce about 1 cm3 of alkaline KI solution to the same
Neutralisation- Enthalpy Change
Measurement of Enthalpy change Of Neutralisation Calculations: Formula for calculating enthalpy change is [Q = M x C x change in T] * Q is enthalpy change, which is measured in Joules * M is the mass of liquid used, which is measured in grams * C is the specific heat capacity of the liquid, which is constantly 4.2 J deg¯¹g¯¹ * Change in T is the temperature change, which is the highest temperature minus the initial temperature Formula for calculating moles is [N = C x V] * N is the number of moles * C is the concentration of liquid used, which is measured in moles/dm³ * V is the volume of liquid used, which is measured in dm³ (d) Enthalpy change of HCL: Enthalpy change (Q) of HCL = M x C x change in T = (25 + 21.5) x 4.2 x 6.9 = 46.5 x 4.2 x 6.9 = 1347.57 J No of moles = C x V = 1 x 25/1000 = 0.025 /0.025 = 40 40 x 1347.57 = 53902.8 53902.8/1000=53.9028 ENTHALPY CHANGE = -53.90 kJ mol¯¹ Enthalpy change of HNO3: Q = M x C x change in T = (25 + 21) x 4.2 x 7.2 = 46 x 4.2 x 7.2 = 1391.04 J No of moles = C x V = 1 x 25/1000 = 0.025 /0.025 = 40 40 x 1391.04 = 55641.6 55641.6/1000=55.64kJ ENTHALPY CHANGE = -55.64 kJ mol¯¹ Enthalpy change of H2SO4: Q = M x C x change in T = (25 + 24) x 4.2 x 8.1 = 49 x 4.2 x 8.1 = 1666.98 J No of moles = C x V = 1 x 25/1000 = 0.025 /0.025 = 40 40 x 1666.98 = 133358.4 33358.4/1000=133.36kJ