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Standardisation Of 0.02moldm-3 Potassium Permanganate Solution Using Ferrous Ammonium Sulphate (Ammonium Iron (Ii) Sulphate) - Lab Report Example

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"Standardisation of 0.02moldm-3 Potassium Permanganate Solution Using Ferrous Ammonium Sulphate" paper finds the exact concentration of Potassium Permanganate Solution. The concentration is expressed in Moles per liter (M). A 0.02moldm-3 is prepared and its exact molarity is determined by titration…
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Standardisation Of 0.02moldm-3 Potassium Permanganate Solution Using Ferrous Ammonium Sulphate (Ammonium Iron (Ii) Sulphate)
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STANDARDISATION OF 0.02MOLDM-3 POTASSIUM PERMANGANATE SOLUTION USING FERROUS AMMONIUM SULPHATE (AMMONIUM IRON (II) SULPHATE) By Instructor’s name September 7, 2016 Title Standardisation of 0.02moldm-3 Potassium Permanganate Solution using Ferrous Ammonium Sulphate (Ammonium Iron (II) Sulphate) Introduction The technique for determining the concentration of an acid or a base is known as titration. This is the process which involves carefully adding one reactant to another while at the same time measuring how much is needed for the reaction to be complete. In this experiment, a base (Iron Ammonium Sulphate – FeSO4.(NH4)2SO4.6H2O Mr=392g/mole) will be added to 0.02moldm-3 Potassium Permanganate Solution. In this experiment Manganese reacts with Iron through a redox process by which electrons are transferred from one to the other. In this experiment, Mn goes from a +7 state (MN+7) to a +2 state (Mn+2) – i.e. each Mn+7 picks up 5 electrons – Mn is reduced. Fe goes from +2 (Fe+2) to +3 (Fe+3) i.e. each Fe+2 loses 1 electron – Fe is oxidized As a result the mole ratio in the reaction is Mn: Fe – 1:5 The point at which the reactants will have completely neutralized each other is the equivalence point. In this experiment, this will be determined by noting the color change of the reactants. Mn+7 are purple. When Fe2+ is added, redox reaction starts making the purple color to disappear due to formation of Mn2+ which is colorless. In the process, the Fe2+ is converted to Fe3+. This happens only when Fe2+ is present. When Fe2+ is used up (at endpoint) the solution will have a permanent pink tinge and this is the point where there is no further addition of permanganate. The main objective of this experiment is to standardize i.e. find the exact concentration of Potassium Permanganate Solution. The concentration is expressed in Moles per litre (M). A 0.02moldm-3 is prepared and its exact molarity determined by titration. Method Weigh out accurately about 9.8g of the A.R. salt. Transfer to a 250ml volumetric flask, add 2/3 (approx. 150ml) bench sulphuric acid (2M). Shake until the solid has dissolved and make up to the mark with deionised water. Shake well. Remove 25ml of the solution with a pipette and titrate with the permanganate solution to the first permanent pink colour. Results and calculations The results are as shown in table 1.0 below Titration 1 2 3 4 Start 1.60 1.0 4.00 1.0 End 24.16 23.20 25.30 22.10 Difference 22.50 22.20 21.20 21.10 21.20 + 21.30 = 21.25cm3 2 Mass of the iron salt weighed = 9g Mr = 392g/mole No. of moles of the iron salt weighed = 9.8/392 = 0.025moles Concentration of iron salt M = Moles weighed Vol. of solution made Therefore Molarity = 0.025/25 Hence M= 0.001 Molarity of the acid, M = 25*0.001 21.25 M = 0.001176 moles/cm3 Discussion The Concentration of 0.02moldm-3 Potassium Permanganate Solution was established to be 0.001176moles/cm3. This value may not be the exact concentration though it’s approximately close to the exact concentration. This is because of errors that may have arisen during the experiment. These errors include parallax errors, inaccurate instrument calibration and changes in the room temperature. Conclusion The experiment was successful the concentration of 0.02moldm-3 Potassium Permanganate determined to be 0.001176moles/cm3 Investigation of the Beer – Lambert Law By Name Course Instructor’s name September 7, 2016 Investigation of the Beer – Lambert Law Introduction For any uniformly absorbing medium, both solution: solvent and solute molecules that absorb light, there is a proportion of light radiation which passes through. This proportion of light is known as transmittance abbreviated by T (it’s normally expressed as percent transmittance). The proportion of the light which is absorbed by the molecules in the medium is referred to as absorbance which is abbreviated as A. Transmittance is calculated as: T = I/Io where: I = intensity of the transmitted radiation leaving the medium. Io= intensity of the incident radiation entering the medium Percentage transmittance is related with absorbance as: A = 2-log (%T). Absorbance has no units and it ranges from 0 – 2. According to the Beer – Lambert law, the absorbance is proportional to the concentration of absorbing molecules the length of light – path through the medium and molar extinction coefficient i.e. A = εcl Where ε= molar extinction coefficient for the absorbing material at Wavelength in units of 1/ (molx cm) c= concentration of the absorbing solution (molar) l= light path in the absorbing material (l=1 cm for our purposes) The main objective of this experiment is to get an understanding of spectrometry is used qualitatively and quantitatively in the analysis of biological samples. Method 1. Solutions were made in test tubes using 0.0004M potassium permanganate solution and de - ionized water. This is illustrated as in the table 2.0 shown below: 2. The absorbance of tube 1 over the range 400 – 600nm was recorded using the cuvettes of 1 cm path length and de – ionized water in reference cell. 3. A graph of absorbance against wavelength and the wavelength that has maximum absorbance for the subsequent measurements of tubes 2 to 5 4. The transmittance value, T and the absorbance value (A), of solutions 1 to 5 and recorded in the table 2.1 below. T 100/T Log(100/T) A Conc. (c) Tube No. 13.5 7.41 0.87 0.871 0.0004 1 20.2 4.95 0.69 0.679 0.0032 2 29.7 3.37 0.53 0.528 0.00024 3 43.0 2.33 0.37 0.366 0.00016 4 69.3 1.44 0.16 0.157 0.00008 5 Results and calculations The results obtained for transmittance and absorbance were measured and recorded as in the table 2.1 below. T 100/T Log(100/T) A Conc. (c) Tube No. 13.5 7.41 0.87 0.871 0.0004 1 20.2 4.95 0.69 0.679 0.0032 2 29.7 3.37 0.53 0.528 0.00024 3 43.0 2.33 0.37 0.366 0.00016 4 69.3 1.44 0.16 0.157 0.00008 5 From the data of transmittance and absorbance, it was possible to use calculate 100/T and log (100/T).With this data a graph of T against C was drawn as on graph 2.3 below Also, a graph of 100/T against c was constructed from the data as shown in graph 2.4 shown below: The figure 2.5 below shows the graph of log (100/T) against C Discussion From the graphs we can deduce that absorbance is directly proportional to the product of concentration and path length. Conclusion From the experiment, we learned how to use spectrometry on analysis of samples quantitatively and qualitatively. Also, it was found that absorption is directly proportional to the product of molar absorption coefficient, path length and concentration Read More
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